1. Definitions

Linear equation:
An equation in which each variable occurs only to the first power and no product of variables appears is called a linear equation.
General form: $a_1x_1+a_2x_2+\cdots+a_nx_n=b$ where (a_1,a_2,\dots,a_n,b) are constants and not all (a_i) are zero.

Example: $2x+3y=7$

System of linear equations:
A collection of two or more linear equations involving the same variables is called a system of linear equations.

Example:

$$\begin{cases} x+y=5\\ 2x-y=1 \end{cases}$$

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2. Consistency of the given system

The given system is

$$\begin{cases} x+3y-2z=3\\ 2x-3y+z=2\\ 4x-3y+2z=2 \end{cases}$$

Write the augmented matrix:

$$[A|B]= \begin{bmatrix} 1 & 3 & -2 & | & 3\\ 2 & -3 & 1 & | & 2\\ 4 & -3 & 2 & | & 2 \end{bmatrix}$$

Apply Gaussian elimination.

Step 1:
(R_2 \to R_2-2R_1)

$$\begin{bmatrix} 1 & 3 & -2 & | & 3\\ 0 & -9 & 5 & | & -4\\ 4 & -3 & 2 & | & 2 \end{bmatrix}$$

Step 2:
(R_3 \to R_3-4R_1)

$$\begin{bmatrix} 1 & 3 & -2 & | & 3\\ 0 & -9 & 5 & | & -4\\ 0 & -15 & 10 & | & -10 \end{bmatrix}$$

Step 3:
(R_3 \to R_3-\frac{5}{3}R_2)

$$\begin{bmatrix} 1 & 3 & -2 & | & 3\\ 0 & -9 & 5 & | & -4\\ 0 & 0 & \frac{5}{3} & | & -\frac{10}{3} \end{bmatrix}$$

This is in row-echelon form.

Here,

$$\operatorname{rank}(A)=3,\qquad \operatorname{rank}([A|B])=3$$

Since

$$\operatorname{rank}(A)=\operatorname{rank}([A|B])=3=\text{number of variables},$$

the system is consistent and has a unique solution.

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3. Solution by back substitution

From the last row:

$$\frac{5}{3}z=-\frac{10}{3}$$ $$z=-2$$

From the second row:

$$-9y+5z=-4$$ $$-9y+5(-2)=-4$$ $$-9y-10=-4$$ $$-9y=6$$ $$y=-\frac{2}{3}$$

From the first row:

$$x+3y-2z=3$$ $$x+3\left(-\frac{2}{3}\right)-2(-2)=3$$ $$x-2+4=3$$ $$x+2=3$$ $$x=1$$

Therefore,

$$\boxed{x=1,\quad y=-\frac{2}{3},\quad z=-2}$$

So the system is consistent and its unique solution is

$$\boxed{(x,y,z)=\left(1,-\frac{2}{3},-2\right)}$$

First alternative

Let
[ A=\begin{pmatrix}1&2\3&4\0&0\end{pmatrix}. ]

1. (T:\mathbb R^2\to \mathbb R^3), (T(x)=Ax), is one-to-one

Let (x=(x_1,x_2)^T,\ y=(y_1,y_2)^T). Suppose
[ T(x)=T(y). ] Then [ A(x-y)=0. ] Put (u=x-y=(u_1,u_2)^T). Then [ \begin{pmatrix}1&2\3&4\0&0\end{pmatrix} \begin{pmatrix}u_1\u_2\end{pmatrix}

\begin{pmatrix}0\0\0\end{pmatrix}. ] So [ u_1+2u_2=0,\qquad 3u_1+4u_2=0. ] From (u_1+2u_2=0), (u_1=-2u_2). Substituting in the second equation: [ 3(-2u_2)+4u_2=0 ] [ -6u_2+4u_2=0 ] [ -2u_2=0\implies u_2=0. ] Hence (u_1=0). Therefore (u=0), so (x=y). Thus (T) is one-to-one.

2. (T) is not onto

For any (x=(x_1,x_2)^T), [ T(x)=Ax= \begin{pmatrix} x_1+2x_2\ 3x_1+4x_2\ 0 \end{pmatrix}. ] The third coordinate is always (0). Hence the vector [ e_3=\begin{pmatrix}0\0\1\end{pmatrix}\in \mathbb R^3 ] is not in the image of (T). Therefore (T) is not onto.

So (T(x)=Ax) is one-to-one but not onto.

3. The row-vector map (x\mapsto xA) is onto but not one-to-one

Interpreting the second statement as the row-vector map [ S:\mathbb R^3\to \mathbb R^2,\qquad S(x)=xA, ] which is equivalent to the column map (x\mapsto A^T x), we have [ A^T=\begin{pmatrix}1&3&0\2&4&0\end{pmatrix}. ] For (x=(x_1,x_2,x_3)^T), [ S(x)=A^T x= \begin{pmatrix} x_1+3x_2\ 2x_1+4x_2 \end{pmatrix}. ] The first two columns of (A^T) are [ \begin{pmatrix}1\2\end{pmatrix},\qquad \begin{pmatrix}3\4\end{pmatrix}, ] and [ \det\begin{pmatrix}1&3\2&4\end{pmatrix}=4-6=-2\ne 0. ] So these columns are linearly independent. Hence (\operatorname{rank}(A^T)=2), and therefore the image of (S) is all of (\mathbb R^2). Thus (S) is onto.

But [ S(0,0,1)=A^T\begin{pmatrix}0\0\1\end{pmatrix}

\begin{pmatrix}0\0\end{pmatrix} =S(0,0,0). ] So two different vectors have the same image. Hence (S) is not one-to-one.

Therefore the map (x\mapsto xA) is onto but not one-to-one.

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OR

1. Condition for a square matrix to possess an inverse

A square matrix (A) has an inverse if and only if [ \det(A)\ne 0. ] Equivalently, (A) is non-singular.

2. Determine (B) when (A) and (B) are inverse of each other

Given [ A=\begin{pmatrix}3&-1\5&-2\end{pmatrix}. ] First, [ \det(A)=3(-2)-(-1)(5)=-6+5=-1\ne 0. ] So (A^{-1}) exists. Since (A) and (B) are inverses, [ B=A^{-1}. ] For a (2\times 2) matrix [ A=\begin{pmatrix}a&b\c&d\end{pmatrix}, ] [ A^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d&-b\-c&a\end{pmatrix}. ] Here (a=3,\ b=-1,\ c=5,\ d=-2), and (ad-bc=-1). Hence [ B=A^{-1}

\frac{1}{-1} \begin{pmatrix} -2&1\ -5&3 \end{pmatrix}

\begin{pmatrix} 2&-1\ 5&-3 \end{pmatrix}. ] Thus [ p=2,\quad q=-1,\quad r=5,\quad s=-3. ]

3. Verify that (A^2) and (B^2) are inverse of each other

Now [ A^2= \begin{pmatrix}3&-1\5&-2\end{pmatrix} \begin{pmatrix}3&-1\5&-2\end{pmatrix}

\begin{pmatrix} 4&-1\ 5&-1 \end{pmatrix}. ] Also [ B^2= \begin{pmatrix}2&-1\5&-3\end{pmatrix} \begin{pmatrix}2&-1\5&-3\end{pmatrix}

\begin{pmatrix} -1&1\ -5&4 \end{pmatrix}. ] Then [ A^2B^2= \begin{pmatrix} 4&-1\ 5&-1 \end{pmatrix} \begin{pmatrix} -1&1\ -5&4 \end{pmatrix}

\begin{pmatrix} 1&0\ 0&1 \end{pmatrix} =I. ] Similarly, [ B^2A^2=I. ] Therefore (A^2) and (B^2) are inverses of each other.

Proof of the Rank–Nullity Theorem

Let (V) be an (n)-dimensional vector space over a field (F), and let
[ T: V \to W ] be a linear transformation. Put [ \ker T = {v \in V : T(v)=0}. ] Since (\ker T) is a subspace of the finite-dimensional space (V), it is finite-dimensional. Let [ \dim(\ker T)=k. ] Choose a basis of (\ker T): [ {u_1,u_2,\dots,u_k}. ]

If (k=n), then (\ker T=V), so (T(v)=0) for every (v\in V). Hence (\operatorname{Im}T={0}), and therefore [ \dim(\ker T)+\dim(\operatorname{Im}T)=n+0=n. ] So assume (k<n). Extend the basis of (\ker T) to a basis of (V): [ {u_1,\dots,u_k,v_1,\dots,v_{n-k}}. ] Let (m=n-k). We claim that [ {T(v_1),T(v_2),\dots,T(v_m)} ] is a basis of (\operatorname{Im}T).

1. Spanning.
Take any (y\in \operatorname{Im}T). Then (y=T(v)) for some (v\in V). Since the above set is a basis of (V), we can write [ v=a_1u_1+\cdots+a_ku_k+b_1v_1+\cdots+b_mv_m. ] Applying (T), [ T(v)=a_1T(u_1)+\cdots+a_kT(u_k)+b_1T(v_1)+\cdots+b_mT(v_m). ] But each (u_i\in \ker T), so (T(u_i)=0). Thus [ y=b_1T(v_1)+\cdots+b_mT(v_m). ] Hence ({T(v_1),\dots,T(v_m)}) spans (\operatorname{Im}T).

2. Linear independence.
Suppose [ c_1T(v_1)+\cdots+c_mT(v_m)=0. ] Then [ T(c_1v_1+\cdots+c_mv_m)=0, ] so [ c_1v_1+\cdots+c_mv_m\in \ker T. ] Since ({u_1,\dots,u_k}) is a basis of (\ker T), there exist scalars (d_1,\dots,d_k) such that [ c_1v_1+\cdots+c_mv_m=d_1u_1+\cdots+d_ku_k. ] Rearranging, [ c_1v_1+\cdots+c_mv_m-d_1u_1-\cdots-d_ku_k=0. ] But ({u_1,\dots,u_k,v_1,\dots,v_m}) is a basis of (V), hence linearly independent. Therefore all coefficients must be zero: [ c_1=\cdots=c_m=0. ] So ({T(v_1),\dots,T(v_m)}) is linearly independent.

Thus ({T(v_1),\dots,T(v_m)}) is a basis of (\operatorname{Im}T), and therefore [ \dim(\operatorname{Im}T)=m=n-k. ] Hence [ \dim(\ker T)+\dim(\operatorname{Im}T)=k+(n-k)=n. ] [ \boxed{\dim(\ker T)+\dim(\operatorname{Im}T)=n} ] This proves the rank–nullity theorem.

Let
[ v_1=(1,1,1),\quad v_2=(0,1,1),\quad v_3=(1,0,1). ]

1. Orthogonal basis by Gram–Schmidt process

Take
[ u_1=v_1=(1,1,1). ]

Now, [ u_2=v_2-\frac{v_2\cdot u_1}{u_1\cdot u_1}u_1. ] Here, [ v_2\cdot u_1=0+1+1=2,\qquad u_1\cdot u_1=1+1+1=3. ] So, [ u_2=(0,1,1)-\frac{2}{3}(1,1,1) =\left(-\frac23,\frac13,\frac13\right). ] Multiplying by (3), we may take [ u_2=(-2,1,1). ]

Next, [ u_3=v_3-\frac{v_3\cdot u_1}{u_1\cdot u_1}u_1-\frac{v_3\cdot u_2}{u_2\cdot u_2}u_2. ] Now, [ v_3\cdot u_1=1+0+1=2, ] and [ v_3\cdot u_2=(1)(-2)+0(1)+1(1)=-1, ] while [ u_2\cdot u_2=(-2)^2+1^2+1^2=6. ] Thus, [ u_3=(1,0,1)-\frac23(1,1,1)-\frac{-1}{6}(-2,1,1). ] So, [ u_3=(1,0,1)-\left(\frac23,\frac23,\frac23\right)+\left(-\frac13,\frac16,\frac16\right) =\left(0,-\frac12,\frac12\right). ] Multiplying by (2), we may take [ u_3=(0,-1,1). ]

Hence an orthogonal basis of (\mathbb R^3) is [ \boxed{{(1,1,1),\ (-2,1,1),\ (0,-1,1)}}. ]

Check: [ (1,1,1)\cdot(-2,1,1)=0, ] [ (1,1,1)\cdot(0,-1,1)=0, ] [ (-2,1,1)\cdot(0,-1,1)=0. ]

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2. Orthonormal basis

Divide each orthogonal vector by its norm.

[ |u_1|=\sqrt{1^2+1^2+1^2}=\sqrt3, ] [ |u_2|=\sqrt{(-2)^2+1^2+1^2}=\sqrt6, ] [ |u_3|=\sqrt{0^2+(-1)^2+1^2}=\sqrt2. ]

Therefore, [ e_1=\frac{u_1}{|u_1|} =\left(\frac1{\sqrt3},\frac1{\sqrt3},\frac1{\sqrt3}\right), ] [ e_2=\frac{u_2}{|u_2|} =\left(-\frac2{\sqrt6},\frac1{\sqrt6},\frac1{\sqrt6}\right), ] [ e_3=\frac{u_3}{|u_3|} =\left(0,-\frac1{\sqrt2},\frac1{\sqrt2}\right). ]

Thus an orthonormal basis of (\mathbb R^3) is [ \boxed{ \left{ \left(\frac1{\sqrt3},\frac1{\sqrt3},\frac1{\sqrt3}\right), \left(-\frac2{\sqrt6},\frac1{\sqrt6},\frac1{\sqrt6}\right), \left(0,-\frac1{\sqrt2},\frac1{\sqrt2}\right) \right} } ]

These vectors are mutually perpendicular and each has unit length, so they form an orthonormal basis of (\mathbb R^3).

Let
[ L_1={u+tv:t\in\mathbb R},\qquad L_2={w+sz:s\in\mathbb R} ] be two lines in (\mathbb R^3), where (v,z\neq 0).

1. Prove: (L_1=L_2) iff (u-w) and (v) are multiples of (z)

Necessity

Assume (L_1=L_2).

Since (u\in L_1), we have (u\in L_2). Hence there exists (s_0\in\mathbb R) such that [ u=w+s_0z. ] Therefore [ u-w=s_0z, ] so (u-w) is a multiple of (z).

Also, since (u+v\in L_1=L_2), there exists (s_1\in\mathbb R) such that [ u+v=w+s_1z. ] Subtracting (u=w+s_0z), we get [ v=(s_1-s_0)z. ] Thus (v) is also a multiple of (z).

Hence, if (L_1=L_2), then both (u-w) and (v) are multiples of (z).

Sufficiency

Conversely, suppose [ u-w=\alpha z,\qquad v=\beta z ] for some (\alpha,\beta\in\mathbb R). Since (v\neq 0), we have (\beta\neq 0).

Take any point (x\in L_1). Then for some (t\in\mathbb R), [ x=u+tv. ] Using (u=w+\alpha z) and (v=\beta z), [ x=w+\alpha z+t\beta z =w+(\alpha+t\beta)z. ] Thus (x\in L_2), so (L_1\subseteq L_2).

Now take any point (y\in L_2). Then for some (s\in\mathbb R), [ y=w+sz. ] Since (u-w=\alpha z), we have (w=u-\alpha z). Hence [ y=u-\alpha z+sz =u+(s-\alpha)z. ] But (v=\beta z), so (z=\frac1\beta v). Therefore [ y=u+\frac{s-\alpha}{\beta}v\in L_1. ] Thus (L_2\subseteq L_1).

Therefore (L_1=L_2).

Hence, [ \boxed{L_1=L_2 \iff u-w \text{ and } v \text{ are multiples of } z.} ]

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2. Show that the two lines are parallel and distinct

Suppose the direction vectors are parallel, i.e. [ v=\lambda z,\qquad \lambda\neq 0, ] but (u-w) is not a multiple of (z). Then the lines are parallel.

To show they are distinct, suppose they intersect. Then there exist (t,s\in\mathbb R) such that [ u+tv=w+sz. ] Using (v=\lambda z), [ u+t\lambda z=w+sz. ] Hence [ u-w=(s-t\lambda)z, ] which means (u-w) is a multiple of (z), a contradiction.

Therefore the lines do not intersect. Since they are parallel and do not intersect, they are parallel and distinct.

[ \boxed{\text{If } v\parallel z \text{ but } u-w\notin \operatorname{span}{z}, \text{ then } L_1\parallel L_2 \text{ and } L_1\cap L_2=\varnothing.} ]

Example

Let [ L_1={(0,0,0)+t(1,0,0):t\in\mathbb R} ] and [ L_2={(0,1,0)+s(2,0,0):s\in\mathbb R}. ] Here (v=(1,0,0)) and (z=(2,0,0)) are parallel, but [ u-w=(0,0,0)-(0,1,0)=(0,-1,0) ] is not a multiple of (z). Hence (L_1) and (L_2) are parallel and distinct.

1. Kernel of a matrix/linear transformation

Let (A) be an (m\times n) matrix. The kernel (or null space) of (A) is the set of all vectors (x\in \mathbb R^n) such that

[ Ax=0. ]

It is denoted by (\ker(A)) or (N(A)).

[ \ker(A)={x\in \mathbb R^n:Ax=0}. ]

The dimension of the kernel is called the nullity of (A).

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2. Span of vectors

Let (v_1,v_2,\dots,v_k) be vectors in a vector space (V). The span of these vectors is the set of all possible linear combinations of them:

[ \operatorname{span}{v_1,v_2,\dots,v_k}

{c_1v_1+c_2v_2+\cdots+c_kv_k:c_i\in \mathbb R}. ]

It is the smallest subspace containing all the given vectors.

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3. Rank of a matrix and consistency

The rank of a matrix (A) is the number of linearly independent rows or columns of (A). Equivalently, it is the dimension of the row space or column space of (A).

A system of linear equations

[ AX=B ]

is not always consistent. It is consistent if and only if

[ \operatorname{rank}(A)=\operatorname{rank}([A\mid B]). ]

If (\operatorname{rank}(A)<\operatorname{rank}([A\mid B])), the system has no solution.
The homogeneous system (AX=0) is always consistent, because (X=0) is always a solution.

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4. Matrix whose kernel is spanned by ((1,3,2)) and ((-2,0,4))

Let

[ v_1=(1,3,2),\qquad v_2=(-2,0,4). ]

These two vectors are linearly independent. Therefore their span is a plane through the origin in (\mathbb R^3). We need a matrix (A) such that

[ \ker(A)=\operatorname{span}{v_1,v_2}. ]

Take (A) as a (1\times 3) matrix whose row is perpendicular to both (v_1) and (v_2). Let the row be

[ (a,b,c). ]

Then

[ (a,b,c)\cdot (1,3,2)=0 ]

[ a+3b+2c=0 \quad (1) ]

and

[ (a,b,c)\cdot (-2,0,4)=0 ]

[ -2a+4c=0. \quad (2) ]

From (2),

[ a=2c. ]

Putting (a=2c) in (1),

[ 2c+3b+2c=0 ]

[ 4c+3b=0 ]

[ b=-\frac{4c}{3}. ]

Choose (c=3). Then

[ a=6,\qquad b=-4,\qquad c=3. ]

So one such matrix is

[ A= \begin{pmatrix} 6 & -4 & 3 \end{pmatrix}. ]

For this matrix,

[ Ax=0 ]

means

[ 6x_1-4x_2+3x_3=0. ]

Both (v_1=(1,3,2)) and (v_2=(-2,0,4)) satisfy this equation:

[ 6(1)-4(3)+3(2)=6-12+6=0, ]

[ 6(-2)-4(0)+3(4)=-12+0+12=0. ]

Since (v_1,v_2) are independent and the solution space is a plane in (\mathbb R^3), we get

[ \ker(A)=\operatorname{span}{(1,3,2),(-2,0,4)}. ]

If a (3\times 3) matrix is required, we may take

[ A= \begin{pmatrix} 6 & -4 & 3\ 12 & -8 & 6\ 18 & -12 & 9 \end{pmatrix}, ]

since all rows are multiples of ((6,-4,3)), so the kernel is the same.

[ \boxed{A=\begin{pmatrix}6&-4&3\end{pmatrix}} ]

is the simplest required matrix.

Note: As written, (T:\mathbb R^2\to \mathbb R^2) cannot satisfy (T(1,1)=3) and (T(0,1)=-2), because (3) and (-2) are scalars, not vectors in (\mathbb R^2). The standard interpretation is
[ T:\mathbb R^2\to \mathbb R. ] I solve the problem under this natural interpretation.

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1. Existence and uniqueness of (T) (2 marks)

Let
[ B={(1,1),(0,1)}. ] These two vectors are linearly independent, since
[ c_1(1,1)+c_2(0,1)=(0,0) ] gives (c_1=0) and (c_1+c_2=0), hence (c_2=0). Therefore (B) is a basis of (\mathbb R^2).

Any linear map is uniquely determined by its values on a basis. Define (T) on (B) by
[ T(1,1)=3,\qquad T(0,1)=-2. ] For any ((x,y)\in\mathbb R^2), write
[ (x,y)=x(1,1)+(y-x)(0,1). ] Then linearity forces
[ T(x,y)=xT(1,1)+(y-x)T(0,1) =3x-2(y-x) =5x-2y. ] Thus (T(x,y)=5x-2y) is the only possible linear map satisfying the given conditions. Hence the linear map is unique.

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2. Find (T(x,y)) (2 marks)

From above,
[ \boxed{T(x,y)=5x-2y}. ]

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3. Find (T^{-1}(6)) (2 marks)

Here (T^{-1}(6)) means the preimage of (6), i.e. all ((x,y)\in\mathbb R^2) such that (T(x,y)=6).
[ 5x-2y=6. ] Solving for (y),
[ 2y=5x-6 ] [ y=\frac{5x-6}{2}. ] Therefore
[ \boxed{T^{-1}(6)=\left{(x,y)\in\mathbb R^2:5x-2y=6\right}} ] or equivalently
[ \boxed{T^{-1}(6)=\left{\left(x,\frac{5x-6}{2}\right):x\in\mathbb R\right}}. ] This is a straight line in (\mathbb R^2).

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4. Is (T) one-to-one? (1 mark)

No. For example,
[ T(2,5)=5(2)-2(5)=10-10=0=T(0,0), ] but
[ (2,5)\neq (0,0). ] So (T) is not one-to-one. Equivalently,
[ \ker T={(x,y):5x-2y=0}\neq {(0,0)}. ]

[ \boxed{\text{Therefore } T \text{ is not one-to-one.}} ]

Relation of determinant with area and volume

For two vectors
[ \vec u=(x_1,y_1),\qquad \vec v=(x_2,y_2), ] the determinant [ \det(\vec u,\vec v)= \begin{vmatrix} x_1 & y_1\ x_2 & y_2 \end{vmatrix} =x_1y_2-y_1x_2 ] gives the signed area of the parallelogram with adjacent sides (\vec u) and (\vec v).
Hence the area of the triangle formed by them is [ \frac12|\det(\vec u,\vec v)|. ]

For three vectors (\vec u,\vec v,\vec w), the scalar triple product [ \det(\vec u,\vec v,\vec w)=\vec u\cdot(\vec v\times \vec w) ] gives the signed volume of the parallelepiped with coterminous edges (\vec u,\vec v,\vec w).
The actual volume is the absolute value of this determinant. The volume of the corresponding tetrahedron is (\frac16) of this absolute value.

In coordinate form, the area of a triangle with vertices ((x_1,y_1),(x_2,y_2),(x_3,y_3)) is [ A=\frac12 \left| \begin{vmatrix} x_1 & y_1 & 1\ x_2 & y_2 & 1\ x_3 & y_3 & 1 \end{vmatrix} \right|. ] The volume of a parallelepiped with edge vectors (\vec u,\vec v,\vec w) is [ V=|\det(\vec u,\vec v,\vec w)|. ]

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Area of the triangle

Vertices: [ A(-2,-3),\quad B(3,2),\quad C(-1,8). ] Using the determinant formula, [ A=\frac12 \left| \begin{vmatrix} -2 & -3 & 1\ 3 & 2 & 1\ -1 & 8 & 1 \end{vmatrix} \right|. ] Now, [ \begin{vmatrix} -2 & -3 & 1\ 3 & 2 & 1\ -1 & 8 & 1 \end{vmatrix}

-2(2-8)-(-3)(3+1)+1(24+2) ] [ =12+12+26=50. ] Therefore, [ A=\frac12(50)=25. ] So the area of the triangle is (\boxed{25\text{ square units}}).

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Volume of the parallelepiped

Taking the given points as the three coterminous vertices from the origin, the edge vectors are [ \vec u=(1,-2,1),\quad \vec v=(1,0,-1),\quad \vec w=(1,1,1). ] The volume is [ V= \left| \begin{vmatrix} 1 & -2 & 1\ 1 & 0 & -1\ 1 & 1 & 1 \end{vmatrix} \right|. ] Expanding, [ V=|1(0+1)-(-2)(1+1)+1(1-0)| ] [ =|1+4+1|=6. ] Hence the volume of the parallelepiped is (\boxed{6\text{ cubic units}}).

For
[ A=\begin{pmatrix} 1&-2&2\ -2&1&-2\ 2&-2&1 \end{pmatrix}, ] the characteristic equation is obtained from (\det(A-\lambda I)=0).

[ A-\lambda I= \begin{pmatrix} 1-\lambda&-2&2\ -2&1-\lambda&-2\ 2&-2&1-\lambda \end{pmatrix} ]

[ \det(A-\lambda I) =(1-\lambda)^3-12(1-\lambda)+16 =-(\lambda+1)^2(\lambda-5) ]

Hence, [ (\lambda+1)^2(\lambda-5)=0 ]

So the eigenvalues are
[ \boxed{\lambda=-1,,-1,,5} ]

Now find an eigenvector for (\lambda=5). Solve
[ (A-5I)\begin{pmatrix}x\y\z\end{pmatrix}=0. ]

[ A-5I= \begin{pmatrix} -4&-2&2\ -2&-4&-2\ 2&-2&-4 \end{pmatrix} ]

This gives [ -4x-2y+2z=0 ] [ -2x-4y-2z=0 ] [ 2x-2y-4z=0 ]

From the first equation, [ z=2x+y. ] Using the second equation, [ x+2y+z=0 ] [ x+2y+2x+y=0 ] [ 3x+3y=0 ] [ x=-y. ] Then [ z=2(-y)+y=-y. ]

Taking (y=1), we get [ x=-1,\quad y=1,\quad z=-1. ]

Therefore, one eigenvector corresponding to (\lambda=5) is
[ \boxed{\begin{pmatrix}-1\1\-1\end{pmatrix}} ] or equivalently, [ \boxed{\begin{pmatrix}1\-1\1\end{pmatrix}}. ]

Check: [ A\begin{pmatrix}-1\1\-1\end{pmatrix}

\begin{pmatrix}-5\5\-5\end{pmatrix}

5\begin{pmatrix}-1\1\-1\end{pmatrix}. ]

So the required eigenvalues are (\boxed{-1,-1,5}), and one eigenvector is (\boxed{(-1,1,-1)^T}).

Let the direction vectors be (v) and (z). Since (L_1) and (L_2) are lines, (v\neq 0) and (z\neq 0).

Proof

1. If (L_1=L_2), then (u-w) and (v) are multiples of (z)

Since (L_1=L_2), the point (u\in L_1) also lies on (L_2). Hence for some (s_0\in\mathbb R), [ u=w+s_0z, ] so [ u-w=s_0z. ] Thus (u-w) is a multiple of (z).

Also, (u+v\in L_1=L_2), so for some (s_1\in\mathbb R), [ u+v=w+s_1z. ] Subtracting (u=w+s_0z), we get [ v=(s_1-s_0)z. ] Thus (v) is also a multiple of (z).

2. Conversely, suppose (u-w) and (v) are multiples of (z)

Then there exist scalars (\lambda,\mu\in\mathbb R) such that [ u-w=\lambda z,\qquad v=\mu z. ] Since (v\neq 0) and (z\neq 0), we have (\mu\neq 0).

Now [ L_1={u+tv:t\in\mathbb R}. ] Using (u=w+\lambda z) and (v=\mu z), [ u+tv=w+\lambda z+t(\mu z) =w+(\lambda+t\mu)z. ] As (t) ranges over all real numbers and (\mu\neq 0), the number [ s=\lambda+t\mu ] also ranges over all real numbers. Therefore, [ L_1={w+sz:s\in\mathbb R}=L_2. ] Hence the two lines are the same.

So, [ \boxed{L_1=L_2 \iff u-w \text{ and } v \text{ are multiples of } z.} ]

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Check the given lines

Given: [ L_1=(-3,-3,-6)+t(9,15,12), ] [ L_2=(3,7,2)+s(3,5,4). ] So [ u=(-3,-3,-6),\quad v=(9,15,12), ] [ w=(3,7,2),\quad z=(3,5,4). ]

Check (v): [ (9,15,12)=3(3,5,4)=3z. ] So (v) is a multiple of (z).

Check (u-w): [ u-w=(-3,-3,-6)-(3,7,2)=(-6,-10,-8). ] But [ (-6,-10,-8)=-2(3,5,4)=-2z. ] So (u-w) is also a multiple of (z).

Therefore, by the above theorem, the two lines are the same.

Indeed, [ L_1=(-3,-3,-6)+t(9,15,12) ] [ =(3,7,2)-2(3,5,4)+t\cdot 3(3,5,4) ] [ =(3,7,2)+(3t-2)(3,5,4). ] Let (s=3t-2). As (t) ranges over (\mathbb R), (s) also ranges over (\mathbb R). Hence [ L_1=(3,7,2)+s(3,5,4)=L_2. ]

[ \boxed{\text{Yes, the two lines are the same.}} ]

(a) The linear mapping with matrix $\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$

Let $A=\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$. For any vector $(x,y)$,

$T(x,y)=\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}=(-y,\;x).$

Effect. The image of $(x,y)$ is $(-y,x)$, i.e. the point is turned through $90^\circ$ anticlockwise about the origin. Hence $A$ represents a rotation of the plane through $90^\circ$ counterclockwise about the origin.

Effect on the basis vectors:

$T(1,0)=(0,1),\qquad T(0,1)=(-1,0).$

So the $x$-axis is carried to the $y$-axis and the $y$-axis to the negative $x$-axis.

Effect on the unit square. The unit square $OABC$ with $O(0,0)$, $A(1,0)$, $B(1,1)$, $C(0,1)$ is mapped to the square $OA'B'C'$ with

$O(0,0),\quad A'(0,1),\quad B'(-1,1),\quad C'(-1,0).$

Figure:

        y                          y
        ^                          ^
  C(0,1)--B(1,1)            B'(-1,1)--A'(0,1)
   |      |                  |        |
   |      |        ---->     |        |
  O(0,0)--A(1,0)-->x     C'(-1,0)--O(0,0)-->x
   original square            image square

The square is rotated bodily through $90^\circ$; its shape and size are unchanged. Indeed $\det A=0\cdot 0-(-1)(1)=1$ and $A^{T}A=I$, so the mapping is orthogonal: it preserves lengths, angles and areas. Every circle centred at the origin is mapped onto itself.

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(b) The linear mapping with matrix $\begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$

Let $B=\begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}=2I$. For any vector $(x,y)$,

$T(x,y)=\begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}=(2x,\;2y).$

Effect. Every point is pushed radially outward from the origin to twice its distance along the same ray. Hence $B$ represents a uniform dilation (homothety) of the plane with centre the origin and scale factor $2$.

Effect on the basis vectors:

$T(1,0)=(2,0),\qquad T(0,1)=(0,2).$

Both axes are stretched by the factor $2$ and each line through the origin is mapped onto itself (every non-zero vector is an eigenvector with eigenvalue $2$).

Effect on the unit square. The unit square $OABC$ is mapped to the square $OA'B'C'$ of side $2$:

$O(0,0),\quad A'(2,0),\quad B'(2,2),\quad C'(0,2).$

Figure:

        y                          y
        ^                          ^
  C(0,1)--B(1,1)          C'(0,2)------B'(2,2)
   |      |                  |            |
   |      |        ---->     |            |
  O(0,0)--A(1,0)-->x      O(0,0)------A'(2,0)-->x
   unit square                image square (side 2)

The square is enlarged twice in every direction; lengths are multiplied by $2$ and areas by $\det B=2\cdot 2-0=4$. The mapping is not orthogonal (it does not preserve length), but it is a similarity — it preserves shape and angles.

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OR

Definitions

Vector space. Let $F$ be a field (here $F=\mathbb{R}$). A non-empty set $V$ is called a vector space over $F$ if two operations — vector addition $+$ and scalar multiplication $\cdot$ — are defined on $V$ such that for all $u,v,w\in V$ and all $\alpha,\beta\in F$:

1. $u+v\in V$ (closure under addition)
2. $u+v=v+u$ (commutativity)
3. $(u+v)+w=u+(v+w)$ (associativity)
4. there exists $0\in V$ with $u+0=u$ (additive identity)
5. there exists $-u\in V$ with $u+(-u)=0$ (additive inverse)
6. $\alpha u\in V$ (closure under scalar multiplication)
7. $\alpha(u+v)=\alpha u+\alpha v$
8. $(\alpha+\beta)u=\alpha u+\beta u$
9. $\alpha(\beta u)=(\alpha\beta)u$
10. $1\cdot u=u$, where $1$ is the unity of $F$.

Linearly dependent set. A set of vectors $\{v_1,v_2,\dots,v_n\}$ in $V$ is linearly dependent if there exist scalars $c_1,c_2,\dots,c_n$, not all zero, such that

$c_1v_1+c_2v_2+\dots+c_nv_n=0.$

Equivalently, at least one vector of the set can be written as a linear combination of the others.

Linearly independent set. The set is linearly independent if the only scalars satisfying

$c_1v_1+c_2v_2+\dots+c_nv_n=0$

are $c_1=c_2=\dots=c_n=0$; that is, no vector of the set is a linear combination of the remaining ones.

Test for the given vectors

Given $a_1=(2,-3,1)$, $a_2=(3,-5,2)$, $a_3=(4,-5,1)$. Suppose

$c_1a_1+c_2a_2+c_3a_3=0.$

Then, comparing components,

$2c_1+3c_2+4c_3=0 \quad (1)$ $-3c_1-5c_2-5c_3=0 \quad (2)$ $c_1+2c_2+c_3=0 \quad (3)$

From (3): $c_1=-2c_2-c_3$. Substituting in (1):

$2(-2c_2-c_3)+3c_2+4c_3=-c_2+2c_3=0\;\Rightarrow\;c_2=2c_3 .$

Substituting in (2):

$-3(-2c_2-c_3)-5c_2-5c_3=c_2-2c_3=0\;\Rightarrow\;c_2=2c_3,$

which is the same condition. Hence $c_2=2c_3$ and $c_1=-2(2c_3)-c_3=-5c_3$.

Taking $c_3=1$ (a non-zero choice) gives $c_1=-5,\;c_2=2,\;c_3=1$, and indeed

$-5(2,-3,1)+2(3,-5,2)+1(4,-5,1)=(-10+6+4,\;15-10-5,\;-5+4+1)=(0,0,0).$

Since the scalars are not all zero, the given vectors are linearly dependent.

Check by determinant. Forming the matrix with these vectors as rows,

$$\begin{vmatrix} 2 & -3 & 1 \\ 3 & -5 & 2 \\ 4 & -5 & 1 \end{vmatrix} =2(-5+10)+3(3-8)+1(-15+20)=10-15+5=0 .$$

A zero determinant confirms linear dependence. In fact, from the relation above,

$a_3=5a_1-2a_2,$

so $a_3$ is a linear combination of $a_1$ and $a_2$; the three vectors are therefore linearly dependent (while $a_1$ and $a_2$ alone are linearly independent).

Definitions

Vector subspace: A non-empty subset $S$ of a vector space $V$ over a field $F$ is called a subspace if it is closed under addition and scalar multiplication, i.e. for all $u,v\in S$ and $\alpha\in F$,

$$u+v\in S,\qquad \alpha u\in S.$$

Equivalently, $S$ contains the zero vector and is closed under all linear combinations.

Inverse image: If $f:X\to Y$ is a map and $W\subseteq Y$, then the inverse image of $W$ under $f$ is

$$f^{-1}(W)=\{x\in X: f(x)\in W\}.$$

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Proof: $f^{-1}(W)$ is a subspace of $X$

Let $f:X\to Y$ be linear and let $W$ be a subspace of $Y$.

Take $u,v\in f^{-1}(W)$ and a scalar $\alpha$. Then

$$f(u)\in W,\qquad f(v)\in W.$$

Since $W$ is a subspace,

$$f(u)+f(v)\in W.$$

Because $f$ is linear,

$$f(u+v)=f(u)+f(v)\in W.$$

Hence $u+v\in f^{-1}(W)$.

Also, since $W$ is closed under scalar multiplication,

$$\alpha f(u)\in W.$$

Again using linearity,

$$f(\alpha u)=\alpha f(u)\in W,$$

so $\alpha u\in f^{-1}(W)$.

Finally, $f(0)=0\in W$, so $0\in f^{-1}(W)$. Therefore $f^{-1}(W)$ is non-empty and closed under addition and scalar multiplication. Hence $f^{-1}(W)$ is a subspace of $X$.

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Computation of $f^{-1}(W)$

The displayed formula for $f$ is incomplete. Taking the intended map as

$$f(x_1,x_2,x_3,x_4)=(2x_1+x_2+x_3+x_4,\; x_1+2x_2+x_3+x_4,\; x_1+x_2+2x_3+x_4),$$

we find $f^{-1}(W)$.

Let

$$z=(1,1,0),\qquad w=(0,1,1).$$

A vector $y=(y_1,y_2,y_3)\in W=\operatorname{Span}\{z,w\}$ has the form

$$y=a z+b w=(a,a+b,b).$$

Thus

$$y\in W \iff y_2=y_1+y_3.$$

So $x=(x_1,x_2,x_3,x_4)\in f^{-1}(W)$ iff

$$f_2(x)=f_1(x)+f_3(x).$$

Here

$$f_1=2x_1+x_2+x_3+x_4,$$ $$f_2=x_1+2x_2+x_3+x_4,$$ $$f_3=x_1+x_2+2x_3+x_4.$$

Therefore,

$$x_1+2x_2+x_3+x_4 = (2x_1+x_2+x_3+x_4)+(x_1+x_2+2x_3+x_4).$$

Simplifying,

$$2x_1+2x_3+x_4=0,$$

or

$$x_4=-2x_1-2x_3.$$

Hence

$$f^{-1}(W)=\{(x_1,x_2,x_3,-2x_1-2x_3): x_1,x_2,x_3\in\mathbb R\}.$$

A basis is

$$(1,0,0,-2),\qquad (0,1,0,0),\qquad (0,0,1,-2).$$

So $f^{-1}(W)$ is a $3$-dimensional subspace of $\mathbb R^4$, namely the hyperplane

$$2x_1+2x_3+x_4=0.$$

In simple terms, $f^{-1}(W)$ consists of all points in $\mathbb R^4$ whose image under $f$ lies in the plane $W$ in $\mathbb R^3$.

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OR

Definition: Basis of a vector space

A subset $B$ of a vector space $V$ is called a basis if:

1. $B$ is linearly independent.
2. $B$ spans $V$, i.e. every vector in $V$ can be written as a linear combination of vectors in $B$.

Equivalently, every vector in $V$ has a unique representation as a linear combination of the basis vectors.

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Show that $\{(1,1,1),(0,1,0),(1,2,3)\}$ is a basis for $\mathbb R^3$

Let

$$v_1=(1,1,1),\qquad v_2=(0,1,0),\qquad v_3=(1,2,3).$$

Suppose

$$a v_1+b v_2+c v_3=(0,0,0).$$

Then

$$a(1,1,1)+b(0,1,0)+c(1,2,3)=(0,0,0).$$

Comparing components,

$$a+c=0,$$ $$a+b+2c=0,$$ $$a+3c=0.$$

From $a+c=0$, we get $a=-c$. From $a+3c=0$, we get $-c+3c=0$, so $2c=0$, hence $c=0$. Then $a=0$. Finally, from $a+b+2c=0$, we get $b=0$.

Thus

$$a=b=c=0.$$

So the only linear combination of $v_1,v_2,v_3$ equal to zero is the trivial one, i.e. the three vectors are linearly independent.

Since $\dim\mathbb R^3=3$, any three linearly independent vectors in $\mathbb R^3$ also span $\mathbb R^3$. This can be checked directly with the determinant:

$\begin{vmatrix}1&1&1\\0&1&0\\1&2&3\end{vmatrix}=1(3-0)-1(0-0)+1(0-1)=2\neq 0,$

so the matrix with these rows is invertible and every $(x,y,z)\in\mathbb R^3$ can be written as a combination of $v_1,v_2,v_3$.

Hence $\{(1,1,1),(0,1,0),(1,2,3)\}$ is linearly independent and spans $\mathbb R^3$, so it is a basis of $\mathbb R^3$.

1. Definition of Upper Triangular Matrix

A square matrix (A=[a_{ij}]_{n\times n}) is called an upper triangular matrix if all the elements below its principal diagonal are zero, i.e.

[ a_{ij}=0 \quad \text{for all } i>j. ]

Elements on and above the principal diagonal may be any numbers.

Example:

[ A= \begin{pmatrix} 2 & -1 & 4\ 0 & 3 & 5\ 0 & 0 & -2 \end{pmatrix} ]

Here, all elements below the principal diagonal are zero, so (A) is an upper triangular matrix.

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2. Prove that the product of two upper triangular matrices of the same order is also upper triangular

Let

[ A=[a_{ij}]{n\times n}, \qquad B=[b{ij}]_{n\times n} ]

be two upper triangular matrices of the same order (n).

Then,

[ a_{ij}=0 \quad \text{for } i>j ]

and

[ b_{ij}=0 \quad \text{for } i>j. ]

Let

[ C=AB=[c_{ij}]_{n\times n}. ]

By the definition of matrix multiplication,

[ c_{ij}=\sum_{k=1}^{n} a_{ik}b_{kj}. ]

We have to prove that (C) is also upper triangular, i.e.

[ c_{ij}=0 \quad \text{for all } i>j. ]

Take (i>j). For any (k),

• If (k<i), then (i>k), so (a_{ik}=0).
• If (k\ge i), then (k>j), so (b_{kj}=0).

Thus, in every term (a_{ik}b_{kj}), at least one factor is zero. Hence,

[ a_{ik}b_{kj}=0 ]

for every (k). Therefore,

[ c_{ij}=\sum_{k=1}^{n} a_{ik}b_{kj}=0. ]

So,

[ c_{ij}=0 \quad \text{for all } i>j. ]

Hence, (C=AB) is also an upper triangular matrix.

Therefore, the product of two upper triangular matrices of the same order is also an upper triangular matrix.

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3. Verification by an Example

Let

[ A= \begin{pmatrix} 1 & 2 & 3\ 0 & 4 & 5\ 0 & 0 & 6 \end{pmatrix}, \qquad B= \begin{pmatrix} 2 & 1 & 0\ 0 & 3 & 1\ 0 & 0 & 4 \end{pmatrix}. ]

Both (A) and (B) are upper triangular matrices.

Now,

[ AB= \begin{pmatrix} 1 & 2 & 3\ 0 & 4 & 5\ 0 & 0 & 6 \end{pmatrix} \begin{pmatrix} 2 & 1 & 0\ 0 & 3 & 1\ 0 & 0 & 4 \end{pmatrix} ]

[

\begin{pmatrix} 1\cdot2+2\cdot0+3\cdot0 & 1\cdot1+2\cdot3+3\cdot0 & 1\cdot0+2\cdot1+3\cdot4\ 0\cdot2+4\cdot0+5\cdot0 & 0\cdot1+4\cdot3+5\cdot0 & 0\cdot0+4\cdot1+5\cdot4\ 0\cdot2+0\cdot0+6\cdot0 & 0\cdot1+0\cdot3+6\cdot0 & 0\cdot0+0\cdot1+6\cdot4 \end{pmatrix} ]

[

\begin{pmatrix} 2 & 7 & 14\ 0 & 12 & 24\ 0 & 0 & 24 \end{pmatrix}. ]

In (AB), all elements below the principal diagonal are zero. Hence, (AB) is also an upper triangular matrix.

Thus, the result is verified.

Proof of (\det(AB)=\det(A)\det(B))

Let (A) and (B) be square matrices of order (n).

Case 1: (A) or (B) is singular

If (A) is singular, then (\det A=0). Also,
[ \operatorname{rank}(AB)\le \operatorname{rank}(A)<n, ] so (AB) is singular. Hence (\det(AB)=0). Therefore, [ \det(AB)=0=\det A\det B. ]

Similarly, if (B) is singular, then (\det B=0). Since (Bx=0) for some nonzero (x), we get [ ABx=A(Bx)=0, ] so (AB) is also singular. Hence (\det(AB)=0=\det A\det B).

Case 2: (A) and (B) are both non-singular

Every non-singular matrix can be expressed as a product of elementary matrices. So let [ A=E_1E_2\cdots E_r, ] where each (E_i) is an elementary matrix.

For an elementary matrix (E), we have [ \det(EB)=\det(E)\det(B). ] This is true because:

1. If (E) is obtained by interchanging two rows, then (\det E=-1), and (EB) is obtained by interchanging the same two rows of (B). Hence
   [ \det(EB)=-\det B=\det E\det B. ]

2. If (E) is obtained by multiplying a row by a nonzero scalar (k), then (\det E=k), and (EB) multiplies the same row of (B) by (k). Hence
   [ \det(EB)=k\det B=\det E\det B. ]

3. If (E) is obtained by adding a multiple of one row to another row, then (\det E=1), and this row operation leaves the determinant unchanged. Hence
   [ \det(EB)=\det B=\det E\det B. ]

Now, [ \begin{aligned} \det(AB) &=\det\bigl(E_1E_2\cdots E_rB\bigr)\ &=\det E_1\det\bigl(E_2\cdots E_rB\bigr)\ &=\det E_1\det E_2\cdots \det E_r\det B\ &=\det(E_1E_2\cdots E_r)\det B\ &=\det A\det B. \end{aligned} ]

Therefore, for square matrices of the same size, [ \boxed{\det(AB)=\det(A)\det(B)}. ]

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Verification

Given [ A=\begin{bmatrix} 1&0&2\ 3&-1&4\ 1&5&1 \end{bmatrix}, \qquad B=\begin{bmatrix} 2&2&3\ -1&1&3\ 5&1&4 \end{bmatrix}. ]

Determinant of (A)

[ \begin{aligned} \det A &=\begin{vmatrix} 1&0&2\ 3&-1&4\ 1&5&1 \end{vmatrix}\ &=1\bigl[(-1)(1)-4(5)\bigr]-0+2\bigl[(3)(5)-(-1)(1)\bigr]\ &=1(-1-20)+2(15+1)\ &=-21+32\ &=11. \end{aligned} ]

Determinant of (B)

[ \begin{aligned} \det B &=\begin{vmatrix} 2&2&3\ -1&1&3\ 5&1&4 \end{vmatrix}\ &=2\bigl[(1)(4)-3(1)\bigr]-2\bigl[(-1)(4)-3(5)\bigr]+3\bigl[(-1)(1)-1(5)\bigr]\ &=2(4-3)-2(-4-15)+3(-1-5)\ &=2+38-18\ &=22. \end{aligned} ]

Thus, [ \det A\det B=11\times 22=242. ]

Product (AB)

[ \begin{aligned} AB &=\begin{bmatrix} 1&0&2\ 3&-1&4\ 1&5&1 \end{bmatrix} \begin{bmatrix} 2&2&3\ -1&1&3\ 5&1&4 \end{bmatrix}\ &=\begin{bmatrix} 12&4&11\ 27&9&22\ 2&8&22 \end{bmatrix}. \end{aligned} ]

Determinant of (AB)

[ \begin{aligned} \det(AB) &=\begin{vmatrix} 12&4&11\ 27&9&22\ 2&8&22 \end{vmatrix}\ &=12(9\cdot22-22\cdot8)-4(27\cdot22-22\cdot2)+11(27\cdot8-9\cdot2)\ &=12(198-176)-4(594-44)+11(216-18)\ &=12(22)-4(550)+11(198)\ &=264-2200+2178\ &=242. \end{aligned} ]

Hence, [ \det(AB)=242=\det A\det B. ]

So the result (\det(AB)=\det(A)\det(B)) is verified.

1. Proof: Eigenvectors corresponding to distinct eigenvalues are linearly independent

Theorem. Let (A) be an (n\times n) matrix. If
[ v_1,v_2,\dots,v_k ] are eigenvectors of (A) corresponding to distinct eigenvalues
[ \lambda_1,\lambda_2,\dots,\lambda_k, ] then (v_1,v_2,\dots,v_k) are linearly independent.

Proof. We prove by induction on (k).

For (k=1), (v_1\neq 0), so (v_1) is linearly independent. Thus the result is true for (k=1).

Assume the result is true for (k-1) eigenvectors. Suppose [ c_1v_1+c_2v_2+\cdots+c_kv_k=0. \quad (1) ] Apply (A) to both sides of (1): [ c_1Av_1+c_2Av_2+\cdots+c_kAv_k=0. ] Since (Av_i=\lambda_i v_i), [ c_1\lambda_1v_1+c_2\lambda_2v_2+\cdots+c_k\lambda_kv_k=0. \quad (2) ] Multiply (1) by (\lambda_k): [ c_1\lambda_kv_1+c_2\lambda_kv_2+\cdots+c_k\lambda_kv_k=0. \quad (3) ] Subtract (3) from (2): [ c_1(\lambda_1-\lambda_k)v_1+c_2(\lambda_2-\lambda_k)v_2+\cdots+c_{k-1}(\lambda_{k-1}-\lambda_k)v_{k-1}=0. ] By the induction hypothesis, (v_1,v_2,\dots,v_{k-1}) are linearly independent. Hence [ c_i(\lambda_i-\lambda_k)=0,\qquad i=1,2,\dots,k-1. ] Since the eigenvalues are distinct, (\lambda_i-\lambda_k\neq 0). Therefore [ c_1=c_2=\cdots=c_{k-1}=0. ] Substituting these values in (1), we get [ c_kv_k=0. ] Since (v_k\neq 0), we have (c_k=0). Thus all (c_i=0), proving that [ v_1,v_2,\dots,v_k ] are linearly independent. (\square)

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2. Eigenvalue and eigenvector of

[ A=\begin{bmatrix}3&-1\1&1\end{bmatrix} ]

The characteristic equation is [ \det(A-\lambda I)=0. ] Now, [ A-\lambda I= \begin{bmatrix} 3-\lambda&-1\ 1&1-\lambda \end{bmatrix}. ] So, [ \det(A-\lambda I) =(3-\lambda)(1-\lambda)-(-1)(1) ] [ =(3-\lambda)(1-\lambda)+1 ] [ =\lambda^2-4\lambda+4 =(\lambda-2)^2. ] Hence, [ (\lambda-2)^2=0 ] so [ \lambda=2,\quad \lambda=2. ] Thus the matrix has the repeated eigenvalue [ \boxed{\lambda=2} ] with algebraic multiplicity (2).

Now find the corresponding eigenvector. Solve [ (A-2I)X=0. ] [ A-2I= \begin{bmatrix} 1&-1\ 1&-1 \end{bmatrix}. ] Let [ X=\begin{bmatrix}x\y\end{bmatrix}. ] Then [ \begin{bmatrix} 1&-1\ 1&-1 \end{bmatrix} \begin{bmatrix}x\y\end{bmatrix}

\begin{bmatrix}0\0\end{bmatrix}. ] This gives [ x-y=0 \implies x=y. ] Taking (x=1), we get (y=1). Therefore one eigenvector is [ \boxed{X=\begin{bmatrix}1\1\end{bmatrix}}. ] The general eigenvector corresponding to (\lambda=2) is [ \boxed{X=t\begin{bmatrix}1\1\end{bmatrix},\qquad t\neq 0.} ]

Since the eigenvalue (\lambda=2) is repeated, there is only one linearly independent eigenvector for this matrix.

Main Question

1. Definition of Orthogonal Complement

Let (V) be an inner product space and (S) be any non-empty subset of (V). The orthogonal complement of (S), denoted by (S^\perp), is defined as

[ S^\perp = { x \in V : \langle x, s \rangle = 0 \text{ for every } s \in S }. ]

Thus (S^\perp) consists of all vectors in (V) which are orthogonal to every vector of (S).

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2. Check whether (x=(3,7,1)) belongs to ({u,v}^\perp)

Given:

[ u=(5,-2,-1), \qquad v=(2,-3,15), \qquad x=(3,7,1). ]

Using the usual dot product in (\mathbb{R}^3),

[ \langle x,u \rangle = 3(5)+7(-2)+1(-1) =15-14-1=0. ]

Also,

[ \langle x,v \rangle = 3(2)+7(-3)+1(15) =6-21+15=0. ]

Since (x) is orthogonal to both (u) and (v), it is orthogonal to every linear combination of (u) and (v). Hence

[ x \in {u,v}^\perp. ]

Therefore, the vector (x=(3,7,1)) lies in the orthogonal complement of the pair ({u,v}).

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3. Prove that the orthogonal complement of any subset is a subspace

Let (S) be any subset of an inner product space (V). Consider

[ S^\perp = { x \in V : \langle x,s \rangle = 0 \text{ for all } s \in S }. ]

We show that (S^\perp) is a subspace of (V).

1. Zero vector:
   For every (s \in S),

   [ \langle 0,s \rangle = 0. ]

   Hence (0 \in S^\perp), so (S^\perp) is non-empty.

2. Closed under addition:
   Let (x,y \in S^\perp). Then for every (s \in S),

   [ \langle x,s \rangle = 0, \qquad \langle y,s \rangle = 0. ]

   Therefore,

   [ \langle x+y,s \rangle = \langle x,s \rangle + \langle y,s \rangle = 0+0=0. ]

   Hence (x+y \in S^\perp).

3. Closed under scalar multiplication:
   Let (x \in S^\perp) and (\alpha) be a scalar. Then for every (s \in S),

   [ \langle \alpha x,s \rangle = \alpha \langle x,s \rangle = \alpha \cdot 0 = 0. ]

   Hence (\alpha x \in S^\perp).

Since (S^\perp) contains the zero vector and is closed under addition and scalar multiplication, it is a subspace of (V).

[ \boxed{S^\perp \text{ is a subspace of } V.} ]

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OR

Gram-Schmidt Process

Given vectors in (\mathbb{R}^3):

[ u_1=(1,1,1), \qquad u_2=(0,1,1), \qquad u_3=(0,0,1). ]

Let the orthogonal basis be (v_1,v_2,v_3).

Step 1: First orthogonal vector

[ v_1=u_1=(1,1,1). ]

Step 2: Second orthogonal vector

[ v_2=u_2-\frac{\langle u_2,v_1 \rangle}{\langle v_1,v_1 \rangle}v_1. ]

Now,

[ \langle u_2,v_1 \rangle = 0(1)+1(1)+1(1)=2, ]

[ \langle v_1,v_1 \rangle = 1^2+1^2+1^2=3. ]

So,

[ v_2=(0,1,1)-\frac{2}{3}(1,1,1) =\left(-\frac{2}{3},\frac{1}{3},\frac{1}{3}\right). ]

Multiplying by (3), we may take

[ v_2=(-2,1,1). ]

Step 3: Third orthogonal vector

[ v_3=u_3-\frac{\langle u_3,v_1 \rangle}{\langle v_1,v_1 \rangle}v_1 -\frac{\langle u_3,v_2 \rangle}{\langle v_2,v_2 \rangle}v_2. ]

Now,

[ \langle u_3,v_1 \rangle = 0(1)+0(1)+1(1)=1, ]

[ \langle u_3,v_2 \rangle = 0(-2)+0(1)+1(1)=1, ]

[ \langle v_2,v_2 \rangle = (-2)^2+1^2+1^2=6. ]

Thus,

[ v_3=(0,0,1)-\frac{1}{3}(1,1,1)-\frac{1}{6}(-2,1,1). ]

[ v_3=(0,0,1)-\left(\frac13,\frac13,\frac13\right) -\left(-\frac13,\frac16,\frac16\right) ]

[ v_3=\left(0,-\frac12,\frac12\right). ]

Multiplying by (2), we may take

[ v_3=(0,-1,1). ]

Hence an orthogonal basis is

[ \boxed{ v_1=(1,1,1), \qquad v_2=(-2,1,1), \qquad v_3=(0,-1,1) } ]

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Orthonormal Basis

Normalize each orthogonal vector.

[ |v_1|=\sqrt{1^2+1^2+1^2}=\sqrt3. ]

[ e_1=\frac{v_1}{|v_1|} =\left(\frac1{\sqrt3},\frac1{\sqrt3},\frac1{\sqrt3}\right). ]

[ |v_2|=\sqrt{(-2)^2+1^2+1^2}=\sqrt6. ]

[ e_2=\frac{v_2}{|v_2|} =\left(-\frac2{\sqrt6},\frac1{\sqrt6},\frac1{\sqrt6}\right). ]

[ |v_3|=\sqrt{0^2+(-1)^2+1^2}=\sqrt2. ]

[ e_3=\frac{v_3}{|v_3|} =\left(0,-\frac1{\sqrt2},\frac1{\sqrt2}\right). ]

Therefore, the orthonormal basis is

[ \boxed{ e_1=\left(\frac1{\sqrt3},\frac1{\sqrt3},\frac1{\sqrt3}\right), \quad e_2=\left(-\frac2{\sqrt6},\frac1{\sqrt6},\frac1{\sqrt6}\right), \quad e_3=\left(0,-\frac1{\sqrt2},\frac1{\sqrt2}\right) } ]

These vectors are mutually orthogonal and each has unit length.

First alternative

1. Definition of self-adjoint mapping

Let (V) be an inner product space over (\mathbb{R}) or (\mathbb{C}), and let (T:V\to V) be a linear operator. Then (T) is called self-adjoint if

[ \langle T(x),y\rangle=\langle x,T(y)\rangle ]

for all (x,y\in V). In matrix form, if (A) is the matrix of (T), then (T) is self-adjoint when

[ A=A^\dagger ]

where (A^\dagger) is the conjugate transpose of (A). Over (\mathbb{R}), this means (A=A^T), i.e. (A) is symmetric.

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2. Prove that all eigenvalues of a self-adjoint operator are real

Let (\lambda) be an eigenvalue of the self-adjoint operator (T), and let (v\neq 0) be a corresponding eigenvector. Then

[ T(v)=\lambda v. ]

Now consider

[ \lambda\langle v,v\rangle

\langle \lambda v,v\rangle

\langle T(v),v\rangle. ]

Since (T) is self-adjoint,

[ \langle T(v),v\rangle=\langle v,T(v)\rangle. ]

Again, (T(v)=\lambda v), so

[ \langle v,T(v)\rangle=\langle v,\lambda v\rangle=\overline{\lambda}\langle v,v\rangle. ]

Thus

[ \lambda\langle v,v\rangle=\overline{\lambda}\langle v,v\rangle. ]

Since (v\neq 0), we have (\langle v,v\rangle>0). Therefore,

[ \lambda=\overline{\lambda}. ]

A number equal to its conjugate is real. Hence every eigenvalue of a self-adjoint operator is real.

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3. Prove that eigenvectors corresponding to different eigenvalues are mutually orthogonal

Let (\lambda) and (\mu) be two distinct eigenvalues of the self-adjoint operator (T), and let (u) and (v) be corresponding nonzero eigenvectors:

[ T(u)=\lambda u,\qquad T(v)=\mu v,\qquad \lambda\neq \mu. ]

Now,

[ \lambda\langle u,v\rangle

\langle \lambda u,v\rangle

\langle T(u),v\rangle. ]

Since (T) is self-adjoint,

[ \langle T(u),v\rangle=\langle u,T(v)\rangle. ]

Using (T(v)=\mu v),

[ \langle u,T(v)\rangle=\langle u,\mu v\rangle=\overline{\mu}\langle u,v\rangle. ]

But from part 2, (\mu) is real, so (\overline{\mu}=\mu). Hence

[ \lambda\langle u,v\rangle=\mu\langle u,v\rangle. ]

Therefore,

[ (\lambda-\mu)\langle u,v\rangle=0. ]

Since (\lambda\neq \mu), we must have

[ \langle u,v\rangle=0. ]

Thus (u) and (v) are orthogonal. Hence any two eigenvectors of a self-adjoint operator associated with different eigenvalues are mutually orthogonal.

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OR

Cayley-Hamilton theorem

The Cayley-Hamilton theorem states that every square matrix satisfies its own characteristic equation. That is, if (A) is an (n\times n) matrix and

[ p(\lambda)=\det(A-\lambda I) ]

is its characteristic polynomial, then

[ p(A)=O, ]

where (O) is the zero matrix.

---

Verification for

[ A=\begin{bmatrix}0&3\2&-1\end{bmatrix} ]

The characteristic polynomial is

[ p(\lambda)=\det(A-\lambda I)

\begin{vmatrix} -\lambda & 3\ 2 & -1-\lambda \end{vmatrix}. ]

So,

[ p(\lambda)=(-\lambda)(-1-\lambda)-6 =\lambda^2+\lambda-6. ]

Thus the characteristic equation is

[ A^2+A-6I=O. ]

Now,

[ A^2= \begin{bmatrix}0&3\2&-1\end{bmatrix} \begin{bmatrix}0&3\2&-1\end{bmatrix}

\begin{bmatrix}6&-3\-2&7\end{bmatrix}. ]

Therefore,

[ A^2+A-6I

\begin{bmatrix}6&-3\-2&7\end{bmatrix} + \begin{bmatrix}0&3\2&-1\end{bmatrix}

\begin{bmatrix}6&0\0&6\end{bmatrix}. ]

[

\begin{bmatrix}6&0\0&6\end{bmatrix}

\begin{bmatrix}6&0\0&6\end{bmatrix}

\begin{bmatrix}0&0\0&0\end{bmatrix} =O. ]

Hence the Cayley-Hamilton theorem is verified for the given matrix (A).