Statement. 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), with (v,z\neq 0). Then (L_1=L_2) if and only if both (u-w) and (v) are scalar multiples of (z).

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Proof

1. Suppose (L_1=L_2). We show (u-w) and (v) are multiples of (z).

Since (u\in L_1=L_2), there exists (s_0\in\mathbb R) such that [ u=w+s_0z. ] Hence [ 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).

So if the lines are the same, then (u-w) and (v) are multiples of (z).

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

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

Now take any point of (L_1): [ x=u+tv. ] Using (u=w+az) and (v=bz), [ x=w+az+t(bz)=w+(a+tb)z. ] Thus (x\in L_2). Hence (L_1\subseteq L_2).

Conversely, take any point of (L_2): [ y=w+sz. ] Since (b\neq 0), we can write [ z=\frac1b v. ] Also from (u-w=az), [ w=u-az=u-\frac{a}{b}v. ] Therefore [ y=w+sz =u-\frac{a}{b}v+s\left(\frac1b v\right) =u+\left(\frac{s-a}{b}\right)v. ] So (y\in L_1). Hence (L_2\subseteq L_1).

Thus (L_1=L_2).

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

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Given lines

[ L_1=(-3,-3,-6)+t(9,15,12), ] [ L_2=(3,7,2)+s(3,5,4). ]

Here, [ u=(-3,-3,-6),\qquad v=(9,15,12), ] [ w=(3,7,2),\qquad z=(3,5,4). ]

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

Also, [ 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).

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

Indeed, comparing coordinates, [ (-3+9t,,-3+15t,,-6+12t)=(3+3s,,7+5s,,2+4s) ] gives [ s=3t-2. ] So every point of (L_1) lies on (L_2), and vice versa. Therefore, [ \boxed{L_1=L_2} ] and the two given lines are the same line in (\mathbb R^3).

First alternative

(a) Matrix $\begin{bmatrix}0&-1\\1&0\end{bmatrix}$

The mapping is
[ T(x,y)=\begin{bmatrix}0&-1\1&0\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=(-y,x). ]

This is a rotation through $90^\circ$ anticlockwise about the origin.

Effect on the unit square:
The vertices [ (0,0),\ (1,0),\ (1,1),\ (0,1) ] map to [ (0,0),\ (0,1),\ (-1,1),\ (-1,0). ] So the unit square is turned through a right angle anticlockwise.

Figure idea:

Before rotation:                 After rotation:
y                                y
↑                                ↑
|  C(0,1)----B(1,1)              |  B'(-1,1)----A'(0,1)
|  |          |                  |  |            |
|  O(0,0)----A(1,0)              |  C'(-1,0)----O(0,0)
+----------------→ x             +----------------→ x

Properties:

• Lengths are preserved: $\|(-y,x)\|=\|(x,y)\|$.
• Angles are preserved.
• Area is preserved because $\det = 1$.
• It is invertible; the inverse is rotation by $90^\circ$ clockwise.
• It has no real eigenvector except the zero vector.

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(b) Matrix $\begin{bmatrix}2&0\\0&2\end{bmatrix}$

The mapping is
[ T(x,y)=\begin{bmatrix}2&0\0&2\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=(2x,2y). ]

This is a uniform scaling or dilation by factor $2$ about the origin.

Effect on the unit square:
The vertices [ (0,0),\ (1,0),\ (1,1),\ (0,1) ] map to [ (0,0),\ (2,0),\ (2,2),\ (0,2). ] So the unit square becomes a square of side $2$.

Figure idea:

Before scaling:                  After scaling:
y                                y
↑                                ↑
|  C(0,1)----B(1,1)              |  C'(0,2)----B'(2,2)
|  |          |                  |  |            |
|  O(0,0)----A(1,0)              |  O(0,0)----A'(2,0)
+----------------→ x             +----------------→ x

Properties:

• Every vector is stretched radially outward by factor $2$.
• Directions are preserved.
• Lengths are multiplied by $2$.
• Areas are multiplied by $4$ because $\det = 4$.
• Eigenvalues are $2,2$; every nonzero vector is an eigenvector.
• The inverse is scaling by factor $\frac12$.

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OR alternative

Definitions

Vector space:
A nonempty set $V$ over a field $F$ is called a vector space if two operations, vector addition and scalar multiplication, are defined on $V$ and satisfy the following axioms:

1. Closure under addition: $u+v\in V$.
2. Commutativity: $u+v=v+u$.
3. Associativity: $(u+v)+w=u+(v+w)$.
4. Additive identity: there exists $0\in V$ such that $v+0=v$.
5. Additive inverse: for each $v\in V$, there exists $-v\in V$ such that $v+(-v)=0$.
6. Closure under scalar multiplication: $cv\in V$.
7. Distributivity over vector addition: $c(u+v)=cu+cv$.
8. Distributivity over scalar addition: $(c+d)v=cv+dv$.
9. Associativity of scalar multiplication: $c(dv)=(cd)v$.
10. Identity scalar: $1v=v$.

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Linearly dependent set:
A set of vectors $\{v_1,v_2,\dots,v_n\}$ in a vector space $V$ 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+\cdots+c_nv_n=0. ]

Linearly independent set:
A set of vectors $\{v_1,v_2,\dots,v_n\}$ is linearly independent if
[ c_1v_1+c_2v_2+\cdots+c_nv_n=0 ] implies that [ c_1=c_2=\cdots=c_n=0. ]

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Determination for $\{a_1,a_2,a_3\}$

Given: [ a_1=(2,-3,1),\quad a_2=(3,-5,2),\quad a_3=(4,-5,1). ]

Let [ c_1a_1+c_2a_2+c_3a_3=0. ] Then [ c_1(2,-3,1)+c_2(3,-5,2)+c_3(4,-5,1)=(0,0,0). ]

Equating 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. ] Substitute in (1): [ 2(-2c_2-c_3)+3c_2+4c_3=0 ] [ -c_2+2c_3=0 ] [ c_2=2c_3. ] Then [ c_1=-2(2c_3)-c_3=-5c_3. ] Taking $c_3=1$, we get [ c_1=-5,\quad c_2=2,\quad c_3=1. ] Thus [ -5a_1+2a_2+a_3=0. ] Since the scalars are not all zero, the given set of vectors is linearly dependent.

Equivalently, [ \det\begin{bmatrix} 2&-3&1\ 3&-5&2\ 4&-5&1 \end{bmatrix}=0, ] so the vectors are linearly dependent.

Alternative 1

1. Definitions

Vector subspace:
A nonempty subset (W) of a vector space (V) over a field (F) is called a subspace of (V) if it is closed under vector addition and scalar multiplication, i.e. for all (u,v\in W) and (c\in F), [ u+v\in W,\qquad cu\in W. ] Equivalently, (W) is itself a vector space under the operations inherited from (V).

Inverse image:
Let (f:X\to Y) be a function and (W\subseteq Y). The inverse image of (W) under (f) is [ f^{-1}(W)={x\in X: f(x)\in W}. ] It is a subset of (X), not the inverse function of (f).

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2. Proof that (f^{-1}(W)) is a subspace

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

We show that (f^{-1}(W)) is a subspace of (X).

1. Nonempty:
   Since (f) is linear, [ f(0_X)=0_Y. ] Also (0_Y\in W) because (W) is a subspace. Hence [ f(0_X)\in W, ] so (0_X\in f^{-1}(W)). Thus (f^{-1}(W)\neq\varnothing).

2. Closed under addition:
   Let (x_1,x_2\in f^{-1}(W)). Then [ f(x_1),f(x_2)\in W. ] Since (W) is a subspace, [ f(x_1)+f(x_2)\in W. ] But (f) is linear, so [ f(x_1+x_2)=f(x_1)+f(x_2)\in W. ] Therefore, [ x_1+x_2\in f^{-1}(W). ]

3. Closed under scalar multiplication:
   Let (x\in f^{-1}(W)) and (c\in F). Then (f(x)\in W). Since (W) is a subspace, [ c f(x)\in W. ] Again, (f) is linear, so [ f(cx)=c f(x)\in W. ] Hence, [ cx\in f^{-1}(W). ]

Therefore (f^{-1}(W)) is nonempty and closed under addition and scalar multiplication. Hence (f^{-1}(W)) is a subspace of (X). (\square)

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3. Computation of (f^{-1}(W))

Let [ f:\mathbb R^4\to \mathbb R^3 ] be defined by [ f(x_1,x_2,x_3,x_4)=(2x_1-x_2+x_4,\ x_1+x_2-x_3,\ x_3+x_4). ] Let [ W=\operatorname{Span}{z,w}, ] where [ z=(1,1,0),\qquad w=(0,1,1). ]

A vector (y=(y_1,y_2,y_3)\in W) iff [ y=a z+b w ] for some (a,b\in\mathbb R). Thus [ (y_1,y_2,y_3)=a(1,1,0)+b(0,1,1)=(a,a+b,b). ] So [ y_1=a,\qquad y_2=a+b,\qquad y_3=b. ] Hence [ y_2=y_1+y_3, ] or equivalently, [ y_1-y_2+y_3=0. ]

Now (x=(x_1,x_2,x_3,x_4)\in f^{-1}(W)) iff (f(x)\in W). Therefore, [ f_1-f_2+f_3=0, ] where [ f_1=2x_1-x_2+x_4,\quad f_2=x_1+x_2-x_3,\quad f_3=x_3+x_4. ] Thus, [ (2x_1-x_2+x_4)-(x_1+x_2-x_3)+(x_3+x_4)=0. ] Simplifying, [ x_1-2x_2+2x_3+2x_4=0. ]

Therefore, [ \boxed{ f^{-1}(W)={(x_1,x_2,x_3,x_4)\in\mathbb R^4: x_1-2x_2+2x_3+2x_4=0} } ]

This is a hyperplane through the origin in (\mathbb R^4). Equivalently, [ x_1=2x_2-2x_3-2x_4. ] So [ f^{-1}(W)={(2s-2t-2u,\ s,\ t,\ u): s,t,u\in\mathbb R}. ] A basis for (f^{-1}(W)) is [ {(2,1,0,0),\ (-2,0,1,0),\ (-2,0,0,1)}. ] Hence (\dim f^{-1}(W)=3).

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OR

1. Definition of basis

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

1. (B) is linearly independent, and
2. (B) spans (V), i.e. every vector in (V) can be expressed as a linear combination of vectors in (B).

Equivalently, every vector in (V) can be written uniquely as a finite linear combination of elements of (B).

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2. 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). ] Consider the matrix whose columns are (v_1,v_2,v_3): [ A= \begin{pmatrix} 1&0&1\ 1&1&2\ 1&0&3 \end{pmatrix}. ] Its determinant is [ \det A

1(1\cdot 3-2\cdot 0) -0(1\cdot 3-2\cdot 1) +1(1\cdot 0-1\cdot 1) ] [ =3-0-1=2\neq 0. ] Since (\det A\neq 0), the vectors (v_1,v_2,v_3) are linearly independent.

Also, (\mathbb R^3) has dimension (3). Therefore any set of (3) linearly independent vectors in (\mathbb R^3) automatically spans (\mathbb R^3). Hence [ {(1,1,1),(0,1,0),(1,2,3)} ] is a basis for (\mathbb R^3).

For completeness, let ((a,b,c)\in\mathbb R^3). Suppose [ (a,b,c)=\alpha(1,1,1)+\beta(0,1,0)+\gamma(1,2,3). ] Then [ \alpha+\gamma=a, ] [ \alpha+\beta+2\gamma=b, ] [ \alpha+3\gamma=c. ] Solving, [ \gamma=\frac{c-a}{2},\qquad \alpha=\frac{3a-c}{2},\qquad \beta=\frac{2b-a-c}{2}. ] Thus every vector ((a,b,c)\in\mathbb R^3) can be expressed uniquely as a linear combination of the given three vectors. Therefore the set spans (\mathbb R^3) and is linearly independent. Hence it is a basis for (\mathbb R^3). (\square)

Definition:
A square matrix (A=[a_{ij}]) of order (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. ] Thus its general form is [ A=\begin{pmatrix} a_{11}&a_{12}&\cdots&a_{1n}\ 0&a_{22}&\cdots&a_{2n}\ \vdots&\vdots&\ddots&\vdots\ 0&0&\cdots&a_{nn} \end{pmatrix}. ]

Example:
[ A=\begin{pmatrix} 1&2&3\ 0&4&5\ 0&0&6 \end{pmatrix} ] is an upper triangular matrix of order (3).

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Proof 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 upper triangular, i.e. [ c_{ij}=0 \quad \text{for all } i>j. ]

Take any (i>j). Consider a term (a_{ik}b_{kj}) in the sum.

• Since (A) is upper triangular, (a_{ik}=0) whenever (i>k). Hence (a_{ik}\neq 0) only when (i\le k).
• Since (B) is upper triangular, (b_{kj}=0) whenever (k>j). Hence (b_{kj}\neq 0) only when (k\le j).

Therefore, for the product (a_{ik}b_{kj}) to be nonzero, we must have [ i\le k \quad \text{and} \quad k\le j, ] which gives [ i\le j. ] But we are considering (i>j), which contradicts (i\le j). Hence every term (a_{ik}b_{kj}) is zero for (i>j). Therefore, [ c_{ij}=0 \quad \text{for all } i>j. ]

Thus (C=AB) is also an upper triangular matrix. Hence the product of two upper triangular matrices of the same order is also upper triangular.

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Verification by example:

Take [ 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 of order (3).

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}. ]

Computing the entries: [ c_{11}=1\cdot2+2\cdot0+3\cdot0=2, ] [ c_{12}=1\cdot1+2\cdot3+3\cdot0=7, ] [ c_{13}=1\cdot0+2\cdot1+3\cdot4=14, ] [ c_{21}=0\cdot2+4\cdot0+5\cdot0=0, ] [ c_{22}=0\cdot1+4\cdot3+5\cdot0=12, ] [ c_{23}=0\cdot0+4\cdot1+5\cdot4=24, ] [ c_{31}=0,\qquad c_{32}=0, ] [ c_{33}=0\cdot0+0\cdot1+6\cdot4=24. ]

Hence, [ AB= \begin{pmatrix} 2&7&14\ 0&12&24\ 0&0&24 \end{pmatrix}. ] This matrix has all entries below the principal diagonal equal to zero. Therefore, (AB) is also an upper triangular matrix. Hence the result is verified.

Proof of (\operatorname{Det}(AB)=\operatorname{Det}(A)\operatorname{Det}(B))

Let (A,B) be (n\times n) matrices over a field.

If (A) or (B) is singular, then [ \operatorname{rank}(AB)\le \min(\operatorname{rank}A,\operatorname{rank}B)<n, ] so (AB) is also singular. Hence [ \operatorname{Det}(AB)=0=\operatorname{Det}(A)\operatorname{Det}(B). ]

Now assume (A,B) are nonsingular. Every nonsingular matrix is a product of elementary matrices. Write [ A=E_kE_{k-1}\cdots E_1. ] For an elementary matrix (E), we have [ \operatorname{Det}(EX)=\operatorname{Det}(E)\operatorname{Det}(X) ] for any matrix (X), because:

• If (E) interchanges two rows, then (\operatorname{Det}(E)=-1), and (EX) interchanges the same two rows of (X), so the determinant changes sign.
• If (E) multiplies a row by (c), then (\operatorname{Det}(E)=c), and (EX) multiplies that row of (X) by (c), so the determinant is multiplied by (c).
• If (E) adds a multiple of one row to another, then (\operatorname{Det}(E)=1), and this row operation leaves the determinant unchanged.

Therefore, [ \begin{aligned} \operatorname{Det}(AB) &=\operatorname{Det}(E_kE_{k-1}\cdots E_1B)\ &=\operatorname{Det}(E_k)\operatorname{Det}(E_{k-1}\cdots E_1B)\ &=\cdots\ &=\operatorname{Det}(E_k)\operatorname{Det}(E_{k-1})\cdots \operatorname{Det}(E_1)\operatorname{Det}(B)\ &=\operatorname{Det}(A)\operatorname{Det}(B). \end{aligned} ] Hence, [ \boxed{\operatorname{Det}(AB)=\operatorname{Det}(A)\operatorname{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}. ]

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

Also, [ \begin{aligned} \operatorname{Det}(B) &=\begin{vmatrix} 2&2&3\ -1&1&3\ 5&1&4 \end{vmatrix}\ &=2\begin{vmatrix}1&3\1&4\end{vmatrix} -2\begin{vmatrix}-1&3\5&4\end{vmatrix} +3\begin{vmatrix}-1&1\5&1\end{vmatrix}\ &=2(4-3)-2(-4-15)+3(-1-5)\ &=2(1)-2(-19)+3(-6)\ &=2+38-18=22. \end{aligned} ]

Now, [ \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} ]

Thus, [ \begin{aligned} \operatorname{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} ]

Also, [ \operatorname{Det}(A)\operatorname{Det}(B)=11\cdot22=242. ]

Therefore, [ \boxed{\operatorname{Det}(AB)=\operatorname{Det}(A)\operatorname{Det}(B)} ] is verified for the given matrices.

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

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

Proof. We use induction on $k$.

Base case: For $k=1$, since $\mathbf v_1\neq \mathbf 0$ is an eigenvector, the set $\{\mathbf v_1\}$ is linearly independent.

Induction hypothesis: Suppose the result is true for any set of $k-1$ eigenvectors corresponding to distinct eigenvalues.

Now consider $k$ eigenvectors $\mathbf v_1,\dots,\mathbf v_k$ corresponding to distinct eigenvalues $\lambda_1,\dots,\lambda_k$. Let

$$c_1\mathbf v_1+c_2\mathbf v_2+\cdots+c_k\mathbf v_k=\mathbf 0. \tag{1}$$

Apply $A$ to both sides of (1). Since $A\mathbf v_i=\lambda_i\mathbf v_i$,

$$c_1\lambda_1\mathbf v_1+c_2\lambda_2\mathbf v_2+\cdots+c_k\lambda_k\mathbf v_k=\mathbf 0. \tag{2}$$

Multiply (1) by $\lambda_k$:

$$c_1\lambda_k\mathbf v_1+c_2\lambda_k\mathbf v_2+\cdots+c_k\lambda_k\mathbf v_k=\mathbf 0. \tag{3}$$

Subtract (3) from (2):

$$c_1(\lambda_1-\lambda_k)\mathbf v_1+\cdots+c_{k-1}(\lambda_{k-1}-\lambda_k)\mathbf v_{k-1}=\mathbf 0.$$

By the induction hypothesis, $\mathbf v_1,\dots,\mathbf v_{k-1}$ 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) gives

$$c_k\mathbf v_k=\mathbf 0.$$

Since $\mathbf v_k\neq \mathbf 0$, we get $c_k=0$. Thus all $c_i=0$, so $\mathbf v_1,\dots,\mathbf v_k$ are linearly independent. Hence, by induction, the result is true for any number of eigenvectors corresponding to distinct eigenvalues. $\square$

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2. Eigenvalues and eigenvectors 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

$$\det(A-\lambda I) =(3-\lambda)(1-\lambda)-(-1)(1)$$ $$=(3-\lambda)(1-\lambda)+1$$ $$=\lambda^2-4\lambda+4 =(\lambda-2)^2.$$

Thus

$$(\lambda-2)^2=0 \quad\Rightarrow\quad \lambda=2,\;2.$$

So the only eigenvalue is

$$\boxed{\lambda=2}$$

with algebraic multiplicity $2$.

Eigenvector corresponding to $\lambda=2$

Solve

$$(A-2I)\mathbf v=\mathbf 0.$$

Now

$$A-2I= \begin{bmatrix} 1&-1\\ 1&-1 \end{bmatrix}.$$

Let

$$\mathbf v=\begin{bmatrix}x\\y\end{bmatrix}.$$

Then

$$\begin{bmatrix} 1&-1\\ 1&-1 \end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix}.$$

This gives

$$x-y=0 \quad\Rightarrow\quad x=y.$$

Taking $x=1$, we get $y=1$. Hence one eigenvector is

$$\boxed{\mathbf v=\begin{bmatrix}1\\1\end{bmatrix}}.$$

Therefore,

$$\boxed{\lambda=2,\qquad \mathbf v=\begin{bmatrix}1\\1\end{bmatrix}}$$

or any non-zero scalar multiple of $\begin{bmatrix}1\\1\end{bmatrix}$.

Since the eigenvalue $\lambda=2$ is repeated and there is only one independent eigenvector, the matrix is not diagonalizable.

Option A

1. Definition of orthogonal complement

Let (S) be a subset of an inner product space (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).

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

Given
[ u=(5,-2,1),\qquad v=(2,-3,15),\qquad x=(3,7,1). ] For (x) to belong to the orthogonal complement of ({u,v}), we must have
[ x\cdot u=0 \quad\text{and}\quad x\cdot v=0. ]

Now, [ x\cdot u=(3)(5)+(7)(-2)+(1)(1)=15-14+1=2\neq 0. ] Also, [ x\cdot v=(3)(2)+(7)(-3)+(1)(15)=6-21+15=0. ] Since (x\cdot u\neq 0), (x) is not orthogonal to (u). Therefore, [ x\notin {u,v}^\perp. ]

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

Let (S\subseteq V), where (V) is an inner product space. Put [ W=S^\perp={x\in V:\langle x,s\rangle=0\ \forall s\in S}. ] We show that (W) is a subspace of (V).

1. Zero vector:
   For every (s\in S), [ \langle 0,s\rangle=0. ] Hence (0\in W), so (W\neq\varnothing).

2. Closed under addition:
   Let (x,y\in W). 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 W).

3. Closed under scalar multiplication:
   Let (x\in W) and (\alpha\in\mathbb R). Then for every (s\in S), [ \langle \alpha x,s\rangle=\alpha\langle x,s\rangle=\alpha\cdot 0=0. ] Hence (\alpha x\in W).

Thus (W=S^\perp) is non-empty and closed under addition and scalar multiplication. Therefore, the orthogonal complement of any subset is a subspace.

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Option B (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 vectors be (v_1,v_2,v_3).

Step 1

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

Step 2

[ 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(-\frac23,\frac13,\frac13\right). ] Multiplying by (3), we may take [ w_2=(-2,1,1). ]

Step 3

[ v_3=u_3-\frac{\langle u_3,v_1\rangle}{\langle v_1,v_1\rangle}v_1 -\frac{\langle u_3,w_2\rangle}{\langle w_2,w_2\rangle}w_2. ] Now, [ \langle u_3,v_1\rangle=(0)(1)+(0)(1)+(1)(1)=1, ] [ \langle u_3,w_2\rangle=(0)(-2)+(0)(1)+(1)(1)=1, ] [ \langle w_2,w_2\rangle=(-2)^2+1^2+1^2=6. ] Thus, [ v_3=(0,0,1)-\frac13(1,1,1)-\frac16(-2,1,1). ] [ v_3=\left(0,-\frac12,\frac12\right). ] Multiplying by (2), we may take [ w_3=(0,-1,1). ]

Orthogonal basis

[ \boxed{{(1,1,1),\ (-2,1,1),\ (0,-1,1)}} ] Check: [ (1,1,1)\cdot(-2,1,1)=-2+1+1=0, ] [ (1,1,1)\cdot(0,-1,1)=0-1+1=0, ] [ (-2,1,1)\cdot(0,-1,1)=0-1+1=0. ]

Orthonormal basis

The norms are [ |(1,1,1)|=\sqrt3,\qquad |(-2,1,1)|=\sqrt6,\qquad |(0,-1,1)|=\sqrt2. ] Hence the orthonormal basis is [ \boxed{ \left{ \frac{(1,1,1)}{\sqrt3},\ \frac{(-2,1,1)}{\sqrt6},\ \frac{(0,-1,1)}{\sqrt2} \right} } ] i.e. [ \boxed{ \left{ \left(\frac1{\sqrt3},\frac1{\sqrt3},\frac1{\sqrt3}\right), \left(\frac{-2}{\sqrt6},\frac1{\sqrt6},\frac1{\sqrt6}\right), \left(0,\frac{-1}{\sqrt2},\frac1{\sqrt2}\right) \right} } ]

First alternative

1. Self-adjoint mapping
Let (V) be an inner product space over (\mathbb{R}) or (\mathbb{C}). A linear operator (T:V\to V) is called self-adjoint if
[ \langle T(x),y\rangle=\langle x,T(y)\rangle ] for all (x,y\in V). In matrix form, a self-adjoint operator corresponds to a Hermitian matrix, i.e. (A^*=A).

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2. Eigenvalues of a self-adjoint operator are real
Let (\lambda) be an eigenvalue of a self-adjoint operator (T), and let (v\neq 0) be a corresponding eigenvector. Then
[ T(v)=\lambda v. ] Now, [ \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 =\langle v,\lambda v\rangle =\overline{\lambda}\langle v,v\rangle. ] Hence, [ \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}. ] Thus (\lambda) is real. Hence every eigenvalue of a self-adjoint operator is real.

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3. Eigenvectors corresponding to different eigenvalues are orthogonal
Let (\lambda) and (\mu) be two distinct eigenvalues of a self-adjoint operator (T), and let (v) and (w) be corresponding nonzero eigenvectors: [ T(v)=\lambda v,\qquad T(w)=\mu w. ] Then [ \lambda\langle v,w\rangle =\langle \lambda v,w\rangle =\langle T(v),w\rangle. ] Since (T) is self-adjoint, [ \langle T(v),w\rangle=\langle v,T(w)\rangle =\langle v,\mu w\rangle =\mu\langle v,w\rangle. ] Therefore, [ \lambda\langle v,w\rangle=\mu\langle v,w\rangle ] so [ (\lambda-\mu)\langle v,w\rangle=0. ] Since (\lambda\neq\mu), we get [ \langle v,w\rangle=0. ] Hence (v) and (w) are mutually orthogonal. Thus any two eigenvectors of a self-adjoint operator associated with different eigenvalues are orthogonal.

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OR

Cayley-Hamilton theorem
Every square matrix satisfies its own characteristic equation. That is, if the characteristic polynomial of a square matrix (A) is
[ f(\lambda)=\det(\lambda I-A), ] then
[ f(A)=0. ]

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

Characteristic polynomial: [ \det(\lambda I-A)

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

\lambda(\lambda+1)-6

\lambda^2+\lambda-6. ] So the characteristic equation is [ \lambda^2+\lambda-6=0. ] By Cayley-Hamilton theorem, [ A^2+A-6I=0. ]

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}0&0\0&0\end{bmatrix}. ] Hence, [ A^2+A-6I=0. ] So the Cayley-Hamilton theorem is verified for the given matrix (A).