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 in variables (x_1,x_2,\dots,x_n): [ 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+z=4. ]

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=3,\ 2x-y=0. \end{cases} ]

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Given system

[ \begin{aligned} x+3y-2z&=3\ 2x-3y+z&=2\ 4x-3y+2z&=2 \end{aligned} ]

Augmented matrix: [ \left[\begin{array}{ccc|c} 1 & 3 & -2 & 3\ 2 & -3 & 1 & 2\ 4 & -3 & 2 & 2 \end{array}\right] ]

Apply Gaussian elimination.

[ R_2 \to R_2-2R_1,\qquad R_3 \to R_3-4R_1 ]

[ \left[\begin{array}{ccc|c} 1 & 3 & -2 & 3\ 0 & -9 & 5 & -4\ 0 & -15 & 10 & -10 \end{array}\right] ]

Now eliminate the (y)-term from (R_3): [ R_3 \to R_3-\frac{5}{3}R_2 ]

[ \left[\begin{array}{ccc|c} 1 & 3 & -2 & 3\ 0 & -9 & 5 & -4\ 0 & 0 & \frac{5}{3} & -\frac{10}{3} \end{array}\right] ]

This is in row-echelon form.

The coefficient matrix has rank (3), and the augmented matrix also has rank (3). Since [ \text{rank}(A)=\text{rank}([A|B])=3=\text{number of unknowns}, ] the system is consistent and has a unique solution.

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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 ]

Hence the solution is [ \boxed{x=1,\quad y=-\frac{2}{3},\quad z=-2} ]

So the system is consistent and has the unique solution [ \boxed{(x,y,z)=\left(1,-\frac{2}{3},-2\right)}. ]

Part 1: $T:\mathbb{R}^2\to\mathbb{R}^3$, $T(x)=Ax$

Here $A=\begin{pmatrix}1&2\\3&4\\0&0\end{pmatrix}$, so for $x=\begin{pmatrix}x_1\\x_2\end{pmatrix}$,

$T(x)=Ax=\begin{pmatrix}x_1+2x_2\\3x_1+4x_2\\0\end{pmatrix}.$

$T$ is one-to-one. Let $T(x)=T(y)$ for $x,y\in\mathbb{R}^2$. Then $A(x-y)=0$. Put $u=x-y=\begin{pmatrix}u_1\\u_2\end{pmatrix}$. Then

$u_1+2u_2=0,\qquad 3u_1+4u_2=0.$

From the first, $u_1=-2u_2$. Substituting in the second, $-6u_2+4u_2=0$, i.e. $-2u_2=0$, so $u_2=0$ and hence $u_1=0$. Thus $u=0$, i.e. $x=y$. Therefore $T$ is one-to-one.

(Equivalently, the two columns $\begin{pmatrix}1\\3\\0\end{pmatrix},\begin{pmatrix}2\\4\\0\end{pmatrix}$ are linearly independent, since $\det\begin{pmatrix}1&2\\3&4\end{pmatrix}=-2\neq0$; so $\ker T=\{0\}$.)

$T$ is not onto. Every image vector has third coordinate $0$:

$\text{Range}(T)=\left\{\begin{pmatrix}a\\b\\0\end{pmatrix}:a,b\in\mathbb{R}\right\}=\{(x,y,z)\in\mathbb{R}^3:z=0\},$

the $xy$-plane. For instance $\begin{pmatrix}0\\0\\1\end{pmatrix}\in\mathbb{R}^3$ has no pre-image, since $T(x)$ can never have third coordinate $1$. Hence $T$ is not onto. (Indeed $\dim(\text{Range}\,T)=\text{rank}\,A=2<3=\dim\mathbb{R}^3$.)

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Part 2: $S:\mathbb{R}^3\to\mathbb{R}^2$, $S(x)=A^{T}x$

(As printed the map $S(x)=Ax$ would coincide with $T$; the intended statement is the transpose map $S:\mathbb{R}^3\to\mathbb{R}^2$, $S(x)=A^{T}x$, which is onto but not one-to-one.)

$A^{T}=\begin{pmatrix}1&3&0\\2&4&0\end{pmatrix},\qquad S(x)=A^{T}x=\begin{pmatrix}x_1+3x_2\\2x_1+4x_2\end{pmatrix}.$

$S$ is onto. Let $\begin{pmatrix}a\\b\end{pmatrix}\in\mathbb{R}^2$ be arbitrary. We need $x_1,x_2$ with

$x_1+3x_2=a,\qquad 2x_1+4x_2=b.$

The coefficient matrix $\begin{pmatrix}1&3\\2&4\end{pmatrix}$ has determinant $4-6=-2\neq0$, so the system has a unique solution for every $(a,b)$. Explicitly, from the first equation $x_1=a-3x_2$; putting this in the second, $2a-6x_2+4x_2=b$, so $x_2=a-\tfrac b2$ and $x_1=-2a+\tfrac{3b}{2}$. Hence

$S\left(-2a+\tfrac{3b}{2},\;a-\tfrac b2,\;0\right)=\begin{pmatrix}a\\b\end{pmatrix},$

so every element of $\mathbb{R}^2$ is an image. Thus $S$ is onto. (Here $\text{rank}\,A^{T}=2=\dim\mathbb{R}^2$.)

$S$ is not one-to-one. Take $u=\begin{pmatrix}0\\0\\1\end{pmatrix}\neq\begin{pmatrix}0\\0\\0\end{pmatrix}$, but

$S(u)=\begin{pmatrix}0+0\\0+0\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}=S(0).$

So two distinct vectors have the same image; $S$ is not one-to-one. In fact

$\ker S=\{(0,0,t):t\in\mathbb{R}\}\neq\{0\},$

so $\text{nullity}=3-2=1$.

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OR

Condition for the existence of $A^{-1}$

A square matrix $A$ of order $n$ possesses an inverse if and only if $A$ is non-singular, i.e.

$\det A\neq 0\qquad(\text{equivalently, rank }A=n).$

When this holds, $A^{-1}$ is unique and is given by

$A^{-1}=\frac{1}{\det A}\,\text{adj}\,A .$

If $\det A=0$, then $A$ is singular and has no inverse.

Determination of $B$

Given $A=\begin{pmatrix}3&-1\\5&-2\end{pmatrix}$ and $B=\begin{pmatrix}p&q\\r&s\end{pmatrix}$, with $A$ and $B$ inverses of each other, i.e. $AB=BA=I$.

$\det A=3(-2)-(-1)(5)=-6+5=-1\neq0,$

so $A^{-1}$ exists. For a $2\times2$ matrix $\begin{pmatrix}a&b\\c&d\end{pmatrix}$,

$\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}.$

Hence

$B=A^{-1}=\frac{1}{-1}\begin{pmatrix}-2&1\\-5&3\end{pmatrix}=\begin{pmatrix}2&-1\\5&-3\end{pmatrix}.$

So

$\boxed{p=2,\quad q=-1,\quad r=5,\quad s=-3}$

Check:

$AB=\begin{pmatrix}3&-1\\5&-2\end{pmatrix}\begin{pmatrix}2&-1\\5&-3\end{pmatrix}=\begin{pmatrix}6-5&-3+3\\10-10&-5+6\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}=I,$

and similarly $BA=I$. So $A$ and $B$ are indeed inverses.

Verification that $A^2$ and $B^2$ are inverses

$A^2=\begin{pmatrix}3&-1\\5&-2\end{pmatrix}\begin{pmatrix}3&-1\\5&-2\end{pmatrix}=\begin{pmatrix}9-5&-3+2\\15-10&-5+4\end{pmatrix}=\begin{pmatrix}4&-1\\5&-1\end{pmatrix},$

$B^2=\begin{pmatrix}2&-1\\5&-3\end{pmatrix}\begin{pmatrix}2&-1\\5&-3\end{pmatrix}=\begin{pmatrix}4-5&-2+3\\10-15&-5+9\end{pmatrix}=\begin{pmatrix}-1&1\\-5&4\end{pmatrix}.$

Now

$A^2B^2=\begin{pmatrix}4&-1\\5&-1\end{pmatrix}\begin{pmatrix}-1&1\\-5&4\end{pmatrix}=\begin{pmatrix}-4+5&4-4\\-5+5&5-4\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}=I .$

Similarly,

$B^2A^2=\begin{pmatrix}-1&1\\-5&4\end{pmatrix}\begin{pmatrix}4&-1\\5&-1\end{pmatrix}=\begin{pmatrix}-4+5&1-1\\-20+20&5-4\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}=I .$

(Alternatively, $A^2B^2=(AB)^2=I^2=I$.)

Since $A^2B^2=B^2A^2=I$, the matrices $A^2$ and $B^2$ are inverses of each other. $\blacksquare$

Proof: Let (V) be a vector space over a field (F). Suppose (V) has a finite basis
[ B={v_1,v_2,\dots,v_n}. ] Let
[ B'={w_1,w_2,\dots,w_m} ] be any other basis of (V). We prove that (m=n).

Since (B) spans (V), each (w_j) can be written as a linear combination of the vectors (v_i): [ w_j=a_{1j}v_1+a_{2j}v_2+\cdots+a_{nj}v_n,\qquad j=1,2,\dots,m. ] Form the (n\times m) matrix (A=(a_{ij})).

If (m>n), then the homogeneous system [ A c=0 ] has a nontrivial solution (c=(c_1,c_2,\dots,c_m)^T\neq 0), because the number of unknowns (m) is greater than the number of equations (n). Equivalently, [ \operatorname{rank}(A)\le n<m, ] so the nullity of (A) is positive.

For this nonzero vector (c), [ \sum_{j=1}^m c_j w_j

\sum_{j=1}^m c_j\left(\sum_{i=1}^n a_{ij}v_i\right)

\sum_{i=1}^n\left(\sum_{j=1}^m a_{ij}c_j\right)v_i

] Since (c\neq 0), this gives a nontrivial linear relation among (w_1,w_2,\dots,w_m), contradicting the fact that (B') is linearly independent. Hence [ m\le n. ]

Now interchange the roles of (B) and (B'). Since (B') spans (V) and (B) is linearly independent, the same argument gives [ n\le m. ] Therefore, [ m=n. ]

Thus every basis of a finite-dimensional vector space has the same number of elements. This common number is called the dimension of (V).

Let
[ v_1=(1,1,1),\quad v_2=(0,1,1),\quad v_3=(1,0,1). ] Use the Gram–Schmidt process.

1. Orthogonal basis

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

Now orthogonalize (v_2) with respect to (u_1): [ 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=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). ]

Now orthogonalize (v_3) with respect to (u_1) and (u_2): [ 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. ] Compute: [ v_3\cdot u_1=1+0+1=2, ] [ v_3\cdot u_2=(1)(-2)+0(1)+1(1)=-1, ] [ u_2\cdot u_2=4+1+1=6. ] Thus [ u_3=(1,0,1)-\frac{2}{3}(1,1,1)-\frac{-1}{6}(-2,1,1). ] [ u_3=(1,0,1)-\left(\frac23,\frac23,\frac23\right)+\frac16(-2,1,1) ] [ u_3=\left(0,-\frac12,\frac12\right). ] Multiplying by (2), take [ u_3=(0,-1,1). ]

Hence an orthogonal basis is [ \boxed{{(1,1,1),\ (-2,1,1),\ (0,-1,1)}}. ]

2. Orthonormal basis

Divide each vector by its norm.

[ |(1,1,1)|=\sqrt{1^2+1^2+1^2}=\sqrt3, ] [ |(-2,1,1)|=\sqrt{4+1+1}=\sqrt6, ] [ |(0,-1,1)|=\sqrt{0+1+1}=\sqrt2. ]

Therefore an orthonormal basis is [ \boxed{ \left{ \frac1{\sqrt3}(1,1,1),\ \frac1{\sqrt6}(-2,1,1),\ \frac1{\sqrt2}(0,-1,1) \right} } ] or explicitly, [ \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 three 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}, ] where (v,z\neq 0) are direction vectors.

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

((\Rightarrow)) Suppose (L_1=L_2).

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

((\Leftarrow)) Suppose (u-w) and (v) are multiples of (z).

Then for some (\alpha,\beta\in\mathbb R), [ u-w=\alpha z,\qquad v=\beta z. ] 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. ] This is a point of (L_2). Hence (L_1\subseteq L_2).

Conversely, take any point (y\in L_2). Then for some (s\in\mathbb R), [ y=w+sz. ] Since (v=\beta z), we have (z=\frac1\beta v). Also, [ w=u-\alpha z=u-\frac{\alpha}{\beta}v. ] Therefore, [ y=u-\frac{\alpha}{\beta}v+\frac{s}{\beta}v =u+\left(\frac{s-\alpha}{\beta}\right)v. ] This is a point of (L_1). Hence (L_2\subseteq L_1).

Thus (L_1=L_2).

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

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2. Parallel and distinct lines in (\mathbb R^3)

Two parametrically given lines [ L_1={u+tv},\qquad L_2={w+sz} ] are parallel iff their direction vectors are scalar multiples: [ v=\lambda z,\qquad \lambda\neq 0. ]

If in addition (u-w) is not a multiple of (z), then the lines are parallel and distinct.

Indeed, suppose they met. Then for some (t,s\in\mathbb R), [ u+tv=w+sz. ] Since (v=\lambda z), [ u-w=(s-t\lambda)z, ] which contradicts the assumption that (u-w) is not a multiple of (z). Hence they do not intersect. Since their directions are the same, they are parallel and distinct.

Example

[ L_1:(0,0,0)+t(1,2,3), ] [ L_2:(1,0,0)+s(2,4,6). ] Here [ (2,4,6)=2(1,2,3), ] so the lines are parallel. Also, [ u-w=(-1,0,0) ] is not a multiple of ((2,4,6)), so the lines are distinct. Thus (L_1) and (L_2) are parallel and distinct.

Kernel:
Let (T:V\to W) be a linear transformation. The kernel of (T) is [ \ker T={v\in V:T(v)=0}. ] For a matrix (A), the kernel (null space) is [ \ker A={x:Ax=0}. ] It is a subspace of the domain.

Span of vectors:
If (S={v_1,v_2,\dots,v_n}) is a set of vectors in a vector space (V), then [ \operatorname{span}(S)={c_1v_1+c_2v_2+\cdots+c_nv_n:c_i\in \mathbb R}. ] It is the smallest subspace containing all the given vectors.

Rank and consistency:
The rank of a matrix (A) is the dimension of its row space or column space, i.e. the number of nonzero rows in its row-echelon form.
A system (AX=B) is not always consistent. It is consistent if and only if [ \operatorname{rank}(A)=\operatorname{rank}(A|B). ] If (\operatorname{rank}(A)<\operatorname{rank}(A|B)), the system has no solution. The homogeneous system (AX=0) is always consistent, since (X=0) is a solution.

Required matrix:
Let [ v_1=(1,-2,3),\qquad v_2=(-2,0,4). ] A matrix whose kernel is (\operatorname{span}{v_1,v_2}) must have row vectors orthogonal to both (v_1) and (v_2). Compute [ v_1\times v_2= \begin{vmatrix} i&j&k\ 1&-2&3\ -2&0&4 \end{vmatrix} =(-8,-10,-4)\sim (4,5,2). ] So take [ A=\begin{pmatrix}4&5&2\end{pmatrix}. ] Then [ \ker A={(x,y,z)^T:4x+5y+2z=0}. ] Check: [ A v_1=4(1)+5(-2)+2(3)=0, ] [ A v_2=4(-2)+5(0)+2(4)=0. ] Since (v_1,v_2) are independent, (\dim(\ker A)=2), and hence [ \ker A=\operatorname{span}{(1,-2,3),(-2,0,4)}. ]

If a square matrix is required, we may write [ A=\begin{pmatrix} 4&5&2\ 0&0&0\ 0&0&0 \end{pmatrix}, ] which has the same kernel. Its rank is (1).

Let (T:\mathbb{R}^2\to \mathbb{R}) be linear and suppose

[ T(1,1)=3,\qquad T(0,1)=-2. ]

1. Uniqueness of (T)

The vectors ((1,1)) and ((0,1)) form a basis of (\mathbb{R}^2), since

[ \det\begin{pmatrix}1&0\1&1\end{pmatrix}=1\ne 0. ]

Any linear map is uniquely determined by its values on a basis. Hence there can be at most one such linear map (T).

So the map is unique.

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2. Formula for (T(x,y))

Express ((x,y)) as a linear combination of ((1,1)) and ((0,1)):

[ (x,y)=a(1,1)+b(0,1). ]

Then

[ a=x,\qquad a+b=y. ]

So

[ b=y-x. ]

Therefore,

[ (x,y)=x(1,1)+(y-x)(0,1). ]

Using linearity,

[ T(x,y)=xT(1,1)+(y-x)T(0,1). ]

Substitute the given values:

[ T(x,y)=x(3)+(y-x)(-2) ]

[ =3x-2y+2x ]

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

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3. Is (T) one-to-one?

No. For (T) to be one-to-one, its kernel must contain only ((0,0)).

Find the kernel:

[ T(x,y)=0 ]

[ 5x-2y=0 ]

[ 2y=5x ]

[ y=\frac{5x}{2}. ]

Thus, for example,

[ T(2,5)=5(2)-2(5)=10-10=0. ]

So ((2,5)\neq (0,0)) but (T(2,5)=T(0,0)=0). Hence (T) is not one-to-one.

[ \boxed{\text{No, } T \text{ is not one-to-one.}} ]

Also, since (\dim(\mathbb{R}^2)=2) and (\dim(\mathbb{R})=1), a linear map from (\mathbb{R}^2) to (\mathbb{R}) cannot be injective.

Determinants, area and volume. The determinant of a $2\times2$ matrix gives the (signed) area of the parallelogram spanned by its row vectors, so half of it is the area of a triangle. The determinant of a $3\times3$ matrix gives the (signed) volume of the parallelepiped spanned by its three row vectors. The absolute value is taken because area and volume are positive.

Area of the triangle with vertices $(-2,-3)$, $(3,2)$, $(-1,8)$:

$\Delta=\frac12\left|\begin{vmatrix}-2&-3&1\\3&2&1\\-1&8&1\end{vmatrix}\right|=\frac12\times|50|$

Expanding along the first row, $-2(2-8)-(-3)(3+1)+1(24+2)=12+12+26=50$, so

$\Delta=\frac12\times 50=25\ \text{square units}.$

Volume of the parallelepiped with edge vectors $(1,-2,1)$, $(1,0,-1)$, $(1,1,1)$ from the origin:

$V=\left|\begin{vmatrix}1&-2&1\\1&0&-1\\1&1&1\end{vmatrix}\right|=\left|1(0+1)+2(1+1)+1(1-0)\right|=|1+4+1|=6\ \text{cubic units}.$

OR

Determinant vs matrix.

Proof. Let $d=b-a$ and let the determinant be $\Delta$ with rows $R_1=(a,b,b,b)$, $R_2=(a,b,a,a)$, $R_3=(a,a,b,a)$, $R_4=(b,b,b,a)$.

Apply $R_1\to R_1-R_2$, $R_2\to R_2-R_3$, $R_3\to R_3-R_4$ (using the original rows):

$\Delta=\begin{vmatrix}0&0&d&d\\0&d&-d&0\\-d&-d&0&0\\b&b&b&a\end{vmatrix}=d^3\begin{vmatrix}0&0&1&1\\0&1&-1&0\\-1&-1&0&0\\b&b&b&a\end{vmatrix}$

Apply $C_2\to C_2-C_1$:

$\Delta=d^3\begin{vmatrix}0&0&1&1\\0&1&-1&0\\-1&0&0&0\\b&0&b&a\end{vmatrix}$

Expand along the third row, whose only non-zero entry is $-1$ in position $(3,1)$, with sign $(-1)^{3+1}=+1$:

$\Delta=d^3\cdot(-1)\begin{vmatrix}0&1&1\\1&-1&0\\0&b&a\end{vmatrix}=d^3\cdot(-1)\left[-1(a-0)+1(b-0)\right]=-d^3(b-a)$

$\therefore\ \Delta=-(b-a)^4.$

Characteristic equation:

$$|A-\lambda I|= \begin{vmatrix} 1-\lambda & -2 & 2\\ -2 & 1-\lambda & -2\\ 2 & -2 & 1-\lambda \end{vmatrix}=0$$

Expanding,

$(1-\lambda)^3-12(1-\lambda)+16=0$

$\Rightarrow -\lambda^3+3\lambda^2+9\lambda+5=0$

$\Rightarrow (\lambda+1)^2(\lambda-5)=0$

Hence the eigenvalues are

$\boxed{\lambda=-1,\,-1,\,5}$

So the eigenvalues are $\lambda=-1$ with multiplicity $2$, and $\lambda=5$.

Now find an eigenvector for $\lambda=5$:

$(A-5I)\begin{pmatrix}x\\y\\z\end{pmatrix}=0$

$$\begin{pmatrix} -4&-2&2\\ -2&-4&-2\\ 2&-2&-4 \end{pmatrix} \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}0\\0\\0\end{pmatrix}$$

The equations reduce to

$x+2y+z=0,\qquad x-y-2z=0$

Solving gives

$x=-y,\qquad z=-y$

Taking $y=1$, we get

$x=-1,\qquad z=-1$

Therefore one eigenvector corresponding to $\lambda=5$ is

$\boxed{\begin{pmatrix}-1\\1\\-1\end{pmatrix}}$

Equivalently, $\begin{pmatrix}1\\-1\\1\end{pmatrix}$ is also an eigenvector for $\lambda=5$.