# Linear Algebra/Eigenvalues and Eigenvectors/Solutions
## Solutions
### Problem 1

For each, find the characteristic polynomial and the eigenvalues.
* ${\begin{pmatrix}10&-9\\4&-2\end{pmatrix}}$
* ${\begin{pmatrix}1&2\\4&3\end{pmatrix}}$
* ${\begin{pmatrix}0&3\\7&0\end{pmatrix}}$
* ${\begin{pmatrix}0&0\\0&0\end{pmatrix}}$
* ${\begin{pmatrix}1&0\\0&1\end{pmatrix}}$

### Answer

* This simplifies to the characteristic equation $x^{2}-8x+16=0$. Because the equation factors into $(x-4)^{2}$ there is only one eigenvalue $\lambda _{1}=4$.
    $0={\begin{vmatrix}10-x&-9\\4&-2-x\end{vmatrix}}=(10-x)(-2-x)-(-36)$
* $0=(1-x)(3-x)-8=x^{2}-4x-5$; $\lambda _{1}=5$, $\lambda _{2}=-1$
* $x^{2}-21=0$; $\lambda _{1}={\sqrt {21}}$, $\lambda _{2}=-{\sqrt {21}}$
* $x^{2}=0$; $\lambda _{1}=0$
* $x^{2}-2x+1=0$; $\lambda _{1}=1$
    This exercise is recommended for all readers.

### Problem 2

For each matrix, find the characteristic equation, and the eigenvalues and associated eigenvectors.
* ${\begin{pmatrix}3&0\\8&-1\end{pmatrix}}$
* ${\begin{pmatrix}3&2\\-1&0\end{pmatrix}}$

### Answer

* The characteristic equation is $(3-x)(-1-x)=0$. Its roots, the eigenvalues, are $\lambda _{1}=3$ and $\lambda _{2}=-1$. For the eigenvectors we consider this equation. For the eigenvector associated with $\lambda _{1}=3$, we consider the resulting linear system. The eigenspace is the set of vectors whose second component is twice the first component. (Here, the parameter is $b_{2}$ only because that is the variable that is free in the above system.) Hence, this is an eigenvector associated with the eigenvalue $3$. Finding an eigenvector associated with $\lambda _{2}=-1$ is similar. This system leads to the set of vectors whose first component is zero. And so this is an eigenvector associated with $\lambda _{2}$.
    ${\begin{pmatrix}3-x&0\\8&-1-x\end{pmatrix}}{\begin{pmatrix}b_{1}\\b_{2}\end{pmatrix}}={\begin{pmatrix}0\\0\end{pmatrix}}$
    ${\begin{array}{*{2}{rc}r}0\cdot b_{1}&+&0\cdot b_{2}&=&0\\8\cdot b_{1}&+&-4\cdot b_{2}&=&0\end{array}}$
    $\{{\begin{pmatrix}b_{2}/2\\b_{2}\end{pmatrix}}\,{\big |}\,b_{2}\in \mathbb {C} \}\qquad {\begin{pmatrix}3&0\\8&-1\end{pmatrix}}{\begin{pmatrix}b_{2}/2\\b_{2}\end{pmatrix}}=3\cdot {\begin{pmatrix}b_{2}/2\\b_{2}\end{pmatrix}}$
    ${\begin{pmatrix}1/2\\1\end{pmatrix}}$
    ${\begin{array}{*{2}{rc}r}4\cdot b_{1}&+&0\cdot b_{2}&=&0\\8\cdot b_{1}&+&0\cdot b_{2}&=&0\end{array}}$
    $\{{\begin{pmatrix}0\\b_{2}\end{pmatrix}}\,{\big |}\,b_{2}\in \mathbb {C} \}\qquad {\begin{pmatrix}3&0\\8&-1\end{pmatrix}}{\begin{pmatrix}0\\b_{2}\end{pmatrix}}=-1\cdot {\begin{pmatrix}0\\b_{2}\end{pmatrix}}$
    ${\begin{pmatrix}0\\1\end{pmatrix}}$
* The characteristic equation is and so the eigenvalues are $\lambda _{1}=2$ and $\lambda _{2}=1$. To find eigenvectors, consider this system. For $\lambda _{1}=2$ we get leading to this eigenspace and eigenvector. For $\lambda _{2}=1$ the system is leading to this.
    $0={\begin{vmatrix}3-x&2\\-1&-x\end{vmatrix}}=x^{2}-3x+2=(x-2)(x-1)$
    ${\begin{array}{*{2}{rc}r}(3-x)\cdot b_{1}&+&2\cdot b_{2}&=&0\\-1\cdot b_{1}&-&x\cdot b_{2}&=&0\end{array}}$
    ${\begin{array}{*{2}{rc}r}1\cdot b_{1}&+&2\cdot b_{2}&=&0\\-1\cdot b_{1}&-&2\cdot b_{2}&=&0\end{array}}$
    $\{{\begin{pmatrix}-2b_{2}\\b_{2}\end{pmatrix}}\,{\big |}\,b_{2}\in \mathbb {C} \}\qquad {\begin{pmatrix}-2\\1\end{pmatrix}}$
    ${\begin{array}{*{2}{rc}r}2\cdot b_{1}&+&2\cdot b_{2}&=&0\\-1\cdot b_{1}&-&1\cdot b_{2}&=&0\end{array}}$
    $\{{\begin{pmatrix}-b_{2}\\b_{2}\end{pmatrix}}\,{\big |}\,b_{2}\in \mathbb {C} \}\qquad {\begin{pmatrix}-1\\1\end{pmatrix}}$

### Problem 3

Find the characteristic equation, and the eigenvalues and associated eigenvectors for this matrix. Hint. The eigenvalues are complex.
    ${\begin{pmatrix}-2&-1\\5&2\end{pmatrix}}$

### Answer

The characteristic equation
    $0={\begin{vmatrix}-2-x&-1\\5&2-x\end{vmatrix}}=x^{2}+1$
has the complex roots $\lambda _{1}=i$ and $\lambda _{2}=-i$. This system
    ${\begin{array}{*{2}{rc}r}(-2-x)\cdot b_{1}&-&1\cdot b_{2}&=&0\\5\cdot b_{1}&&(2-x)\cdot b_{2}&=&0\end{array}}$
For $\lambda _{1}=i$ Gauss' method gives this reduction.
    ${\begin{array}{*{2}{rc}r}(-2-i)\cdot b_{1}&-&1\cdot b_{2}&=&0\\5\cdot b_{1}&-&(2-i)\cdot b_{2}&=&0\end{array}}{\xrightarrow[]{(-5/(-2-i))\rho _{1}+\rho _{2}}}{\begin{array}{*{2}{rc}r}(-2-i)\cdot b_{1}&-&1\cdot b_{2}&=&0\\&&0&=&0\end{array}}$
(For the calculation in the lower right get a common denominator
    ${\frac {5}{-2-i}}-(2-i)={\frac {5}{-2-i}}-{\frac {-2-i}{-2-i}}\cdot (2-i)={\frac {5-(-5)}{-2-i}}$
to see that it gives a $0=0$ equation.) These are the resulting eigenspace and eigenvector.
    $\{{\begin{pmatrix}(1/(-2-i))b_{2}\\b_{2}\end{pmatrix}}\,{\big |}\,b_{2}\in \mathbb {C} \}\qquad {\begin{pmatrix}1/(-2-i)\\1\end{pmatrix}}$
For $\lambda _{2}=-i$ the system
    ${\begin{array}{*{2}{rc}r}(-2+i)\cdot b_{1}&-&1\cdot b_{2}&=&0\\5\cdot b_{1}&-&(2+i)\cdot b_{2}&=&0\end{array}}{\xrightarrow[]{(-5/(-2+i))\rho _{1}+\rho _{2}}}{\begin{array}{*{2}{rc}r}(-2+i)\cdot b_{1}&-&1\cdot b_{2}&=&0\\&&0&=&0\end{array}}$
leads to this.
    $\{{\begin{pmatrix}(1/(-2+i))b_{2}\\b_{2}\end{pmatrix}}\,{\big |}\,b_{2}\in \mathbb {C} \}\qquad {\begin{pmatrix}1/(-2+i)\\1\end{pmatrix}}$

### Problem 4

Find the characteristic polynomial, the eigenvalues, and the associated eigenvectors of this matrix.
    ${\begin{pmatrix}1&1&1\\0&0&1\\0&0&1\end{pmatrix}}$

### Answer

The characteristic equation is
    $0={\begin{vmatrix}1-x&1&1\\0&-x&1\\0&0&1-x\end{vmatrix}}=(1-x)^{2}(-x)$
and so the eigenvalues are $\lambda _{1}=1$ (this is a repeated root of the equation) and $\lambda _{2}=0$. For the rest, consider this system.
    ${\begin{array}{*{3}{rc}r}(1-x)\cdot b_{1}&+&b_{2}&+&b_{3}&=&0\\&&-x\cdot b_{2}&+&b_{3}&=&0\\&&&&(1-x)\cdot b_{3}&=&0\end{array}}$
When $x=\lambda _{1}=1$ then the solution set is this eigenspace.
    $\{{\begin{pmatrix}b_{1}\\0\\0\end{pmatrix}}\,{\big |}\,b_{1}\in \mathbb {C} \}$
When $x=\lambda _{2}=0$ then the solution set is this eigenspace.
    $\{{\begin{pmatrix}-b_{2}\\b_{2}\\0\end{pmatrix}}\,{\big |}\,b_{2}\in \mathbb {C} \}$
So these are eigenvectors associated with $\lambda _{1}=1$ and $\lambda _{2}=0$.
    ${\begin{pmatrix}1\\0\\0\end{pmatrix}}\qquad {\begin{pmatrix}-1\\1\\0\end{pmatrix}}$
    This exercise is recommended for all readers.

### Problem 5

For each matrix, find the characteristic equation, and the eigenvalues and associated eigenvectors.
* ${\begin{pmatrix}3&-2&0\\-2&3&0\\0&0&5\end{pmatrix}}$
* ${\begin{pmatrix}0&1&0\\0&0&1\\4&-17&8\end{pmatrix}}$

### Answer

* The characteristic equation is and so the eigenvalues are $\lambda _{1}=1$ and also the repeated eigenvalue $\lambda _{2}=5$. To find eigenvectors, consider this system. For $\lambda _{1}=1$ we get leading to this eigenspace and eigenvector. For $\lambda _{2}=5$ the system is leading to this.
    $0={\begin{vmatrix}3-x&-2&0\\-2&3-x&0\\0&0&5-x\end{vmatrix}}=x^{3}-11x^{2}+35x-25=(x-1)(x-5)^{2}$
    ${\begin{array}{*{3}{rc}r}(3-x)\cdot b_{1}&-&2\cdot b_{2}&&&=&0\\-2\cdot b_{1}&+&(3-x)\cdot b_{2}&&&=&0\\&&&&(5-x)\cdot b_{3}&=&0\end{array}}$
    ${\begin{array}{*{3}{rc}r}2\cdot b_{1}&-&2\cdot b_{2}&&&=&0\\-2\cdot b_{1}&+&2\cdot b_{2}&&&=&0\\&&&&4\cdot b_{3}&=&0\end{array}}$
    $\{{\begin{pmatrix}b_{2}\\b_{2}\\0\end{pmatrix}}\,{\big |}\,b_{2}\in \mathbb {C} \}\qquad {\begin{pmatrix}1\\1\\0\end{pmatrix}}$
    ${\begin{array}{*{3}{rc}r}-2\cdot b_{1}&-&2\cdot b_{2}&&&=&0\\-2\cdot b_{1}&-&2\cdot b_{2}&&&=&0\\&&&&0\cdot b_{3}&=&0\end{array}}$
    $\{{\begin{pmatrix}-b_{2}\\b_{2}\\0\end{pmatrix}}+{\begin{pmatrix}0\\0\\b_{3}\end{pmatrix}}\,{\big |}\,b_{2},b_{3}\in \mathbb {C} \}\qquad {\begin{pmatrix}-1\\1\\0\end{pmatrix}},\,{\begin{pmatrix}0\\0\\1\end{pmatrix}}$
* The characteristic equation is and the eigenvalues are $\lambda _{1}=4$ and (by using the quadratic equation) $\lambda _{2}=2+{\sqrt {3}}$ and $\lambda _{3}=2-{\sqrt {3}}$. To find eigenvectors, consider this system. Substituting $x=\lambda _{1}=4$ gives the system leading to this eigenspace and eigenvector. Substituting $x=\lambda _{2}=2+{\sqrt {3}}$ gives the system (the middle coefficient in the third equation equals the number $(-4/(-2-{\sqrt {3}}))-17$; find a common denominator of $-2-{\sqrt {3}}$ and then rationalize the denominator by multiplying the top and bottom of the frsction by $-2+{\sqrt {3}}$) which leads to this eigenspace and eigenvector. Finally, substituting $x=\lambda _{3}=2-{\sqrt {3}}$ gives the system which gives this eigenspace and eigenvector.
    $0={\begin{vmatrix}-x&1&0\\0&-x&1\\4&-17&8-x\end{vmatrix}}=-x^{3}+8x^{2}-17x+4=-1\cdot (x-4)(x^{2}-4x+1)$
    ${\begin{array}{*{3}{rc}r}-x\cdot b_{1}&+&b_{2}&&&=&0\\&&-x\cdot b_{2}&+&b_{3}&=&0\\4\cdot b_{1}&-&17\cdot b_{2}&+&(8-x)\cdot b_{3}&=&0\end{array}}$
    ${\begin{array}{*{3}{rc}r}-4\cdot b_{1}&+&b_{2}&&&=&0\\&&-4\cdot b_{2}&+&b_{3}&=&0\\4\cdot b_{1}&-&17\cdot b_{2}&+&4\cdot b_{3}&=&0\end{array}}{\xrightarrow[]{\rho _{1}+\rho _{3}}}{\begin{array}{*{3}{rc}r}-4\cdot b_{1}&+&b_{2}&&&=&0\\&&-4\cdot b_{2}&+&b_{3}&=&0\\&&-16\cdot b_{2}&+&4\cdot b_{3}&=&0\end{array}}{\xrightarrow[]{-4\rho _{2}+\rho _{3}}}{\begin{array}{*{3}{rc}r}-4\cdot b_{1}&+&b_{2}&&&=&0\\&&-4\cdot b_{2}&+&b_{3}&=&0\\&&&&0&=&0\end{array}}$
    $V_{4}=\{{\begin{pmatrix}(1/16)\cdot b_{3}\\(1/4)\cdot b_{3}\\b_{3}\end{pmatrix}}\,{\big |}\,b_{2}\in \mathbb {C} \}\qquad {\begin{pmatrix}1\\4\\16\end{pmatrix}}$
    ${\begin{array}{*{3}{rc}r}(-2-{\sqrt {3}})\cdot b_{1}&+&b_{2}&&&=&0\\&&(-2-{\sqrt {3}})\cdot b_{2}&+&b_{3}&=&0\\4\cdot b_{1}&-&17\cdot b_{2}&+&(6-{\sqrt {3}})\cdot b_{3}&=&0\end{array}}$
       ${\xrightarrow[]{(-4/(-2-{\sqrt {3}}))\rho _{1}+\rho _{3}}}{\begin{array}{*{3}{rc}r}(-2-{\sqrt {3}})\cdot b_{1}&+&b_{2}&&&=&0\\&&(-2-{\sqrt {3}})\cdot b_{2}&+&b_{3}&=&0\\&+&(-9-4{\sqrt {3}})\cdot b_{2}&+&(6-{\sqrt {3}})\cdot b_{3}&=&0\end{array}}$
    ${\xrightarrow[]{((9+4{\sqrt {3}})/(-2-{\sqrt {3}}))\rho _{2}+\rho _{3}}}{\begin{array}{*{3}{rc}r}(-2-{\sqrt {3}})\cdot b_{1}&+&b_{2}&&&=&0\\&&(-2-{\sqrt {3}})\cdot b_{2}&+&b_{3}&=&0\\&&&&0&=&0\end{array}}$
    $V_{2+{\sqrt {3}}}=\{{\begin{pmatrix}(1/(2+{\sqrt {3}})^{2})\cdot b_{3}\\(1/(2+{\sqrt {3}}))\cdot b_{3}\\b_{3}\end{pmatrix}}\,{\big |}\,b_{3}\in \mathbb {C} \}\qquad {\begin{pmatrix}(1/(2+{\sqrt {3}})^{2})\\(1/(2+{\sqrt {3}}))\\1\end{pmatrix}}$
    ${\begin{array}{*{3}{rc}r}(-2+{\sqrt {3}})\cdot b_{1}&+&b_{2}&&&=&0\\&&(-2+{\sqrt {3}})\cdot b_{2}&+&b_{3}&=&0\\4\cdot b_{1}&-&17\cdot b_{2}&+&(6+{\sqrt {3}})\cdot b_{3}&=&0\end{array}}$
       ${\begin{aligned}&{\xrightarrow[]{(-4/(-2+{\sqrt {3}}))\rho _{1}+\rho _{3}}}{\begin{array}{*{3}{rc}r}(-2+{\sqrt {3}})\cdot b_{1}&+&b_{2}&&&=&0\\&&(-2+{\sqrt {3}})\cdot b_{2}&+&b_{3}&=&0\\&&(-9+4{\sqrt {3}})\cdot b_{2}&+&(6+{\sqrt {3}})\cdot b_{3}&=&0\end{array}}\\&{\xrightarrow[]{((9-4{\sqrt {3}})/(-2+{\sqrt {3}}))\rho _{2}+\rho _{3}}}{\begin{array}{*{3}{rc}r}(-2+{\sqrt {3}})\cdot b_{1}&+&b_{2}&&&=&0\\&&(-2+{\sqrt {3}})\cdot b_{2}&+&b_{3}&=&0\\&&&&0&=&0\end{array}}\end{aligned}}$
    $V_{2-{\sqrt {3}}}=\{{\begin{pmatrix}(1/(2+{\sqrt {3}})^{2})\cdot b_{3}\\(1/(2-{\sqrt {3}}))\cdot b_{3}\\b_{3}\end{pmatrix}}\,{\big |}\,b_{3}\in \mathbb {C} \}\qquad {\begin{pmatrix}(1/(-2+{\sqrt {3}})^{2})\\(1/(-2+{\sqrt {3}}))\\1\end{pmatrix}}$
    This exercise is recommended for all readers.

### Problem 6

Let $t:{\mathcal {P}}_{2}\to {\mathcal {P}}_{2}$ be
    $a_{0}+a_{1}x+a_{2}x^{2}\mapsto (5a_{0}+6a_{1}+2a_{2})-(a_{1}+8a_{2})x+(a_{0}-2a_{2})x^{2}.$
Find its eigenvalues and the associated eigenvectors.

### Answer

With respect to the natural basis $B=\langle 1,x,x^{2}\rangle$ the matrix representation is this.
    ${\rm {Rep}}_{B,B}(t)={\begin{pmatrix}5&6&2\\0&-1&-8\\1&0&-2\end{pmatrix}}$
Thus the characteristic equation
    $0={\begin{pmatrix}5-x&6&2\\0&-1-x&-8\\1&0&-2-x\end{pmatrix}}=(5-x)(-1-x)(-2-x)-48-2\cdot (-1-x)$
is $0=-x^{3}+2x^{2}+15x-36=-1\cdot (x+4)(x-3)^{2}$. To find the associated eigenvectors, consider this system.
    ${\begin{array}{*{3}{rc}r}(5-x)\cdot b_{1}&+&6\cdot b_{2}&+&2\cdot b_{3}&=&0\\&&(-1-x)\cdot b_{2}&-&8\cdot b_{3}&=&0\\b_{1}&&&+&(-2-x)\cdot b_{3}&=&0\end{array}}$
Plugging in $x=\lambda _{1}=4$ gives
    ${\begin{array}{*{3}{rc}r}b_{1}&+&6\cdot b_{2}&+&2\cdot b_{3}&=&0\\&&-5\cdot b_{2}&-&8\cdot b_{3}&=&0\\b_{1}&&&-&6\cdot b_{3}&=&0\end{array}}{\xrightarrow[]{-\rho _{1}+\rho _{2}}}{\begin{array}{*{3}{rc}r}b_{1}&+&6\cdot b_{2}&+&2\cdot b_{3}&=&0\\&&-5\cdot b_{2}&-&8\cdot b_{3}&=&0\\&&-6\cdot b_{2}&-&8\cdot b_{3}&=&0\end{array}}{\xrightarrow[]{-\rho _{1}+\rho _{2}}}{\begin{array}{*{3}{rc}r}b_{1}&+&6\cdot b_{2}&+&2\cdot b_{3}&=&0\\&&-5\cdot b_{2}&-&8\cdot b_{3}&=&0\\&&-6\cdot b_{2}&-&8\cdot b_{3}&=&0\end{array}}$

### Problem 7

Find the eigenvalues and eigenvectors of this map $t:{\mathcal {M}}_{2}\to {\mathcal {M}}_{2}$.
    ${\begin{pmatrix}a&b\\c&d\end{pmatrix}}\mapsto {\begin{pmatrix}2c&a+c\\b-2c&d\end{pmatrix}}$

### Answer

$\lambda =1,{\begin{pmatrix}0&0\\0&1\end{pmatrix}}{\text{ and }}{\begin{pmatrix}2&3\\1&0\end{pmatrix}}$, $\lambda =-2,{\begin{pmatrix}-1&0\\1&0\end{pmatrix}}$, $\lambda =-1,{\begin{pmatrix}-2&1\\1&0\end{pmatrix}}$
    This exercise is recommended for all readers.

### Problem 8

Find the eigenvalues and associated eigenvectors of the differentiation operator $d/dx:{\mathcal {P}}_{3}\to {\mathcal {P}}_{3}$.

### Answer

Fix the natural basis $B=\langle 1,x,x^{2},x^{3}\rangle$. The map's action is $1\mapsto 0$, $x\mapsto 1$, $x^{2}\mapsto 2x$, and $x^{3}\mapsto 3x^{2}$ and its representation is easy to compute.
    $T={\rm {Rep}}_{B,B}(d/dx)={\begin{pmatrix}0&1&0&0\\0&0&2&0\\0&0&0&3\\0&0&0&0\end{pmatrix}}_{B,B}$
We find the eigenvalues with this computation.
    $0=\left|T-xI\right|={\begin{vmatrix}-x&1&0&0\\0&-x&2&0\\0&0&-x&3\\0&0&0&-x\end{vmatrix}}=x^{4}$
Thus the map has the single eigenvalue $\lambda =0$. To find the associated eigenvectors, we solve
    ${\begin{pmatrix}0&1&0&0\\0&0&2&0\\0&0&0&3\\0&0&0&0\end{pmatrix}}_{B,B}{\begin{pmatrix}b_{1}\\b_{2}\\b_{3}\\b_{4}\end{pmatrix}}_{B}=0\cdot {\begin{pmatrix}b_{1}\\b_{2}\\b_{3}\\b_{4}\end{pmatrix}}_{B}\qquad \Longrightarrow \qquad b_{2}=0,b_{3}=0,b_{4}=0$
to get this eigenspace.
    $\{{\begin{pmatrix}b_{1}\\0\\0\\0\end{pmatrix}}_{B}\,{\big |}\,b_{1}\in \mathbb {C} \}=\{b_{1}+0\cdot x+0\cdot x^{2}+0\cdot x^{3}\,{\big |}\,b_{1}\in \mathbb {C} \}=\{b_{1}\,{\big |}\,b_{1}\in \mathbb {C} \}$

### Problem 9

    Prove that
the eigenvalues of a triangular matrix (upper or lower triangular) are the entries on the diagonal.

### Answer

The determinant of the triangular matrix $T-xI$ is the product down the diagonal, and so it factors into the product of the terms $t_{i,i}-x$.
    This exercise is recommended for all readers.

### Problem 10

Find the formula for the characteristic polynomial of a $2\!\times \!2$ matrix.

### Answer

Just expand the determinant of $T-xI$.
    ${\begin{vmatrix}a-x&c\\b&d-x\end{vmatrix}}=(a-x)(d-x)-bc=x^{2}+(-a-d)\cdot x+(ad-bc)$

### Problem 11

Prove that the characteristic polynomial of a transformation is well-defined.

### Answer

Any two representations of that transformation are similar, and similar matrices have the same characteristic polynomial.
    This exercise is recommended for all readers.

### Problem 12

* Can any non-${\vec {0}}$ vector in any nontrivial vector space be a eigenvector? That is, given a ${\vec {v}}\neq {\vec {0}}$ from a nontrivial $V$, is there a transformation $t:V\to V$ and a scalar $\lambda \in \mathbb {R}$ such that $t({\vec {v}})=\lambda {\vec {v}}$?
* Given a scalar $\lambda$, can any non-${\vec {0}}$ vector in any nontrivial vector space be an eigenvector associated with the eigenvalue $\lambda$?

### Answer

* Yes, use $\lambda =1$ and the identity map.
* Yes, use the transformation that multiplies by $\lambda$.
    This exercise is recommended for all readers.

### Problem 13

Suppose that $t:V\to V$ and $T={\rm {Rep}}_{B,B}(t)$. Prove that the eigenvectors of $T$ associated with $\lambda$ are the non-${\vec {0}}$ vectors in the kernel of the map represented (with respect to the same bases) by $T-\lambda I$.

### Answer

If $t({\vec {v}})=\lambda \cdot {\vec {v}}$ then ${\vec {v}}\mapsto {\vec {0}}$ under the map $t-\lambda \cdot {\mbox{id}}$.

### Problem 14

Prove that if $a,\ldots ,\,d$ are all integers and $a+b=c+d$ then
    ${\begin{pmatrix}a&b\\c&d\end{pmatrix}}$
has integral eigenvalues, namely $a+b$ and $a-c$.

### Answer

The characteristic equation
    $0={\begin{vmatrix}a-x&b\\c&d-x\end{vmatrix}}=(a-x)(d-x)-bc$
simplifies to $x^{2}+(-a-d)\cdot x+(ad-bc)$. Checking that the values $x=a+b$ and $x=a-c$ satisfy the equation (under the $a+b=c+d$ condition) is routine.
    This exercise is recommended for all readers.

### Problem 15

Prove that if $T$ is nonsingular and has eigenvalues $\lambda _{1},\dots ,\lambda _{n}$ then $T^{-1}$ has eigenvalues $1/\lambda _{1},\dots ,1/\lambda _{n}$. Is the converse true?

### Answer

Consider an eigenspace $V_{\lambda }$. Any ${\vec {w}}\in V_{\lambda }$ is the image ${\vec {w}}=\lambda \cdot {\vec {v}}$ of some ${\vec {v}}\in V_{\lambda }$ (namely, ${\vec {v}}=(1/\lambda )\cdot {\vec {w}}$). Thus, on $V_{\lambda }$ (which is a nontrivial subspace) the action of $t^{-1}$ is $t^{-1}({\vec {w}})={\vec {v}}=(1/\lambda )\cdot {\vec {w}}$, and so $1/\lambda$ is an eigenvalue of $t^{-1}$.
    This exercise is recommended for all readers.

### Problem 16

Suppose that $T$ is $n\!\times \!n$ and $c,d$ are scalars.
* Prove that if $T$ has the eigenvalue $\lambda$ with an associated eigenvector ${\vec {v}}$ then ${\vec {v}}$ is an eigenvector of $cT+dI$ associated with eigenvalue $c\lambda +d$.
* Prove that if $T$ is diagonalizable then so is $cT+dI$.

### Answer

* We have $(cT+dI){\vec {v}}=cT{\vec {v}}+dI{\vec {v}}=c\lambda {\vec {v}}+d{\vec {v}}=(c\lambda +d)\cdot {\vec {v}}$.
* Suppose that $S=PTP^{-1}$ is diagonal. Then $P(cT+dI)P^{-1}=P(cT)P^{-1}+P(dI)P^{-1}=cPTP^{-1}+dI=cS+dI$ is also diagonal.
    This exercise is recommended for all readers.

### Problem 17

Show that $\lambda$ is an eigenvalue of $T$ if and only if the map represented by $T-\lambda I$ is not an isomorphism.

### Answer

The scalar $\lambda$ is an eigenvalue if and only if the transformation $t-\lambda {\mbox{id}}$ is singular. A transformation is singular if and only if it is not an isomorphism (that is, a transformation is an isomorphism if and only if it is nonsingular).

### Problem 18

* Show that if $\lambda$ is an eigenvalue of $A$ then $\lambda ^{k}$ is an eigenvalue of $A^{k}$.
* What is wrong with this proof generalizing that? "If $\lambda$ is an eigenvalue of $A$ and $\mu$ is an eigenvalue for $B$, then $\lambda \mu$ is an eigenvalue for $AB$, for, if $A{\vec {x}}=\lambda {\vec {x}}$ and $B{\vec {x}}=\mu {\vec {x}}$ then $AB{\vec {x}}=A\mu {\vec {x}}=\mu A{\vec {x}}=\mu \lambda {\vec {x}}$"?

### Answer

* Where the eigenvalue $\lambda$ is associated with the eigenvector ${\vec {x}}$ then $A^{k}{\vec {x}}=A\cdots A{\vec {x}}=A^{k-1}\lambda {\vec {x}}=\lambda A^{k-1}{\vec {x}}=\cdots =\lambda ^{k}{\vec {x}}$. (The full details can be put in by doing induction on $k$.)
* The eigenvector associated with $\lambda$ might not be an eigenvector associated with $\mu$.

### Problem 19

Do matrix-equivalent matrices have the same eigenvalues?

### Answer

No. These are two same-sized, equal rank, matrices with different eigenvalues.
    ${\begin{pmatrix}1&0\\0&1\end{pmatrix}}\qquad {\begin{pmatrix}1&0\\0&2\end{pmatrix}}$

### Problem 20

Show that a square matrix with real entries and an odd number of rows has at least one real eigenvalue.

### Answer

The characteristic polynomial has an odd power and so has at least one real root.

### Problem 21

Diagonalize.
    ${\begin{pmatrix}-1&2&2\\2&2&2\\-3&-6&-6\end{pmatrix}}$

### Answer

The characteristic polynomial $x^{3}-5x^{2}+6x$ has distinct roots $\lambda _{1}=0$, $\lambda _{2}=-2$, and $\lambda _{3}=-3$. Thus the matrix can be diagonalized into this form.
    ${\begin{pmatrix}0&0&0\\0&-2&0\\0&0&-3\end{pmatrix}}$

### Problem 22

Suppose that $P$ is a nonsingular $n\!\times \!n$ matrix. Show that the similarity transformation map $t_{P}:{\mathcal {M}}_{n\!\times \!n}\to {\mathcal {M}}_{n\!\times \!n}$ sending $T\mapsto PTP^{-1}$ is an isomorphism.

### Answer

We must show that it is one-to-one and onto, and that it respects the operations of matrix addition and scalar multiplication.
To show that it is one-to-one, suppose that $t_{P}(T)=t_{P}(S)$, that is, suppose that $PTP^{-1}=PSP^{-1}$, and note that multiplying both sides on the left by $P^{-1}$ and on the right by $P$ gives that $T=S$. To show that it is onto, consider $S\in {\mathcal {M}}_{n\!\times \!n}$ and observe that $S=t_{P}(P^{-1}SP)$.
The map $t_{P}$ preserves matrix addition since $t_{P}(T+S)=P(T+S)P^{-1}=(PT+PS)P^{-1}=PTP^{-1}+PSP^{-1}=t_{P}(T+S)$ follows from properties of matrix multiplication and addition that we have seen. Scalar multiplication is similar: $t_{P}(cT)=P(c\cdot T)P^{-1}=c\cdot (PTP^{-1})=c\cdot t_{P}(T)$.

### ? Problem 23

Show that if $A$ is an $n$ square matrix and each row (column) sums to $c$ then $c$ is a characteristic root of $A$. (Morrison 1967)

### Answer

This is how the answer was given in the cited source.
If the argument of the characteristic function of $A$ is set equal to $c$, adding the first $(n-1)$ rows (columns) to the $n$th row (column) yields a determinant whose $n$th row (column) is zero. Thus $c$ is a characteristic root of $A$.

## References

* Morrison, Clarence C. (proposer) (1967), "Quickie", Mathematics Magazine, 40 (4): 232.
* Strang, Gilbert (1980), Linear Algebra and its Applications (Second ed.), Harcourt Brace Jovanovich.