# Linear Algebra/Eigenvalues and Eigenvectors/Solutions

## SolutionsEdit

Problem 1

For each, find the characteristic polynomial and the eigenvalues.

1. ${\displaystyle {\begin{pmatrix}10&-9\\4&-2\end{pmatrix}}}$
2. ${\displaystyle {\begin{pmatrix}1&2\\4&3\end{pmatrix}}}$
3. ${\displaystyle {\begin{pmatrix}0&3\\7&0\end{pmatrix}}}$
4. ${\displaystyle {\begin{pmatrix}0&0\\0&0\end{pmatrix}}}$
5. ${\displaystyle {\begin{pmatrix}1&0\\0&1\end{pmatrix}}}$
1. This
${\displaystyle 0={\begin{vmatrix}10-x&-9\\4&-2-x\end{vmatrix}}=(10-x)(-2-x)-(-36)}$
simplifies to the characteristic equation ${\displaystyle x^{2}-8x+16=0}$ . Because the equation factors into ${\displaystyle (x-4)^{2}}$  there is only one eigenvalue ${\displaystyle \lambda _{1}=4}$ .
2. ${\displaystyle 0=(1-x)(3-x)-8=x^{2}-4x-5}$ ; ${\displaystyle \lambda _{1}=5}$ , ${\displaystyle \lambda _{2}=-1}$
3. ${\displaystyle x^{2}-21=0}$ ; ${\displaystyle \lambda _{1}={\sqrt {21}}}$ , ${\displaystyle \lambda _{2}=-{\sqrt {21}}}$
4. ${\displaystyle x^{2}=0}$ ; ${\displaystyle \lambda _{1}=0}$
5. ${\displaystyle x^{2}-2x+1=0}$ ; ${\displaystyle \lambda _{1}=1}$
This exercise is recommended for all readers.
Problem 2

For each matrix, find the characteristic equation, and the eigenvalues and associated eigenvectors.

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

Find the characteristic equation, and the eigenvalues and associated eigenvectors for this matrix. Hint. The eigenvalues are complex.

${\displaystyle {\begin{pmatrix}-2&-1\\5&2\end{pmatrix}}}$

The characteristic equation

${\displaystyle 0={\begin{vmatrix}-2-x&-1\\5&2-x\end{vmatrix}}=x^{2}+1}$

has the complex roots ${\displaystyle \lambda _{1}=i}$  and ${\displaystyle \lambda _{2}=-i}$ . This system

${\displaystyle {\begin{array}{*{2}{rc}r}(-2-x)\cdot b_{1}&-&1\cdot b_{2}&=&0\\5\cdot b_{1}&&(2-x)\cdot b_{2}&=&0\end{array}}}$

For ${\displaystyle \lambda _{1}=i}$  Gauss' method gives this reduction.

${\displaystyle {\begin{array}{*{2}{rc}r}(-2-i)\cdot b_{1}&-&1\cdot b_{2}&=&0\\5\cdot b_{1}&-&(2-i)\cdot b_{2}&=&0\end{array}}{\xrightarrow[{}]{(-5/(-2-i))\rho _{1}+\rho _{2}}}{\begin{array}{*{2}{rc}r}(-2-i)\cdot b_{1}&-&1\cdot b_{2}&=&0\\&&0&=&0\end{array}}}$

(For the calculation in the lower right get a common denominator

${\displaystyle {\frac {5}{-2-i}}-(2-i)={\frac {5}{-2-i}}-{\frac {-2-i}{-2-i}}\cdot (2-i)={\frac {5-(-5)}{-2-i}}}$

to see that it gives a ${\displaystyle 0=0}$  equation.) These are the resulting eigenspace and eigenvector.

${\displaystyle \{{\begin{pmatrix}(1/(-2-i))b_{2}\\b_{2}\end{pmatrix}}\,{\big |}\,b_{2}\in \mathbb {C} \}\qquad {\begin{pmatrix}1/(-2-i)\\1\end{pmatrix}}}$

For ${\displaystyle \lambda _{2}=-i}$  the system

${\displaystyle {\begin{array}{*{2}{rc}r}(-2+i)\cdot b_{1}&-&1\cdot b_{2}&=&0\\5\cdot b_{1}&-&(2+i)\cdot b_{2}&=&0\end{array}}{\xrightarrow[{}]{(-5/(-2+i))\rho _{1}+\rho _{2}}}{\begin{array}{*{2}{rc}r}(-2+i)\cdot b_{1}&-&1\cdot b_{2}&=&0\\&&0&=&0\end{array}}}$

${\displaystyle \{{\begin{pmatrix}(1/(-2+i))b_{2}\\b_{2}\end{pmatrix}}\,{\big |}\,b_{2}\in \mathbb {C} \}\qquad {\begin{pmatrix}1/(-2+i)\\1\end{pmatrix}}}$
Problem 4

Find the characteristic polynomial, the eigenvalues, and the associated eigenvectors of this matrix.

${\displaystyle {\begin{pmatrix}1&1&1\\0&0&1\\0&0&1\end{pmatrix}}}$

The characteristic equation is

${\displaystyle 0={\begin{vmatrix}1-x&1&1\\0&-x&1\\0&0&1-x\end{vmatrix}}=(1-x)^{2}(-x)}$

and so the eigenvalues are ${\displaystyle \lambda _{1}=1}$  (this is a repeated root of the equation) and ${\displaystyle \lambda _{2}=0}$ . For the rest, consider this system.

${\displaystyle {\begin{array}{*{3}{rc}r}(1-x)\cdot b_{1}&+&b_{2}&+&b_{3}&=&0\\&&-x\cdot b_{2}&+&b_{3}&=&0\\&&&&(1-x)\cdot b_{3}&=&0\end{array}}}$

When ${\displaystyle x=\lambda _{1}=1}$  then the solution set is this eigenspace.

${\displaystyle \{{\begin{pmatrix}b_{1}\\0\\0\end{pmatrix}}\,{\big |}\,b_{1}\in \mathbb {C} \}}$

When ${\displaystyle x=\lambda _{2}=0}$  then the solution set is this eigenspace.

${\displaystyle \{{\begin{pmatrix}-b_{2}\\b_{2}\\0\end{pmatrix}}\,{\big |}\,b_{2}\in \mathbb {C} \}}$

So these are eigenvectors associated with ${\displaystyle \lambda _{1}=1}$  and ${\displaystyle \lambda _{2}=0}$ .

${\displaystyle {\begin{pmatrix}1\\0\\0\end{pmatrix}}\qquad {\begin{pmatrix}-1\\1\\0\end{pmatrix}}}$
This exercise is recommended for all readers.
Problem 5

For each matrix, find the characteristic equation, and the eigenvalues and associated eigenvectors.

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

Let ${\displaystyle t:{\mathcal {P}}_{2}\to {\mathcal {P}}_{2}}$  be

${\displaystyle a_{0}+a_{1}x+a_{2}x^{2}\mapsto (5a_{0}+6a_{1}+2a_{2})-(a_{1}+8a_{2})x+(a_{0}-2a_{2})x^{2}.}$

Find its eigenvalues and the associated eigenvectors.

With respect to the natural basis ${\displaystyle B=\langle 1,x,x^{2}\rangle }$  the matrix representation is this.

${\displaystyle {\rm {Rep}}_{B,B}(t)={\begin{pmatrix}5&6&2\\0&-1&-8\\1&0&-2\end{pmatrix}}}$

Thus the characteristic equation

${\displaystyle 0={\begin{pmatrix}5-x&6&2\\0&-1-x&-8\\1&0&-2-x\end{pmatrix}}=(5-x)(-1-x)(-2-x)-48-2\cdot (-1-x)}$

is ${\displaystyle 0=-x^{3}+2x^{2}+15x-36=-1\cdot (x+4)(x-3)^{2}}$ . To find the associated eigenvectors, consider this system.

${\displaystyle {\begin{array}{*{3}{rc}r}(5-x)\cdot b_{1}&+&6\cdot b_{2}&+&2\cdot b_{3}&=&0\\&&(-1-x)\cdot b_{2}&-&8\cdot b_{3}&=&0\\b_{1}&&&+&(-2-x)\cdot b_{3}&=&0\end{array}}}$

Plugging in ${\displaystyle x=\lambda _{1}=4}$  gives

${\displaystyle {\begin{array}{*{3}{rc}r}b_{1}&+&6\cdot b_{2}&+&2\cdot b_{3}&=&0\\&&-5\cdot b_{2}&-&8\cdot b_{3}&=&0\\b_{1}&&&-&6\cdot b_{3}&=&0\end{array}}{\xrightarrow[{}]{-\rho _{1}+\rho _{2}}}{\begin{array}{*{3}{rc}r}b_{1}&+&6\cdot b_{2}&+&2\cdot b_{3}&=&0\\&&-5\cdot b_{2}&-&8\cdot b_{3}&=&0\\&&-6\cdot b_{2}&-&8\cdot b_{3}&=&0\end{array}}{\xrightarrow[{}]{-\rho _{1}+\rho _{2}}}{\begin{array}{*{3}{rc}r}b_{1}&+&6\cdot b_{2}&+&2\cdot b_{3}&=&0\\&&-5\cdot b_{2}&-&8\cdot b_{3}&=&0\\&&-6\cdot b_{2}&-&8\cdot b_{3}&=&0\end{array}}}$
Problem 7

Find the eigenvalues and eigenvectors of this map ${\displaystyle t:{\mathcal {M}}_{2}\to {\mathcal {M}}_{2}}$ .

${\displaystyle {\begin{pmatrix}a&b\\c&d\end{pmatrix}}\mapsto {\begin{pmatrix}2c&a+c\\b-2c&d\end{pmatrix}}}$

${\displaystyle \lambda =1,{\begin{pmatrix}0&0\\0&1\end{pmatrix}}{\text{ and }}{\begin{pmatrix}2&3\\1&0\end{pmatrix}}}$ , ${\displaystyle \lambda =-2,{\begin{pmatrix}-1&0\\1&0\end{pmatrix}}}$ , ${\displaystyle \lambda =-1,{\begin{pmatrix}-2&1\\1&0\end{pmatrix}}}$

This exercise is recommended for all readers.
Problem 8

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

Fix the natural basis ${\displaystyle B=\langle 1,x,x^{2},x^{3}\rangle }$ . The map's action is ${\displaystyle 1\mapsto 0}$ , ${\displaystyle x\mapsto 1}$ , ${\displaystyle x^{2}\mapsto 2x}$ , and ${\displaystyle x^{3}\mapsto 3x^{2}}$  and its representation is easy to compute.

${\displaystyle T={\rm {Rep}}_{B,B}(d/dx)={\begin{pmatrix}0&1&0&0\\0&0&2&0\\0&0&0&3\\0&0&0&0\end{pmatrix}}_{B,B}}$

We find the eigenvalues with this computation.

${\displaystyle 0=\left|T-xI\right|={\begin{vmatrix}-x&1&0&0\\0&-x&2&0\\0&0&-x&3\\0&0&0&-x\end{vmatrix}}=x^{4}}$

Thus the map has the single eigenvalue ${\displaystyle \lambda =0}$ . To find the associated eigenvectors, we solve

${\displaystyle {\begin{pmatrix}0&1&0&0\\0&0&2&0\\0&0&0&3\\0&0&0&0\end{pmatrix}}_{B,B}{\begin{pmatrix}b_{1}\\b_{2}\\b_{3}\\b_{4}\end{pmatrix}}_{B}=0\cdot {\begin{pmatrix}b_{1}\\b_{2}\\b_{3}\\b_{4}\end{pmatrix}}_{B}\qquad \Longrightarrow \qquad b_{2}=0,b_{3}=0,b_{4}=0}$

to get this eigenspace.

${\displaystyle \{{\begin{pmatrix}b_{1}\\0\\0\\0\end{pmatrix}}_{B}\,{\big |}\,b_{1}\in \mathbb {C} \}=\{b_{1}+0\cdot x+0\cdot x^{2}+0\cdot x^{3}\,{\big |}\,b_{1}\in \mathbb {C} \}=\{b_{1}\,{\big |}\,b_{1}\in \mathbb {C} \}}$
Problem 9
Prove that

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

The determinant of the triangular matrix ${\displaystyle T-xI}$  is the product down the diagonal, and so it factors into the product of the terms ${\displaystyle t_{i,i}-x}$ .

This exercise is recommended for all readers.
Problem 10

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

Just expand the determinant of ${\displaystyle T-xI}$ .

${\displaystyle {\begin{vmatrix}a-x&c\\b&d-x\end{vmatrix}}=(a-x)(d-x)-bc=x^{2}+(-a-d)\cdot x+(ad-bc)}$
Problem 11

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

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
1. Can any non-${\displaystyle {\vec {0}}}$  vector in any nontrivial vector space be a eigenvector? That is, given a ${\displaystyle {\vec {v}}\neq {\vec {0}}}$  from a nontrivial ${\displaystyle V}$ , is there a transformation ${\displaystyle t:V\to V}$  and a scalar ${\displaystyle \lambda \in \mathbb {R} }$  such that ${\displaystyle t({\vec {v}})=\lambda {\vec {v}}}$ ?
2. Given a scalar ${\displaystyle \lambda }$ , can any non-${\displaystyle {\vec {0}}}$  vector in any nontrivial vector space be an eigenvector associated with the eigenvalue ${\displaystyle \lambda }$ ?
1. Yes, use ${\displaystyle \lambda =1}$  and the identity map.
2. Yes, use the transformation that multiplies by ${\displaystyle \lambda }$ .
This exercise is recommended for all readers.
Problem 13

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

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

Problem 14

Prove that if ${\displaystyle a,\ldots ,\,d}$  are all integers and ${\displaystyle a+b=c+d}$  then

${\displaystyle {\begin{pmatrix}a&b\\c&d\end{pmatrix}}}$

has integral eigenvalues, namely ${\displaystyle a+b}$  and ${\displaystyle a-c}$ .

The characteristic equation

${\displaystyle 0={\begin{vmatrix}a-x&b\\c&d-x\end{vmatrix}}=(a-x)(d-x)-bc}$

simplifies to ${\displaystyle x^{2}+(-a-d)\cdot x+(ad-bc)}$ . Checking that the values ${\displaystyle x=a+b}$  and ${\displaystyle x=a-c}$  satisfy the equation (under the ${\displaystyle a+b=c+d}$  condition) is routine.

This exercise is recommended for all readers.
Problem 15

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

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

This exercise is recommended for all readers.
Problem 16

Suppose that ${\displaystyle T}$  is ${\displaystyle n\!\times \!n}$  and ${\displaystyle c,d}$  are scalars.

1. Prove that if ${\displaystyle T}$  has the eigenvalue ${\displaystyle \lambda }$  with an associated eigenvector ${\displaystyle {\vec {v}}}$  then ${\displaystyle {\vec {v}}}$  is an eigenvector of ${\displaystyle cT+dI}$  associated with eigenvalue ${\displaystyle c\lambda +d}$ .
2. Prove that if ${\displaystyle T}$  is diagonalizable then so is ${\displaystyle cT+dI}$ .
1. We have ${\displaystyle (cT+dI){\vec {v}}=cT{\vec {v}}+dI{\vec {v}}=c\lambda {\vec {v}}+d{\vec {v}}=(c\lambda +d)\cdot {\vec {v}}}$ .
2. Suppose that ${\displaystyle S=PTP^{-1}}$  is diagonal. Then ${\displaystyle P(cT+dI)P^{-1}=P(cT)P^{-1}+P(dI)P^{-1}=cPTP^{-1}+dI=cS+dI}$  is also diagonal.
This exercise is recommended for all readers.
Problem 17

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

The scalar ${\displaystyle \lambda }$  is an eigenvalue if and only if the transformation ${\displaystyle t-\lambda {\mbox{id}}}$  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
1. Show that if ${\displaystyle \lambda }$  is an eigenvalue of ${\displaystyle A}$  then ${\displaystyle \lambda ^{k}}$  is an eigenvalue of ${\displaystyle A^{k}}$ .
2. What is wrong with this proof generalizing that? "If ${\displaystyle \lambda }$  is an eigenvalue of ${\displaystyle A}$  and ${\displaystyle \mu }$  is an eigenvalue for ${\displaystyle B}$ , then ${\displaystyle \lambda \mu }$  is an eigenvalue for ${\displaystyle AB}$ , for, if ${\displaystyle A{\vec {x}}=\lambda {\vec {x}}}$  and ${\displaystyle B{\vec {x}}=\mu {\vec {x}}}$  then ${\displaystyle AB{\vec {x}}=A\mu {\vec {x}}=\mu A{\vec {x}}\mu \lambda {\vec {x}}}$ "?
(Strang 1980)
1. Where the eigenvalue ${\displaystyle \lambda }$  is associated with the eigenvector ${\displaystyle {\vec {x}}}$  then ${\displaystyle A^{k}{\vec {x}}=A\cdots A{\vec {x}}=A^{k-1}\lambda {\vec {x}}=\lambda A^{k-1}{\vec {x}}=\cdots =\lambda ^{k}{\vec {x}}}$ . (The full details can be put in by doing induction on ${\displaystyle k}$ .)
2. The eigenvector associated wih ${\displaystyle \lambda }$  might not be an eigenvector associated with ${\displaystyle \mu }$ .
Problem 19

Do matrix-equivalent matrices have the same eigenvalues?

No. These are two same-sized, equal rank, matrices with different eigenvalues.

${\displaystyle {\begin{pmatrix}1&0\\0&1\end{pmatrix}}\qquad {\begin{pmatrix}1&0\\0&2\end{pmatrix}}}$
Problem 20

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

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

Problem 21

Diagonalize.

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

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

${\displaystyle {\begin{pmatrix}0&0&0\\0&-2&0\\0&0&-3\end{pmatrix}}}$
Problem 22

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

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 ${\displaystyle t_{P}(T)=t_{P}(S)}$ , that is, suppose that ${\displaystyle PTP^{-1}=PSP^{-1}}$ , and note that multiplying both sides on the left by ${\displaystyle P^{-1}}$  and on the right by ${\displaystyle P}$  gives that ${\displaystyle T=S}$ . To show that it is onto, consider ${\displaystyle S\in {\mathcal {M}}_{n\!\times \!n}}$  and observe that ${\displaystyle S=t_{P}(P^{-1}SP)}$ .

The map ${\displaystyle t_{P}}$  preserves matrix addition since ${\displaystyle t_{P}(T+S)=P(T+S)P^{-1}=(PT+PS)P^{-1}=PTP^{-1}+PSP^{-1}=t_{P}(T+S)}$  follows from properties of matrix multiplication and addition that we have seen. Scalar multiplication is similar: ${\displaystyle t_{P}(cT)=P(c\cdot T)P^{-1}=c\cdot (PTP^{-1})=c\cdot t_{P}(T)}$ .

? Problem 23

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

If the argument of the characteristic function of ${\displaystyle A}$  is set equal to ${\displaystyle c}$ , adding the first ${\displaystyle (n-1)}$  rows (columns) to the ${\displaystyle n}$ th row (column) yields a determinant whose ${\displaystyle n}$ th row (column) is zero. Thus ${\displaystyle c}$  is a characteristic root of ${\displaystyle A}$ .