Ch.15 Eigenvectors

$T=\begin{pmatrix}6&-1&-1\\2&11&-1\\-6&-5&7\end{pmatrix}$
is diagonalizable using
$\begin{array}{cc}P=\begin{pmatrix}1/2&1/4&1/4\\-1/2&1/4&1/4\\-1/2&-3/4&1/4\end{pmatrix}&P^{-1}=\begin{pmatrix}1&-1&0\\0&1&-1\\2&1&1\end{pmatrix}\end{array}$
to get this $D=PTP^{-1}$
$D=\begin{pmatrix}4&0&0\\0&8&0\\0&0&12\end{pmatrix}$

Consider if the following is diagonalizable
$N=\begin{pmatrix}0&0\\1&0\end{pmatrix}$
Since $N^2$ is the zero matrix, for any map $n$ represented by $N$ with bases $B,B$, the composition $n\circ n$ is the zero map. This implies that any representation of $n$ with some bases $\hat{B},\hat{B}$ has a square that is the zero map. However, if there was some diagonal matix $D$ similar to $N$, then we must have $D^2$ have diagonal entries be the square of the entries of $D$, so cannot be a zero matrix. Thus, $N$ is not diagonalizable.

If $B=\langle\vec{\beta}_1,...,\vec{\beta}_n\rangle$ is a basis then
$t(\vec{\beta}_i)=\lambda_i\vec{\beta}_i\iff\text{Rep}_B(t(\vec{\beta}_i))=\lambda_i\text{Rep}_B(\vec{\beta}_i)$
which is equivalent to
$\text{Rep}_B(t(\vec{\beta}_i))=\lambda_i\begin{pmatrix}0\\\vdots\\1\\\vdots\\0\end{pmatrix}=\begin{pmatrix}\lambda_1&\cdots&0\\\vdots&\ddots&\vdots\\0&\cdots&\lambda_n\end{pmatrix}\begin{pmatrix}0\\\vdots\\1\\\vdots\\0\end{pmatrix}=\begin{pmatrix}0\\\vdots\\\lambda_i\\\vdots\\0\end{pmatrix}$

$T=\begin{pmatrix}4&1\\0&-1\end{pmatrix}$ is not diagonal, but a similar matrix can be found that is diagonal.
If $T=\text{Rep}_{\mathcal{E}_2,\mathcal{E}_2}(t)$ for $t:\mathbb{R}^2\to\mathbb{R}^2$, we want to find a basis $B=\langle\vec{\beta}_1,\vec{\beta}_2\rangle$ such that $\text{Rep}_{B,B}(t)=\begin{pmatrix}\lambda_1&0\\0&\lambda_2\end{pmatrix}$

We want $\begin{pmatrix}4&1\\0&-1\end{pmatrix}\vec{\beta}_1=\lambda_1\vec{\beta}_1$ and $\begin{pmatrix}4&1\\0&-1\end{pmatrix}\vec{\beta}_2=\lambda_2\vec{\beta}_2$
Or, converting $\vec{\beta}_k$ to a column vector, we want a scalar $x\in\mathbb{C}$ such that
$\begin{pmatrix}4&1\\0&-1\end{pmatrix}\begin{pmatrix}b_1\\b_2\end{pmatrix}=x\begin{pmatrix}b_1\\b_2\end{pmatrix}$
has a solution for $b_1,b_2$ that are not both zeroes.
Rewriting this as a linear system:
$\begin{array}{rrl}(4-x)b_1+&b_2&=0\\&(-1-x)b_2&=0\end{array}$
If $-1-x\ne0$ and $4-x\ne0$, then the only solution is that $b_1=b_2=0$. So, one solution results from $-1-x=0\rightarrow x=-1$, resulting in $b_1=(-1/5)b_2$, and the other solution from $4-x=0\rightarrow x=4$, resulting in $b_2=0$.
Thus, our basis is $B=\langle\begin{pmatrix}-1\\5\end{pmatrix},\begin{pmatrix}1\\0\end{pmatrix}\rangle$, which is diagonalizable to $D=\begin{pmatrix}-1&0\\0&4\end{pmatrix}$

Let $A=\begin{pmatrix}1&2\\-1&4\end{pmatrix}$. Find $A^n$.

We first evaluate $P$ and $D$.
We must find $x\in\mathbb{C}$ such that there are solutions to
$\begin{pmatrix}1&2\\-1&4\end{pmatrix}\begin{pmatrix}b_1\\b_2\end{pmatrix}=x\begin{pmatrix}b_1\\b_2\end{pmatrix}$
that are not $b_1=b_2=0$.
The associated linear system is
$\begin{array}{rcr}(1-x)b_1&+&2b_2=0\\-b_1&+&(4-x)b_2=0\end{array}$
Substituting for $b_1=(4-x)b_2$
$(1-x)(4-x)b_2+2b_2=(x^2-5x+6)b_2=0$
For $b_2$ to be nonzero, we need $x^2-5x+6=0$, which is true when $x=2,3$.
When $x=2$, we have $b_1=2b_2$, and if $x=3$, we have $b_1=b_2$. Thus, our basis is $B=\langle\begin{pmatrix}2\\1\end{pmatrix},\begin{pmatrix}1\\1\end{pmatrix}\rangle$ and diagonal matrix $D=\begin{pmatrix}2&0\\0&3\end{pmatrix}$.
From this, we generate
$\begin{array}{cc}P=\begin{pmatrix}2&1\\1&1\end{pmatrix}&P^{-1}=\begin{pmatrix}1&-1\\-1&2\end{pmatrix}\end{array}$
Thus, we have
$A^n=PD^nP^{-1}=\begin{pmatrix}2&1\\1&1\end{pmatrix}\begin{pmatrix}2^n&0\\0&3^n\end{pmatrix}\begin{pmatrix}1&-1\\-1&2\end{pmatrix}=\begin{pmatrix}2^{n+1}-3^n&-2^{n+1}+2\cdot3^n\\2^n-3^n&-2^n+2\cdot3^n\end{pmatrix}$

The matrix
$D=\begin{pmatrix}4&0\\0&2\end{pmatrix}$
has an eigenvalue $\lambda_1=4$ and $\lambda_2=2$.
The first is associated with eigenvector $\vec{e}_1$
$\begin{pmatrix}4&0\\0&2\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}=4\cdot\begin{pmatrix}1\\0\end{pmatrix}$
The second is asociated with $\vec{e}_2$
$\begin{pmatrix}4&0\\0&2\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix}=2\cdot\begin{pmatrix}0\\1\end{pmatrix}$

Consider the matrix
$T=\begin{pmatrix}0&5&7\\-2&7&7\\-1&1&4\end{pmatrix}$
We want to find scalars $x$ such that $T\vec{\zeta}=x\vec{\zeta}$ for some nonzero $\vec{\zeta}$. Bring to the terms to the left
$\begin{pmatrix}0&5&7\\-2&7&7\\-1&1&4\end{pmatrix}\begin{pmatrix}z_1\\z_2\\z_3\end{pmatrix}-x\begin{pmatrix}z_1\\z_2\\z_3\end{pmatrix}=\vec{0}$
And factor out the vector
$\begin{pmatrix}0-x&5&7\\-2&7-x&7\\-1&1&4-x\end{pmatrix}\begin{pmatrix}z_1\\z_2\\z_3\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}$
This has nonzero solutions iff the matrix is singular, i.e. has a determinant of zero.
$0=\begin{vmatrix}-x&5&7\\-2&7-x&7\\-1&1&4-x\end{vmatrix}=-(x-5)(x-4)(x-2)$
So the eigenvalues of $\lambda_1=5,\lambda_2=4,\lambda_3=2$
To find the associated eigenvectors, substitute each and solve.
For $\lambda_1=5$
$\begin{pmatrix}-5&5&7\\-2&2&7\\-1&1&-1\end{pmatrix}\begin{pmatrix}z_1\\z_2\\z_3\end{pmatrix}=\vec{0}$
Gauss's Method yields the solutions to be
$V_5=\{\begin{pmatrix}1\\1\\0\end{pmatrix}z_2\space|\space z_2\in\mathbb{C}\}$
For $\lambda_2=4$
$\begin{pmatrix}-4&5&7\\-2&3&7\\-1&1&0\end{pmatrix}\begin{pmatrix}z_1\\z_2\\z_3\end{pmatrix}=\vec{0}$
The solutions are
$V_4=\{\begin{pmatrix}-7\\-7\\1\end{pmatrix}z_3\space|\space z_3\in\mathbb{C}\}$
For $\lambda_3=2$
$\begin{pmatrix}-2&5&7\\-2&5&7\\-1&1&2\end{pmatrix}\begin{pmatrix}z_1\\z_2\\z_3\end{pmatrix}=\vec{0}$
gives
$V_2=\{\begin{pmatrix}1\\-1\\1\end{pmatrix}z_3\space|\space z_3\in\mathbb{C}\}$

$\lambda$ is an eigenvalue of $T$ iff there exists a nonzero $\vec{v}$ such that $T\vec{v}=\lambda\vec{v}$, or equivalently, $(T-\lambda I)\vec{v}=0$. Since $\vec{v}\ne0$, we must have $(T-\lambda I)$ be nonintertible, which is equivalent to $|T-\lambda I|=0$

Suppose $T_1$ and $T_2$ are similar and $P$ exists such that $T_2=PT_1P^{-1}$.
Then
$T_2-xI=PT_1P^{-1}-xI=PT_1P^{-1}-xPP^{-1}=PT_1P^{-1}-P(xI)P^{-1}=P(T_1-xI)P^{-1}$
From $T_2-xI=P(T_1-xI)P^{-1}$, we obtain that $|T_2-xI|=|P(T_1-xI)P^{-1}|=|P||T_1-xI||P^{-1}|=|T_1-xI|$

First, it is nonempty, since $t(\vec{0})=\lambda\cdot\vec{0}$. Because $\vec{\zeta}$ cannot be $\vec{0}$, $V_\lambda$ must have more than just $\vec{0}$
Now, we must show closure under addition and multiplication.
$\begin{array}{rcl}t(c_1\vec{\zeta}_1+\cdots+c_n\vec{\zeta}_n)&=&c_1t(\vec{\zeta}_1)+\cdots+c_nt(\vec{\zeta}_n)\\&=&c_1\lambda\vec{\zeta}_1+\cdots+c_n\lambda\vec{\zeta}_n\&=&\lambda(c_1\vec{\zeta}_1+\cdots+c_n\vec{\zeta}_n)\end{array}$

We use induction. The base case is 0 eigenvalues, in which case the set is empty, so is linearly independent.
Assume it holds true for some $k\ge0$ and consider eigenvalues $\lambda_1,...,\lambda_{k+1}$ with associated eigenvectors $\vec{v}_1,...,\vec{v}_{k+1}$.
Suppose $\vec{0}=c_1\vec{v}_1+\cdots+c_k\vec{v}_k+c_{k+1}\vec{v}_{k+1}$
Derive two equations: one from multiplying $\lambda_{k+1}$ on both sides and the other from applying the map
$\vec{0}=c_1\lambda_{k+1}\vec{v}_1+\cdots+c_k\lambda_{k+1}\vec{v}_k+c_{k+1}\lambda_{k+1}\vec{v}_{k+1}$
$\vec{0}=c_1\lambda_1\vec{v}_1+\cdots+c_k\lambda_k\vec{v}_k+v_{k+1}\lambda_{k+1}\vec{v}_{k+1}$
and subtract
$\vec{0}=c_1(\lambda_{k+1}-\lambda_1)\vec{v}_1+\cdots+c_k(\lambda_{k+1}-\lambda_k)\vec{v}_k$
The $\vec{v}_{k+1}$ term vanishes. By the indutive hypothesis, $c_i(\lambda_{k+1}-\lambda_i)=0$ for $i=1,...,k$
However, since the eigenvalues are distinct, we must have $c_1=c_2=\cdots=c_k=0$. From this and the original equation, we have $c_{k+1}\vec{v}_{k+1}$ so $c_{k+1}=0$ as well.

Call the $n$ distinct eigenvalues $\lambda_1,...,\lambda_n$. For $i=1,...,n$ we can find a non-zero eigenvector $\vec{\zeta}_i\in\mathbb{C}^n$ for the eigenvalue $\lambda_i$. From the previous theorem, they are all linearly independent, so they form a basis for $\mathbb{C}^n$. Since this is a basis of eigenvectors, the matrix is diagonalizable.

Diagonalize $A=\begin{pmatrix}4&-3&0\\2&-1&0\\1&-1&1\end{pmatrix}$

First find the characteristic polynomial
$f(x)=\text{det}(A-xI)=(1-x)((4-x)(-1-x)+6)=-(x-1)^2(x-2)$
The eigenvalues are 1 and 2 with multiplicities 2 and 1, respectively.
First compute the $1$-eigenspace:
$\begin{pmatrix}3&-3&0\\2&-2&0\\1&-1&0\end{pmatrix}\vec{v}=0\xrightarrow{\text{Gaussian}}\begin{pmatrix}1&-1&0\\0&0&0\\0&0&0\end{pmatrix}\vec{v}=0$
Therefore, $\vec{v}=\begin{pmatrix}x\\y\\z\end{pmatrix}=y\begin{pmatrix}1\\1\\0\end{pmatrix}+z\begin{pmatrix}0\\0\\1\end{pmatrix}$ and a basis for the $1$-eigenspace is
$B_1=\langle\begin{pmatrix}1\\1\\0\end{pmatrix},\begin{pmatrix}0\\0\\1\end{pmatrix}\rangle$
For the $2$-eigenspace:
$\begin{pmatrix}2&-3&0\\2&-3&0\\1&-1&-1\end{pmatrix}\vec{v}=0\xrightarrow{\text{Gaussian}}\begin{pmatrix}1&0&-3\\0&1&-2\\0&0&0\end{pmatrix}\vec{v}=0$
Therefore, a basis for the $2$-eigenspace is
$B_2=\langle\begin{pmatrix}3\\2\\1\end{pmatrix}\rangle$
The three basis vectors in $B_1$ and $B_2$ are linearly independent, so we have
$A=PDP^{-1}\text{ for }P=\begin{pmatrix}1&0&3\\1&0&2\\0&1&1\end{pmatrix},\space D=\begin{pmatrix}1&0&0\\0&1&0\\0&0&2\end{pmatrix}$

The matrix $A=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ has only one eigenvalue, $1$.
Computing the $1$-eigenspace,
$(A-I)\vec{v}=\begin{pmatrix}0&1\\0&0\end{pmatrix}\vec{v}=0$
The eigenspace only contains multiples of $\begin{pmatrix}1\\0\end{pmatrix}$. Since we need 2 linearly independent vectors for this to be diagonalizable, this is not.

Consider $A=\begin{pmatrix}1&-1\\1&1\end{pmatrix}$
The eigenvalues are $1\pm i$, a conjugate pair.
Let $\lambda=1-i$. We compute an eigenvector $\vec{v}$:
$A-\lambda I=\begin{pmatrix}i&-1\\*&*\end{pmatrix}\rightarrow \vec{v}=\begin{pmatrix}1\\i\end{pmatrix}$
Therefore,
$\begin{array}{cc}P=\begin{pmatrix}\mathscr{R}\begin{pmatrix}1\\i\end{pmatrix}\mathscr{I}\begin{pmatrix}1\\i\end{pmatrix}\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}&C=\begin{pmatrix}\mathscr{R}(\lambda)&\mathscr{I}(\lambda)\\-\mathscr{I}(\lambda)&\mathscr{R}(\lambda)\end{pmatrix}=\begin{pmatrix}1&-1\\1&1\end{pmatrix}\end{array}$

The matrix $C=A$ scales by a factor of $|\lambda|=\sqrt{1^2+1^2}=\sqrt{2}$, and the argument is $-\pi/4$, therefore $C=A$ rotates by $+\pi/4$ (note that $A$ is $\sqrt{2}$ times the rotation matrix by $\pi/4$, so this matches)

If $A\ne C$, then $C$ scales by $|\lambda|$ and rotates by $-\arg{(\lambda)}$ in the standard basis, but $A$ does that in the basis formed by the columns of $P$, i.e. the eigenvectors.
This is similar to the real eigenvalue case, where in $A=PDP^{-1}$, $A$ scales like $D$, but under the basis formed from the columns of $P$

Let $A=\begin{pmatrix}1&-1&0\\1&1&0\\0&0&2\end{pmatrix}$. This is block diagonal. It acts on the $xy$-plane by scaling by $\sqrt{2}$ and rotationg by $\pi/4$. It acts on the $z$-axis by scaling by $2$.