Eigenvalues

Let A be a $n\times n$ matrix (square). A vector $\hat{v}$ is an $\text{eigenvector}$ of $A$ if there is a number $\lambda$ such that

$$A\vec{v}=\lambda \vec{v}.\vec{v}\ne \vec{0}.$$

That number, $\lambda$ is the $\text{eigenvalue}$ of the matrix.

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par exemple,

Let $A=\left[\begin{array}{cc}1& 6\\ 2& 2\end{array}\right]$ and $\vec{v}=\left[\begin{array}{c}3\\ -2\end{array}\right]$, $\vec{w}=\left[\begin{array}{c}6\\ -5\end{array}\right]$.

i) Is $\vec{v}$ an eigenvector of $A$? ii) Is $\vec{w}$ an eigenvector of $A$?

$$A\vec{v}=\left[\begin{array}{cc}1& 6\\ 5& 2\end{array}\right]\left[\begin{array}{c}3\\ -2\end{array}\right]=\left[\begin{array}{c}3-12\\ 15-4\end{array}\right]=\left[\begin{array}{c}-9\\ 11\end{array}\right]\stackrel{?}{=}\lambda \left[\begin{array}{c}3\\ -2\end{array}\right]$$

There is no such $\lambda$, thus $\vec{v}$ is not an eigenvector of $A$.

What about $\vec{w}$?

$$\begin{array}{rl}A\vec{w}=\left[\begin{array}{cc}1& 6\\ 5& 2\end{array}\right]\left[\begin{array}{c}6\\ -5\end{array}\right]=\left[\begin{array}{c}6-30\\ 30-10\end{array}\right]=\left[\begin{array}{c}-24\\ 20\end{array}\right]& \stackrel{?}{=}\lambda \left[\begin{array}{c}6\\ -5\end{array}\right]\\ \left[\begin{array}{c}-24\\ 20\end{array}\right]& =-4\left[\begin{array}{c}6\\ -5\end{array}\right]\end{array}$$

It appears $\vec{w}$ is indeed an eigenvector of $A$ with an eigenvalue of $\lambda =-4$.

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How do we find eigenvalues and eigenvectors of a matrix? We need to find $\lambda$ and $\vec{v}$ such that $\vec{v}\ne \vec{0}$ and $A\vec{v}=\lambda \vec{v}$.

There’s a bunch of math I don’t particularly care to write out, but we yield the following equation:

$$(A-\lambda I)\vec{v}=\vec{0}$$

To find $\vec{v}\ne 0$ such that $(A-\lambda I)\vec{v}=\vec{0}$ we need $A-\lambda I$ to be not invertible. We need

$$\underset{\text{The characteristic equation, if you factor.}}{\underbrace{\det (A-\lambda I)=0}}.$$

If we use solve for $\lambda$ via the above equation, we can then get the eigenvectors from the above-above equation.

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example

$A=\left[\begin{array}{cc}7& 4\\ -3& -1\end{array}\right]$. Find eigenvalues and their eigenspaces.

Step 1: Set up the characteristic polynomial and solve it.

$$\begin{array}{rl}0& =\det (A-\lambda I)\\ 0& =\det (\left[\begin{array}{cc}7& 4\\ -3& -1\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right])\\ 0& =\det (\left[\begin{array}{cc}7& 4\\ -3& -1\end{array}\right]-\left[\begin{array}{cc}\lambda & 0\\ 0& \lambda \end{array}\right])\\ 0& =\det (\left[\begin{array}{cc}7-\lambda & 4\\ -3& -1-\lambda \end{array}\right])\\ 0& =(7-\lambda )(-1-\lambda )\cancel{-}(4)(\cancel{-}3)\\ 0& =-(7-\lambda )(1+\lambda )+(4)(3)\\ 0& =12-(7-\lambda )(1+\lambda )\\ 0& ={\lambda }^2-6\lambda +5\\ & \\ & \therefore \\ & \\ & \lambda =1,\lambda =5\end{array}$$

Step 2: Uhhh do the other thing

$$\begin{array}{rl}\lambda =1& \\ & \\ (A-\lambda I)\vec{v}& =\vec{0}\\ \left[\begin{array}{cc}7-1& 4\\ -3& -1-1\end{array}\right]\left[\begin{array}{c}a\\ b\end{array}\right]& =\left[\begin{array}{c}0\\ 0\end{array}\right]\\ \left[\begin{array}{c}a\\ b\end{array}\right]& =\left[\begin{array}{c}-\frac{2}{3}\\ 1\end{array}\right]b\\ E_1& =\text{span}\{\left[\begin{array}{c}-\frac{2}{3}\\ 1\end{array}\right]\}\end{array}$$ $$\begin{array}{rl}\lambda =5& \\ & \\ (A-\lambda I)\vec{v}& =\vec{0}\\ \left[\begin{array}{cc}7-5& 4\\ -3& -1-5\end{array}\right]\left[\begin{array}{c}a\\ b\end{array}\right]& =\left[\begin{array}{c}0\\ 0\end{array}\right]\\ \left[\begin{array}{c}a\\ b\end{array}\right]& =\left[\begin{array}{c}-2\\ 1\end{array}\right]b\\ E_5& =\text{span}\{\left[\begin{array}{c}-2\\ 1\end{array}\right]\}\end{array}$$

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Theorem: If a matrix is triangular, the eigenvalues are the elements in the main diagonal.

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Nota bene: ${\lambda }_1=\text{conjugate}({\lambda }_2)$ just as $\overrightarrow{v_1}=\text{conjugate}(\overrightarrow{v_2})$

Theorem: Let $\{\overrightarrow{v_1},\overrightarrow{v_2},\overrightarrow{v_3},\dots ,\overrightarrow{v_k}\}$ are eigenvectors of $A$, each associated to a different eigenvalue. Then $\{\overrightarrow{v_1},\overrightarrow{v_2},\overrightarrow{v_3},\dots ,\overrightarrow{v_k}\}$ are linearly independent.

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Diagonal Matrices

A square matrix $A$ is diagonal if anything is not on the main diagonal is zero;

$$A_{ij}=0\text{ if }i\ne j$$

A diagonalizable matrix: A square matrix $A$ is $\text{diagonalizable}$ if there is an invertible matrix $P$ and a diagonal matrix $D$ such that:

$$A=PD{P}^{-1}$$

what?

No matter.

Why are diagonalizable matrices useful? Next week, we’ll see that it’s important to compute powers of a matrix.

$$\begin{array}{rl}A& =PD{P}^{-1}\\ & \\ A^2& =(PD\underset{\text{yields identity}}{\underbrace{P^{-1})(P}}D{P}^{-1})=P{D}^2{P}^{-1}\\ A^3& =P{D}^3{P}^{-1}\\ & \\ \therefore & \\ & \\ A^k& =P{D}^k{P}^{-1}\end{array}$$

Given $A$, the way we find $P$, $D$ and $P^{-}1$, is understanding that $D=\left[\begin{array}{cc}{\lambda }_1& 0\\ 0& {\lambda }_2\end{array}\right]$.

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Recall:

• $P=[\underset{\text{Linearly independent.}}{\underbrace{\text{Eigenvectors are the columns}}}]$
• $D=\left[\begin{array}{cc}{\lambda }_1& 0\\ 0& {\lambda }_2\end{array}\right]$
• The eigenvalue on D corresponds to the eigenvector in the same column in P

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We can find an explicit formula for a function using vectors.

If $x_{n+2}=5x_{n+1}+6x_n$, $x_1=0$ and $x_2=-1$

$$\begin{array}{rl}& {\vec{y}}_n=\left[\begin{array}{c}{x}_{n+1}\\ x_n\end{array}\right]{\vec{y}}_1=\left[\begin{array}{c}{x}_2\\ x_1\end{array}\right]=\left[\begin{array}{c}-1\\ 0\end{array}\right]{\vec{y}}_2=\left[\begin{array}{c}{x}_3\\ x_2\end{array}\right]=\left[\begin{array}{c}-5\\ 1\end{array}\right]\\ & \\ & {\vec{y}}_{n+1}=\left[\begin{array}{c}{x}_{n+1+1}\\ x_{n+1}\end{array}\right]=\left[\begin{array}{c}{x}_{n+2}\\ x_{n+1}\end{array}\right]=\left[\begin{array}{c}5x_{n+1}+6x_n\\ x_{n+1}\end{array}\right]=\left[\begin{array}{cc}5& 6\\ 1& 0\end{array}\right]\left[\begin{array}{c}{x}_{n+1}\\ x_n\end{array}\right]\\ & \\ & {\vec{y}}_{n+1}=\underset{\text{Recursive formula}}{\underbrace{\left[\begin{array}{c}{x}_{n+1}\\ x_n\end{array}\right]{\vec{y}}_n}}\\ & \dots \\ & {\vec{y}}_{100}={\left[\begin{array}{cc}5& 6\\ 1& 0\end{array}\right]}^{99}\left[\begin{array}{c}-1\\ 0\end{array}\right]\end{array}$$

Now we need to find a formula for ${\left[\begin{array}{cc}5& 6\\ 1& 0\end{array}\right]}^{n-1}$ using diagonalization. We have to find $P$, $D$ and $P^{-1}$ such that $\left[\begin{array}{cc}5& 6\\ 1& 0\end{array}\right]=PD{P}^{-1}$. This can be done by finding eigenvalues, eigenvectors

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