# MATH 323 Lecture 25

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## Eigenvalues and Eigenvectors

### Example

${\displaystyle A={\begin{bmatrix}2&-3&1\\1&-2&1\\1&-3&2\end{bmatrix}}}$

${\displaystyle p(\lambda )={\begin{vmatrix}2-\lambda &-3&1\\1&-2-\lambda &1\\1&-3&2-\lambda \end{vmatrix}}=-\lambda (\lambda -1)^{2}}$

Therefore, the eigenvalues are ${\displaystyle \lambda _{1}=0}$, ${\displaystyle \lambda _{2}=\lambda _{3}=1}$

To find eigenvectors, find basis for ${\displaystyle N(A-\lambda \,I)}$:

1. ${\displaystyle N(A-0\cdot I)=N(A)=\left\langle \alpha ,\alpha ,\alpha \right\rangle \quad \alpha \in \mathbb {R} }$
2. ${\displaystyle N(A-1\cdot I)=\left\langle 3\alpha -\beta ,\alpha ,\beta \right\rangle }$

Therefore, ${\displaystyle \left\langle 1,1,1\right\rangle }$ is an eigenvector belonging to ${\displaystyle \lambda _{1}=0}$, and ${\displaystyle \left\langle 3,1,0\right\rangle }$ and ${\displaystyle \left\langle -1,0,1\right\rangle }$ are linearly independent eigenvectors belonging to ${\displaystyle \lambda _{2}=\lambda _{3}=1}$.

### Complex Eigenvalues

For an arbitrary ${\displaystyle n\times n}$ matrix, the characteristic polynomial will be ${\displaystyle (-1)^{n}\lambda ^{n}+\Theta (\lambda ^{n-1})}$.

Polynomials of degree ${\displaystyle n}$ always have exactly ${\displaystyle n}$ complex roots:

{\displaystyle {\begin{aligned}p(\lambda )&=(a_{11}-\lambda )\dots (a_{nn}-\lambda )+\dots \\&=(-1)^{n}(\lambda -\lambda _{1})\dots (\lambda -\lambda _{n})+\Theta (\lambda ^{n-1})&=(\lambda _{1}-\lambda )\dots (\lambda _{n}-\lambda )+\Theta (\lambda ^{n-1})\end{aligned}}}

${\displaystyle p(0)=\prod _{i=1}^{n}\lambda _{i}}$ is the constant term of ${\displaystyle p(\lambda )}$.

The coefficient of the remaining ${\displaystyle \lambda ^{n-1}}$ is given by ${\displaystyle (-1)^{n-1}\sum _{i=1}^{n}a_{ii}}$. This is called the trace of ${\displaystyle A}$.

#### Example

${\displaystyle A={\begin{bmatrix}1&2\\-2&1\end{bmatrix}}}$

{\displaystyle {\begin{aligned}\lambda _{1}&=1+2i\\\lambda _{2}&=1-2i\end{aligned}}}

${\displaystyle (A-\lambda _{1}\,I){\vec {x}}=0\longrightarrow {\begin{bmatrix}1&i\\0&0\end{bmatrix}}}$

Therefore, ${\displaystyle \left\langle -i,1\right\rangle }$ is an eigenvector belonging to ${\displaystyle \lambda _{1}=1+2i}$. If we multiply this eigenvector by ${\displaystyle i}$, we get ${\displaystyle \langle 1,i\rangle }$, which is a much nicer form.

### Theorem 6.1.1

Let ${\displaystyle A}$ and ${\displaystyle B}$ be ${\displaystyle n\times n}$ matrices.

If ${\displaystyle B}$ is similar to ${\displaystyle A}$, then ${\displaystyle A}$ and ${\displaystyle B}$ have the same characteristic polynomials and the same eigenvalues (not eigenvectors).

#### Proof

Recall that ${\displaystyle A\sim B\iff B=S^{-1}AS}$

${\displaystyle p_{B}(\lambda )=|B-\lambda \,I|=|S^{-1}AS-\lambda I|=|S^{-1}(A-\lambda \,I)S|=|S^{-1}||A-\lambda \,I||S|=|A-\lambda \,I|=p_{A}(\lambda )}$

## Review: Complex Numbers

Every complex number ${\displaystyle z=a+b\,i}$ has a conjugate ${\displaystyle {\bar {z}}=a-b\,i}$.

A polynomial that has a root ${\displaystyle z=a+b\,i}$ will also have the conjugate as a root

corollary: if ${\displaystyle \lambda =a+b\,i}$ is an eigenvalue of a real matrix ${\displaystyle A}$, then ${\displaystyle {\bar {\lambda }}=a-b\,i}$ is also an eigenvalue.
if ${\displaystyle z=a+b\,i}$ is an eigenvector of complex eigenvalue ${\displaystyle \lambda _{1}}$, then ${\displaystyle {\bar {z}}=a-b\,i}$ is an eigenvector of the conjugate eigenvalue ${\displaystyle {\bar {\lambda }}_{1}}$.

multiplying a complex by its conjugate will always yield a real number

Conjugate of a Product of complex numbers is just the product of conjugates

## Solving Systems of Linear Differential Equations

A linear differential equation is of the form

{\displaystyle {\begin{aligned}y_{1}'&=a_{11}\,y_{1}+\dots a_{1n}\,y_{n}\\y_{2}'&=a_{21}\,y_{1}+\dots a_{2n}\,y_{n}\\\dots \\y_{n}'&=a_{n1}\,y_{1}+\dots a_{nn}\,y_{n}\end{aligned}}}

Where ${\displaystyle y_{i}}$ are functions in ${\displaystyle C^{1}[a,b]}$.

Which can be written in the form ${\displaystyle {\vec {Y}}'=A{\vec {Y}}}$.

Suppose ${\displaystyle {\vec {Y}}(t)}$ is of the form ${\displaystyle \langle x_{1}\mathrm {e} ^{\lambda t},x_{2}\mathrm {e} ^{\lambda t},\ldots ,x_{n}\mathrm {e} ^{\lambda t}\rangle }$.

Then ${\displaystyle {\vec {Y}}(t)=\mathrm {e} ^{\lambda t}{\vec {x}}}$ and ${\displaystyle {\vec {Y}}'(t)=\lambda {\vec {Y}}(t)=\lambda \mathrm {e} ^{\lambda t}{\vec {x}}}$.

If ${\displaystyle {\vec {x}}}$ is an eigenvector of ${\displaystyle A}$, then ${\displaystyle A{\vec {Y}}=A\mathrm {e} ^{\lambda t}{\vec {x}}=\mathrm {e} ^{\lambda t}A{\vec {x}}=\mathrm {e} ^{\lambda t}\lambda x}$

The set of solutions to ${\displaystyle {\vec {Y}}'=A{\vec {Y}}}$ is an ${\displaystyle n}$-dimensional subspace, and if solutions ${\displaystyle {\vec {Y}}_{1},\ldots ,{\vec {Y}}_{n}}$ are linearly independent solutions, then any linear combination of ${\displaystyle {\vec {Y}}_{i}}$'s are also solutions.

### Example

{\displaystyle {\begin{aligned}y_{1}'&=3y_{1}+4y_{2}\\y_{2}'&=3y_{1}+2y_{2}\end{aligned}}}

${\displaystyle A={\begin{bmatrix}3&4\\3&2\end{bmatrix}}}$

${\displaystyle \lambda _{1}=6,\lambda _{2}=-1}$

{\displaystyle {\begin{aligned}{\vec {x}}_{1}&=\langle 4,3\rangle \\{\vec {x}}_{2}&=\langle 1,-1\rangle \\{\vec {Y}}_{1}(t)&=\mathrm {e} ^{6t}\langle 4,3\rangle \\{\vec {Y}}_{2}(t)&=\mathrm {e} ^{-t}\langle 1,-1\rangle \\{\vec {Y}}&=c_{1}\,{\vec {Y}}_{1}(t)+c_{2}\,{\vec {Y}}_{2}(t)\\&={\begin{bmatrix}4c_{1}\,\mathrm {e} ^{6t}+c_{2}\,\mathrm {e} ^{-t}\\3c_{1}\,\mathrm {e} ^{6t}-c_{2}\,\mathrm {e} ^{-t}\end{bmatrix}}\end{aligned}}}

For initial value problem ${\displaystyle {\vec {Y}}(0)=\langle 6,1\rangle }$, ${\displaystyle c_{1}=1}$ and ${\displaystyle c_{2}=2}$.