MATH 323 Lecture 25

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Eigenvalues and Eigenvectors

Example

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

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

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

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

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

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

Complex Eigenvalues

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

Polynomials of degree $n$ always have exactly $n$ complex roots:

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

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

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

Example

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

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

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

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

Theorem 6.1.1

Let $A$ and $B$ be $n\times n$ matrices.

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

Proof

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

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

Review: Complex Numbers

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

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

corollary: if $\lambda =a+b\,i$ is an eigenvalue of a real matrix $A$, then ${\bar {\lambda }}=a-b\,i$ is also an eigenvalue.

if $z=a+b\,i$ is an eigenvector of complex eigenvalue $\lambda _{1}$, then ${\bar {z}}=a-b\,i$ is an eigenvector of the conjugate eigenvalue ${\bar {\lambda }}_{1}$.

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

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

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

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

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

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

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

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

Example

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

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

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

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

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