MATH 308 Lecture 35

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Lecture Notes

Section 7.8

Exercise 1

What about multiplicity 3?

$X^{\prime }=\left[\begin{array}{ccc}2& 0& 1\\ 0& 2& 1\\ 0& 0& 2\end{array}\right]X$

Eigenvalues: $\lambda \in \{2,2,2\}$

Eigenvectors: $\overrightarrow{v}=⟨\alpha ,\beta ,0⟩$

We can choose any two vectors, e.g. ${\overrightarrow{v}}_1=⟨1,1,0⟩$ and ${\overrightarrow{v}}_2=⟨3,2,0⟩$, but when we add a third, the vectors become linearly dependent.

Thus two solutions are

1. $X_1=t{\mathrm{e}}^{2t}\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]$
2. $X_2=t{\mathrm{e}}^{2t}\left[\begin{array}{c}3\\ 2\\ 0\end{array}\right]$

Here's the trick:

Find $X_3=t{\mathrm{e}}^{2t}w+{\mathrm{e}}^{2t}{\overrightarrow{v}}_3$, where $w$ is related to ${\overrightarrow{v}}_1$ and ${\overrightarrow{v}}_2$.

$\begin{array}{rl}{X}_3& =t{\mathrm{e}}^{2t}w+{\mathrm{e}}^{2t}{\overrightarrow{v}}_3\\ {X_3}^{\prime }& ={\mathrm{e}}^{2t}w+2t{\mathrm{e}}^{2t}w+2{\mathrm{e}}^{2t}{\overrightarrow{v}}_3\\ {X_3}^{\prime }& =A{X}_3\\ {\mathrm{e}}^{2t}w+2t{\mathrm{e}}^{2t}w+2{\mathrm{e}}^{2t}{\overrightarrow{v}}_3& =At{\mathrm{e}}^{2t}w+A{\mathrm{e}}^{2t}\overrightarrow{v_3}\\ w+2tw+2{\overrightarrow{v}}_3& =tAW+A{\overrightarrow{v}}_3\\ \end{array}$

Solving for each term in the polynomial gives:

$\{\begin{array}{l}w+2{\overrightarrow{v}}_3=A{\overrightarrow{v}}_3\\ 2w=Aw\end{array}$

From the second equation (coefficient of $t$), we find $w$ is an eigenvector: $(A-2I)w=0$, so $w=⟨\alpha ,\beta ,0⟩$, and substituting this into the first equation (constant term) gives

$w=\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]=(A-2I){\overrightarrow{v}}_3$

so we have $\{\begin{array}{l}z=\alpha \\ z=\beta \\ 0=0\end{array}$

${\overrightarrow{v}}_3=⟨\alpha ,\beta ,1⟩$

Exercise 2

In the above exercise, we had two eigenvectors, but what if we have only one?

$X^{\prime }=\left[\begin{array}{ccc}2& 1& 1\\ 0& 2& 1\\ 0& 0& 2\end{array}\right]X$

Eigenvalues: $\lambda \in \{2,2,2\}$

Eigenvectors: $\overrightarrow{v}=⟨\alpha ,0,0⟩$

So one solution so far: $X_1={\mathrm{e}}^{2t}\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$

Set $X_2=t{\mathrm{e}}^{2t}\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]+{\mathrm{e}}^{2t}{\overrightarrow{v}}_2$

This ultimately brings us to $(A-2I){\overrightarrow{v}}_2=⟨1,0,0⟩$

$\{\begin{array}{l}y+z=1\\ z=0\\ 0=0\end{array}$

So ${\overrightarrow{v}}_2=⟨\alpha ,1,0⟩$. Let's choose ${\overrightarrow{v}}_2=⟨1,1,0⟩$

Now to find ${\overrightarrow{v}}_3$...

Do the trick again:

$X_3=t^2{\mathrm{e}}^{2t}{\overrightarrow{v}}_1+t{\mathrm{e}}^{2t}{\overrightarrow{v}}_2+{\mathrm{e}}^{2t}{\overrightarrow{v}}_3$

If we differentiate and set equal to $AX$, we're left with a $x^2$ term that doesn't cancel, so we divide that term by 2:

$X_3=\frac{t^2}{2}{\mathrm{e}}^{2t}{\overrightarrow{v}}_1+t{\mathrm{e}}^{2t}{\overrightarrow{v}}_2+{\mathrm{e}}^{2t}{\overrightarrow{v}}_3$

This brings us to the condition: $(A-2I){\overrightarrow{v}}_3={\overrightarrow{v}}_2$

$\{\begin{array}{l}y+z=1\\ z=1\\ 0=0\end{array}$

${\overrightarrow{v}}_3=⟨\alpha ,0,1⟩$