MATH 308 Lecture 35

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Lecture Notes

Section 7.8

Exercise 1

What about multiplicity 3?

$X'={\begin{bmatrix}2&0&1\\0&2&1\\0&0&2\end{bmatrix}}\,X$

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

Eigenvectors: ${\vec {v}}=\left\langle \alpha ,\beta ,0\right\rangle$

We can choose any two vectors, e.g. ${\vec {v}}_{1}=\left\langle 1,1,0\right\rangle$ and ${\vec {v}}_{2}=\left\langle 3,2,0\right\rangle$, but when we add a third, the vectors become linearly dependent.

Thus two solutions are

1. $X_{1}=t\,\mathrm {e} ^{2t}\,{\begin{bmatrix}1\\1\\0\end{bmatrix}}$
2. $X_{2}=t\,\mathrm {e} ^{2t}\,{\begin{bmatrix}3\\2\\0\end{bmatrix}}$

Here's the trick:

Find $X_{3}=t\,\mathrm {e} ^{2t}\,w+\mathrm {e} ^{2t}\,{\vec {v}}_{3}$, where $w$ is related to ${\vec {v}}_{1}$ and ${\vec {v}}_{2}$.

${\begin{aligned}X_{3}&=t\,\mathrm {e} ^{2t}\,w+\mathrm {e} ^{2t}\,{\vec {v}}_{3}\\X_{3}'&=\mathrm {e} ^{2t}\,w+2t\,\mathrm {e} ^{2t}\,w+2\mathrm {e} ^{2t}\,{\vec {v}}_{3}\\X_{3}'&=A\,X_{3}\\\mathrm {e} ^{2t}\,w+2t\,\mathrm {e} ^{2t}\,w+2\mathrm {e} ^{2t}\,{\vec {v}}_{3}&=A\,t\,\mathrm {e} ^{2t}\,w+A\,\mathrm {e} ^{2t}\,{\vec {v_{3}}}\\w+2t\,w+2{\vec {v}}_{3}&=t\,A\,W+A\,{\vec {v}}_{3}\\\end{aligned}}$

Solving for each term in the polynomial gives:

${\begin{cases}w+2{\vec {v}}_{3}=A\,{\vec {v}}_{3}\\2w=A\,w\end{cases}}$

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

$w={\begin{bmatrix}1\\1\\0\end{bmatrix}}=(A-2I){\vec {v}}_{3}$

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

${\vec {v}}_{3}=\left\langle \alpha ,\beta ,1\right\rangle$

Exercise 2

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

$X'={\begin{bmatrix}2&1&1\\0&2&1\\0&0&2\end{bmatrix}}\,X$

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

Eigenvectors: ${\vec {v}}=\left\langle \alpha ,0,0\right\rangle$

So one solution so far: $X_{1}=\mathrm {e} ^{2t}\,{\begin{bmatrix}1\\0\\0\end{bmatrix}}$

Set $X_{2}=t\,\mathrm {e} ^{2t}\,{\begin{bmatrix}1\\0\\0\end{bmatrix}}+\mathrm {e} ^{2t}{\vec {v}}_{2}$

This ultimately brings us to $(A-2I){\vec {v}}_{2}=\left\langle 1,0,0\right\rangle$

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

So ${\vec {v}}_{2}=\left\langle \alpha ,1,0\right\rangle$. Let's choose ${\vec {v}}_{2}=\left\langle 1,1,0\right\rangle$

Now to find ${\vec {v}}_{3}$...

Do the trick again:

$X_{3}=t^{2}\,\mathrm {e} ^{2t}\,{\vec {v}}_{1}+t\,\mathrm {e} ^{2t}\,{\vec {v}}_{2}+\mathrm {e} ^{2t}\,{\vec {v}}_{3}$

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

$X_{3}={\frac {t^{2}}{2}}\,\mathrm {e} ^{2t}\,{\vec {v}}_{1}+t\,\mathrm {e} ^{2t}\,{\vec {v}}_{2}+\mathrm {e} ^{2t}\,{\vec {v}}_{3}$

This brings us to the condition: $(A-2I){\vec {v}}_{3}={\vec {v}}_{2}$

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

${\vec {v}}_{3}=\left\langle \alpha ,0,1\right\rangle$