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Let $D$ denote the linear operator $D =\frac{d}{dt}$. For a real number $r$, let $V$ denote the vector space spanned by the list of functions

$y_k(t) = t^ke^{rt}$, $0\leq\ k < m$

Note that $V$ has dimension $m$. We may restrict $D$ to an operator on $V$.

a. $\{y_0, y_1, . . . , y_{m−1}\}$ is a basis for $V$ . Compute the matrix of $D$ with respect to this basis.

I believe I have done this correctly. I applied $D$ to the first few terms in the basis to see a pattern.

$D[y_0] = re^{rt} = ry_0 + 0y_1 +...+ 0y_{m-1}$

$D[y_1] = tre^{rt} +e^{rt} = 1y_0 + ry_1 +...+ 0y_{m-1}$

$D[y_2] = t^2re^{rt} +2te^{rt} = 0y_0 + 2y_1 + ry_3 + ...+ 0y_{m-1}$

$\vdots$

$D[y_{m-1}] = t^{m-1}re^{rt} +(m-1)t^{m-2}e^{rt} = 0y_0 + 0y_1 +...+ (m-1)y_{m-2} + ry_{m-1}$

From this I put together the matrix for D.

$D= \begin{bmatrix} r & 1 & 0 & ... & 0 & 0\\ 0 & r & 2 & ... & 0 & 0\\ 0 & 0 & r & ... & 0 & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & ... & r & m-1\\ 0 & 0 & 0 & ... & 0 & r \end{bmatrix}$

b. This is where my issue comes. This part asks to find the characteristic polynomial of $D$. The issue is that taking $|A-rI|$ of the matrix I computed for $D$ will create a column of all zeros, so the determinant (and the characteristic polynomial) is $0$. This is obviously not correct as part c) asks to show that $r$ is the only eigenvalue of $D$, which cannot be possible if the characteristic polynomial is $0$. What did I do wrong?

Edit: I see now I used mistakenly used $r$ to compute the characteristic polynomial when I should have used another variable like $x$.

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2 Answers 2

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a) That looks correct

b) The characteristic polynomial is $P(x)=|A-xI| = (r-x)^m$, not $|A-rI|$

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    $\begingroup$ Ahh I see. I think my confusion came from the fact I used the same variable name. $\endgroup$
    – jem do
    Commented Nov 15, 2021 at 21:52
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The characteristic polynomial should be a polynomial not a scalar.

Instead of $|A-rI|$ you should have written $|A-xI|$ where $x$ is the indeterminate of the polynomial.

So, you'd get $(r-x)^m$ which indeed has $x=r$ as only root.

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  • $\begingroup$ How would you find the eigenspace associated to r? I'm thinking that if you take $(A-rI)\textbf{u} = 0$ then $\textbf{u} = 0$ so the eigenspace has dimension 1? $\endgroup$
    – jem do
    Commented Nov 15, 2021 at 22:27
  • $\begingroup$ How did you draw that conclusion? Nevertheless, $y_0$ is an eigenvector. Can you show that there are essentially no more? $\endgroup$
    – Berci
    Commented Nov 15, 2021 at 22:59
  • $\begingroup$ The only vector scaled by $r$ when applying $D$ is $y_0$, all others will be multiplied by some power of $t$ as well. Then the eigenspace of $r$ is the span of $y_0$ so the geometric multiplicity is $1$. $\endgroup$
    – jem do
    Commented Nov 19, 2021 at 3:09

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