FInd the characteristic polynomial of a differential linear operator

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 $$x$$.

a) That looks correct

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

• Ahh I see. I think my confusion came from the fact I used the same variable name. Commented Nov 15, 2021 at 21:52

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.

• 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? Commented Nov 15, 2021 at 22:27
• How did you draw that conclusion? Nevertheless, $y_0$ is an eigenvector. Can you show that there are essentially no more? Commented Nov 15, 2021 at 22:59
• 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 $1$. Commented Nov 19, 2021 at 3:09