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 $|A-rI|$
