# Write down all eigenvalues of $T$ along with their algebraic and geometric multiplicities.

Let $V$ denote the vector space consisting of all polynomials over $\Bbb C$ of degree at most $2018$. Consider the linear operator $T : V → V$ given by $T(f) = f '$ , that is, T maps a polynomial f to its derivative $f'$ . Write down all eigenvalues of $T$ along with their algebraic and geometric multiplicities.

My Answer : first i take$f(x) = a_0 + a_1x + a_2x^2 + a_3x^3 +.....+a_{2018}x^{2018}$

$f' (x) = a_1+ 2a_2 x + 3a_3x^2 +.....+2018a_{2018}x^{2017}$

Now im converting them into matrix

$T=\begin{bmatrix} 0&1&0&0......&0 \\ 0&0&2&0......&0 \\ 0&0&0&3......&0 ....\\........\\......&.....&..&...&2018\\0..&0..&0...&.......&0\end{bmatrix}_{2018\times2018}$

Now im getting characteristic polynomial of $T =(\lambda-0)^{2018}$ where $\lambda$ is the eigenvalue with algebraic multiplicities$= 2018$ and geometric multiplicity $=1$

is my answer is correct or not ?? PLiz verified n rectified my answer

• It is correct, but how did you calculate the characteristic polynomial? I mean, did you expand a certain expand the characteristic determinant along a certain row/column? It would be nice to make it clear, since my logic was that the derivative matrix is nilpotent, so automatically has all its eigenvalues as $0$. – астон вілла олоф мэллбэрг Apr 28 '18 at 9:14
• It's probably the fact that the determinant of a triangular matrix is the product of the diagonal elements. – Antoine Giard Apr 28 '18 at 9:22
• thanks@AntoineGiard – user396850 Apr 28 '18 at 13:08

Your solution is more or less correct - note that the dimension of $V$ is 2019, not 2018. But it's a scary solution; if I'd assigned this the point would have been to force you to think about things in terms of the definitions, instead of just blindly calculating with a huge matrix:
Say $f$ is an eigenvalue: $f'=\lambda f$. It follows that $f(t)=ce^{\lambda t}$ for some constant $c$. (Proof: Let $g(t)=e^{-\lambda t}f(t)$. The product rule shows that $g'=0$, hence $g$ is constant.) Since $f$ is an eigenvector we must have $c\ne0$, and now $f$ is not a polynomial unless $\lambda=0$.
So $\lambda=0$ is the only eigenvalue.
Edit: There's a way to see this with less calculus. Inspired by a comment above: If $f$ is a polynomial of degree no larger than $2018$ then the $2019$-th derivative of $f$ must vanish. So $T^{2019}=0$; hence $\lambda^{2019}=0$, so $\lambda=0$.
Since the space of $f$ with $f'=0$ is one-dimensional, the geometric multiplicity is $1$; since there is only one (complex) eigenvalue the characteristic polynomial must be $\lambda^{2019}$.