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
thanks in advance
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 $\lambda^{2019}$.