# matrix of a linear transformation for $\mathbb{P}^2$

Problem Part (a) statement: a. Given $T$ is a linear transformation from $P_2\to P_2$ as $T(f(X)) = f(x) + f'(x)$,

what is the matrix $A$ of the transformation $T$ in the bases $B,B$, where $B$ is the basis $(1,x,x^2)$?

Note: The reason I find this question confusing is that $f$ is not explicitly given...

My attempt at the solution: is $\begin{bmatrix} T(e_1) & T(e_2) & T(e_3) \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} =Ax$ or $\begin{bmatrix} T(1) & T(x) & T(x^2) \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$ or $\begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$

Problem Part (b). An eigenvector of the transformation $T$ is defined as a noinzero vector $w$ such that $T(w) = \lambda w$ for some $\lambda$, where $\lambda$ is an eigenvalue of $T$.

Explain why 1 is the only eigenvalue for $T$ and find all eigenvectors belonging to $T$!

I should connect my answer with eigenvectors and eigenvalues of $A$...

Attempt at (b):

Using part a above, $A = \begin{bmatrix} 1,1,0\\ 0,1,2\\ 0,0,1 \end{bmatrix} \Rightarrow A-\lambda = \begin{bmatrix} 1-\lambda & 1&0 \\ 0& 1-\lambda& 2 \\ 0 & 0 & 1 \end{bmatrix} \Rightarrow \lambda_1 = \lambda_2 =\lambda_3 = 1$.

I dont know why the corresponding eigenvectors are the columns of the matrix $\begin{bmatrix} 1 & -1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$. Are these the so called "generalized eigenvectors"? Do the eigenvalues/vectors give more information about the vector $A$?

We’re looking for scalars $\lambda$ and polynomials $p[X]$ of degree at most two such that $T(p[X])=p[X]+p'[X]=\lambda p[X]$ or, equivalently, $p'[X]=(\lambda-1)p[X]$. From this we see that the only possibility for an eigenvector of $T$ is a constant (0-degree) polynomial. Otherwise, $p'[X]$ is non-zero and is of a lesser degree than $p[X]$, so can’t be a scalar multiple of $p[X]$. If $p[X]=c$, then $p'[X]=0$, so $\lambda=1$. Thus, the only eigenvalue of $T$ is $1$, with all non-zero constant polynomials as its eigenvectors (by definition, $0$ is never and eigenvector).
To do this with matrices, start with the matrix $A$ that you computed in the first part. Just as above, we seek non-zero vectors $v$ and scalars $\lambda$ such that $Av=\lambda v$, or $(A-\lambda I)v=0$. For this to have non-trivial solutions, $A-\lambda I$ must be singular, so we find its determinant and set it to zero: $$|A-\lambda I|=\left|\matrix{1-\lambda&1&0\\0&1-\lambda&2\\0&0&1-\lambda}\right|=(1-\lambda)^3=0,$$ so $\lambda=1$. (Actually, we didn’t have to do this in this case. The main diagonal elements of an upper-triangular matrix are its eigenvalues, so we could’ve read them directly from $A$.)
To find the associated eigenvectors, we need to find non-trivial solutions to $(A-I)v=0$, but that’s the null space (kernel) of $A-I$. This matrix is $$A-I=\pmatrix{0&1&0\\0&0&2\\0&0&0}.$$ It’s already in row echelon form, so with no further work we can read from it that the kernel is spanned by $(1,0,0)^T$, which corresponds to the polynomial $1$.
The geometric multiplicity of the eigenvalue $1$ is one, while its algebraic multiplicity is three, so if you were being asked to find the Jordan normal form of $A$ you’d have to compute generalized eigenvectors for that eigenvalue, but for this problem, they don’t come into play.