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 $A$?

Thanks in advance


Your answer to the first part is fine. For the second part, it might be easier, or at least somewhat less confusing, to work directly with the definitions of eigenvector and eigenvalue.

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.