I have this problem which I have solved a little bit different from my text book, and hence my answer is not exactly the same. I dont understand why however.
So, the problem is:
"Given the linear map $T:C^{1}(R) \longrightarrow C(R)$ defined by $T(f) = (fg)'$ where g is a fixed function.
b) Let $g(t) = t$ and consider $T:\mathcal{P}_n \longrightarrow > \mathcal{P}_n$. Determine the matrix for $T$ in the $\{1,t,t^2,...,t^n\}$ basis.
c) Calculate all the eigenvalues and eigenvectors to T using (b)."
So, I solve part a without any problems. The matrix is simply $A = diag(1,2,...,n+1)$.
Maybe the next step is completly obvious. I'm a bit confused, however, because now when I want to find all the eigenvectors, I simply choose to solve the linear equation $(A-\lambda_iI)x_i = 0$ $\forall$ $\lambda_i=1,2,...,n+1$ and the corresponding $x_i \in \mathcal{P}_n$. Clearly, the matrix $A-\lambda_iI$ is going to have one row with all zeros with the different lambdas. So when solving this systems of equations one get that all the components of the vector $x_i$ equals zero except the component corresponding to the row of zeros. Specifically, we get the equation $x_{i_j} *0=0$ for the $x_{i_j}$ component of $x_i$. Clearly this equation holds for $x_{i_j} = a$ for some $a\in R$. Thus the eigenvector is e.g. $x_1 = (a,0,0,...,0)$ and $x_2 = (0,a,...,.)$ and so on. Explicitly, $x_i = at^{i-1}$ for some arbitrary $a \in R$, $i=1,2,...,n+1$.
Now, the book says the right answer is the same as mine except $a=1$. Clearly this works, but the questions was to find all the eigenvectors and therefore I'm a bit confused to why they specifically choose $a=1$. Am I wrong in my argument above?
Thanks!
You are right. The eigenvectors are the same up to scaling, so it's fine. If $v$ is an eigenvector trivially $av$ is also an eigenvector.
To find an eigenvector, consider the differential equation $(ft)' = \lambda f$. This has solution $$f = Ct^{\lambda -1 }$$ And we have the restriction that $\lambda \in \{1,\ldots, n+1\}$, which are also the eigenvalues.