# How to find eigenvalues for T without given a matrix

I found it a bit vague on how to find eigenvalues and vectors of $T$ if we do not have a matrix to represent it. Suppose if: $$T: V \rightarrow W, V = W$$ I thought perhaps that the eignevalues might just be from the $det([T]_{\alpha}^{\beta} - \lambda{}I) = 0$ where $\alpha$ is a basis for $V$ and $\beta$ is a basis for $W$. However, I have consistently got these answers wrong in my book and my book does not specify any obvious answers on how to approach problems such as these. In general, given known bases (or just selecting the standard ones, since the change of basis matrix are similar thus having the same characteristic polynomial), how can I find such eigenvalues and vectors?

An example would be, find the eigenvalues and vectors of $T: P_3 \rightarrow P_3$ given that $T(p) = xp' -4p$

• You could always find the answer by finding the matrix with respect to some fixed basis. For example, with $T:P_3 \to P_3$, we can find the eigenvalues of $T$ by looking at the matrix of $T$ with respect to the basis $\{1,x,x^2,x^3\}$. Note: it is important that we use the same basis for both the starting space and the target space. – Omnomnomnom Mar 18 '16 at 3:06
• Sometimes, however, it's easier to find/solve for the eigenvectors directly. For example, in your case we could note that $$xp' - 4p = \lambda p$$ is a differential equation on $p(x)$. We could find a solution $p$ in terms of $x$ and $\lambda$, and then select the $\lambda$ for which the solution might be a polynomial in $P_3$. – Omnomnomnom Mar 18 '16 at 3:09

For your example, you can find the matrix of the transformation with respect to a standard basis, such as $\alpha = \{1,x,x^2,x^3\}$. We then find that $$[T]_{\alpha}^{\alpha} = \pmatrix{ -4 & 0&0&0\\ 0& -3&0&0\\ 0&0&-2&0\\ 0&0&0&-1 }$$ You may notice that it is particularly easy to find the eigenvalues of this matrix.