# Proof verification: Composition of linear transformation preserves eigenvectors.

If $$v$$ is an eigenvector of $$φ$$ and at the same time also an eigenvector of $$ψ : W \rightarrow U$$, then $$v$$ is also an eigenvector of $$ψ\circφ$$.

This is how I approached the problem:

Let $$a,b \in \mathbb{R}$$ be the eigenvalues to the eigenvector $$v$$ for the matrices $$A$$ and $$B$$ associated with the maps $$φ$$ and $$ψ$$, respectively: $$Av=av$$ $$Bv=bv$$ Now we will do some matrix algebra: $$BAv=Bav=aBv=abv=abv$$

So the eigenvector $$v$$ has an associated eigenvalue $$a\cdot b$$.

If there is anything wrong in my proof, I'd very much like to know where I may have done a wrong step.

• I don't understand the specification of $\psi$. Not only do the spaces $W,U$ not occur antwhere else, the notion of an eigenvector only makes sense for a linear map from a space to itself. Sep 10, 2022 at 22:04
• $aBv = ab$ is false, I believe. Sep 11, 2022 at 2:49
• @millsmess I am pretty sure they meant $abv$ and $ab$ was just a typo. Sep 11, 2022 at 4:42
• @MarcvanLeeuwen I think it is enough if $W$ and $U$ have the same dimension, say $n$. Once we fix bases for $W$ and $U$, the linear transformation $\psi$ will be an $n\times n$ matrix and it will make sense to talk about its eigenvalues and eigenvectors. Similarly, $\phi$ also has to be a map between two vector spaces of the same dimension $n$.
– Tony
Sep 11, 2022 at 19:21
• @Tony No. Sure, if the dimensions are the same you can find square matrices to represent the map (for each choice of bases), and each of these matrices may have eigenvalues, but these eigenvalues are not all the same. It may even happen that some of these matrices are diagonalisable and others are not, maybe not having any eigenvalues at all. There is really no sense one can given to eigenvectors of a linear map from a space $W$ to an unrelated (except by dimension) space $U$. Sep 12, 2022 at 11:44