Say $\lambda$ is an eigenvalue of $\textsf{ST}$. There exists $𝑥 \ne \textbf{0}$ such that $\textsf{ST}x= \lambda x$. Multiply both sides by $\textsf T$:

$$\textsf{TST}x=\textsf{T} (\lambda x)$$ $$\textsf{TS}(\textsf{T}x)=\lambda(\textsf{T}x)$$

Thus $\textsf{T}x$ is an eigenvector for $\textsf{TS}$.

  • Why is there such a potent relation between the eigenvectors of $\textsf{ST}$ and $\textsf{TS}$?
  • How come one is just the transform of the other?

Looking for geometric/intuitive explanation. Hints welcome.

  • $\begingroup$ Assuming $T$ is invertible, you can easily show these matrices to be similar. $\endgroup$
    – Aphelli
    Commented Jul 5, 2019 at 11:32
  • 3
    $\begingroup$ This works only if $x\notin\ker T$. $\endgroup$
    – Berci
    Commented Jul 5, 2019 at 11:59

1 Answer 1


$\require{AMScd}$ Maybe more convoluted than necessary, but let me try.

We have the following: $$ S: W \to V\\ T: V \to W $$ where $V, W$ are vector spaces. Thus we have $ST: V \to W \to V$ and $TS: W \to V \to W$. A way to combine all these is to consider the following diagram: \begin{CD} V @>T>> W @>S>> V\\ @V T V V @VV S V @VV T V\\ W @>S>> V @>T>> W \end{CD} Let us however look at the two linear maps $ST:V\to V$ and $TS:W\to W$ then \begin{CD} V @>ST>> V\\ @V T V V @VV T V\\ W @>TS>> W \end{CD} So whatever ''happens'' upstairs ($ST$ map) happens ''downstairs'' ($TS$ map) as long as we consider there is a $T$ map between the two.

So more geometrically, $x$ being the eigenvectors of $ST$ are the only vectors in $V$ that undergo a scaling by $\lambda$ when you apply $ST$, but don't change direction (definition of eigenvector). Similarly if $TSy= \lambda y$, with $y \in W$. Now the previous relation between the vector spaces tells us that the relation between any vector in $V$ and $W$ is related by $T$ and in particular we have that $y = Tx$. So that while the direction of eigenvectors does not change ''upstair'' and ''downstairs'' they need to change when looking at them ''across'' spaces (trying to convey some intuition here, not sure it helps)

Now why would the eigenvector $x$ be mapped through $T$ to and eigenvector $y$. That can be shown by contradiction by assuming that $Tw=y$ but $w$ is not an eigenvector of $ST$.


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