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If $S$ and $T$ are both linear maps then $ST$ and $TS$ have the same eigenvalues. I understand the proof (which can be found at $ST$ and $TS$ have the same eigenvalues.). But why must this be the case intuitively/geometrically? Is there a geometric picture for this?

The statement says that $ST$ and $TS$ stretch the space by exactly the same amount but only in different directions(so just rotating ST by correct amount will give be TS?) So somehow the magnitude gets preserved but not direction.

I am thinking if this tells us that the reason matrix multiplication is not communicative is only due to directional changing.

--Clarification---

Assume T,S are linear operators that map $R^N$ to $R^N$. They are square and invertible.

Rahul Deora
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    Let $S=\pmatrix{4&0\cr0&9\cr}$, $T=\pmatrix{0&16\cr25&0\cr}$. Do ST and TS have the same eigenvectors? – Gerry Myerson Jul 01 '19 at 10:36
  • They would by the proof. Whats your point? – Rahul Deora Jul 01 '19 at 10:39
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    @RahulDeora What are you talking about? The eigenvectors of $ST$ in Gerry Myerson's example are $\pmatrix{\pm8\ 15}$ and the eigenvectors of $TS$ are $\pmatrix{\pm6\ 5}$. In general, $ST$ and $TS$ are only guaranteed to have the same set of nonzero eigenvalues. They can have different eigenvectors. – user1551 Jul 01 '19 at 11:18
  • Yes, Im sorry!, I meant eigenvalues in the first para not vectors. I have edited it. The rest is consistent with eigenvalues being the same – Rahul Deora Jul 01 '19 at 11:22
  • I always thought this phenomenon was algebraic rather than geometric, essentially coming down to the fact that $ST$ and $TS$ have the same characteristic polynomial. I would love to know if it has a geometric interpretation. – Joppy Jul 01 '19 at 13:28
  • @user1551 "only guaranteed to have the same set of nonzero eigenvalues" who told you that? The proof for non-zero eigenvalues is maybe more obvious, but it's a fact that $0$ is an eigenvalue of $ST$ if and only if it's an eigenvalue of $TS$. – David C. Ullrich Jul 01 '19 at 16:03
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    @DavidC.Ullrich By a "set" I meant a multiset (counting multiplicities of eigenvalues). $\pmatrix{1\ 0}\pmatrix{1&0}$ has a zero eigenvalue but $\pmatrix{1& 0}\pmatrix{1\ 0}$ has not any. The OP probably had endomorphisms in mind, but this wasn't stated explicitly. – user1551 Jul 01 '19 at 16:12
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    @user1551 Fine. For no good reason I assumed that $S$ and $T$ were square matrices... – David C. Ullrich Jul 01 '19 at 16:16
  • Lets assume they are square matrices. I am studying operators of the same space – Rahul Deora Jul 01 '19 at 17:00
  • I have also added a clarification in my post at the end which is important. – Rahul Deora Jul 01 '19 at 17:13

1 Answers1

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A partial answer in a special case: If we're talking about square matrices and one of $S$ or $T$ is invertible then $ST$ and $TS$ are similar.

The fact that similar matrices have the same eigenvalues is possibly somewhat "geometric"; the two matrices represent the same transformation in different coordinate systems.

I've been asked to elaborate on "similar matrices represent the same transformation in different coordinate systems". All the assertions below are elementary, and straightforward for many readers; others can find proofs in a book on linear algebra:

Say $C=(b_1,\dots,b_n)$ is an ordered basis for $V$. If $x\in V$ then $[x]_C$ is the coordinate vector for $x$ wrt $C$: $$[x]_C=(c_1,\dots,c_n)$$where $$x=\sum c_jb_j.$$

Now if $T:V\to V$ is linear then $[T]_C$ is the matrix representing $T$ wrt $C$, which by definition means $$[T]_C[x]_C=[Tx]_C\quad(x\in V).$$

True Fact: Suppose $A$ and $B$ are $n\times n$ matrices over the field $F$. Then $A$ is similar to $B$ if and only iif there exists a basis $C$ for $F^n$ with $A=[B]_C$.

David C. Ullrich
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  • Let's assume they are square matrices and both are invertible. Why do you suddenly say ST and TS are similar? Can you elaborate on this: "the two matrices represent the same transformation in different coordinate systems"? – Rahul Deora Jul 01 '19 at 17:04
  • I did try to think of an argument similar to that where I let the coordinate system be in terms of eigenvalues of T. Then Tv just scales the axis. And STv will stretch S's eigenvectors in T's basis. Similarly for TS, we let the space be in terms of S's eigenvectors and apply TS. I'm not sure how to take it forward. I have also added a clarification in my post at the end which is important. – Rahul Deora Jul 01 '19 at 17:13
  • @RahulDeora If just $S$ is invertible then $ST=S(TS)S^{-1}$. For your second question see Edit:. – David C. Ullrich Jul 01 '19 at 17:25
  • This is what I was looking for : =()−1. Could you elaborate on this, and how it will preserve the eigenvalues? – Rahul Deora Jul 02 '19 at 07:25
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    Think about it. – David C. Ullrich Jul 02 '19 at 13:56