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this comes from a comment of this

If and are × and × matrices, and have the same nonzero eigenvalues: is an eigenvector of iff is an eigenvector of , with the same nonzero eigenvalue.

I try to proof this

if is an eigenvector of => is an eigenvector of

$\begin{align} STu &= \lambda u \tag{1} \end{align}$

left multiply by T, then $\begin{align} TSTu &= \lambda Tu \tag{2} \end{align}$

but I'm blocked (2)=>(1) if $TSTu=\lambda Tu$, I can't multiply left with $T^-1$ since T is not square matrix, even T is square matrix, it still may not be invertible.

and also why OP in that thread says have the same nonzero eigenvalues, from (1)=>(2) it seemed the zero eigenvalues are also corresponding?

femto
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2 Answers2

1

$\lambda \ne 0$ is needed to show that $Tu \ne 0$ (remember, an eigenvector is a nonzero vector such that ...).

It's not true that if $Tu$ is an eigenvector of $TS$ then $u$ is an eigenvector of $ST$. For example, consider $$ S = \pmatrix{1 & 1 & 0\cr 1 & 1 & 0\cr},\ T = \pmatrix{1 & 1\cr 1 & 1\cr 0 & 0\cr},\ ST = \pmatrix{2 & 2\cr 2 & 2\cr},\ TS = \pmatrix{2 & 2 & 0\cr 2 & 2 & 0\cr 0 & 0 & 0\cr}$$

Try $u= \pmatrix{1\cr 0\cr}$. Then $Tu = \pmatrix{1\cr 1\cr 0\cr}$ is an eigenvector of $TS$ for eigenvalue $4$, but $u$ is not an eigenvector of $ST$.

Robert Israel
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  • If S and T are both square, would is an eigenvector of iff is an eigenvector of holds? I can only prove when T is invertible, just left multiply by $T^{-1} to (2) and get (1), but blocked when T is not invertible – femto Mar 19 '24 at 08:50
  • If $T$ is a square matrix and not invertible, it has a nontrivial null space. Take any $w \ne 0$ for which $Tw = 0$ and $u$ an eigenvector of $ST$ for $\lambda \ne 0$. Then $T(u+w) = Tu$ is an eigenvector of $TS$, but $u+w$ is not an eigenvector of $ST$. – Robert Israel Mar 19 '24 at 16:07
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Your argument works. If $\lambda\ne0$ and $STu=\lambda u$ for nonzero $u$, then $Tu\ne0$ and $TS(Tu)=\lambda Tu$. This shows that any nonzero eigenvalue of $ST$ is an eigenvalue of $TS$. As the roles can be exchanged, you get that $ST$ and $TS$ have the same nonzero eigenvalues.

The same idea shows that the geometric multiplicity is preserved for nonzero $\lambda$. Indeed, if $STu=\lambda u$ and $STv=\lambda v$ for linearly independent $u,v$, if $aTu+bTv=0$ then $$ au+bv=\frac1\lambda\,(aSTu+bSTv)=\frac1\lambda\,S (aTu+bTv)=0 $$ and so $a=b=0$. As for the algebraic multiplicity, the same ideas allow you to show that the generalized eigenspaces have the same dimension, so the algebraic multiplicity also agrees for nonzero $\lambda$. For an example of this, if $(ST-\lambda I)^2u=0$, then $$STSTu-2\lambda STu+\lambda^2u=0.$$ You cannot have $Tu=0$ for you would have $u=0$. Multiplying by $T$ on the left as before, $(TS-\lambda I)^2Tu=0$, and also as above one can show that the dimensions of the generalized eigenspaces agree.

When $\lambda=0$ the above argument fails because you may have $Tu=0$. But if $S,T$ are square then if $0$ is an eigenvalue of $ST$ it means that $ST$ is not invertible; this forces at least one of $S,T$ to be not invertible (use set) and so $TS$ cannot be invertible and $0$ is an eigenvalue of $TS$.

Things are slightly different if we allow non-square matrices. Take $$ S=\begin{bmatrix}1\\1\end{bmatrix},\qquad\qquad T=\begin{bmatrix}1&1\end{bmatrix}. $$ Then $$ST=\begin{bmatrix}1&1\\1&1\end{bmatrix}$$ with eigenvalues $0,2$ while $TS=2$ with eigenvalue $2$.

Martin Argerami
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  • From your first paragraph, for the same $\lambda$, is multiplicity same? $STu_1 = \lambda u_1 $ & $STu_2 = \lambda u_2$ =>$TSTu_1 = \lambda T u_1 $ & $TSTu_2 = \lambda Tu_2$ (*),the geometric multiplicity may collapse, since $Tu_1$ and $Tu_2$ can become linear dependent vector, but what happen to algebraic multiplicity, do they stay the same? I can't proof it since (*) still doesn't mean $\lambda$ appeared twice as root in characteristic polynominal of TS. – femto Mar 19 '24 at 07:58
  • Also a natural conclusion is if n>m, then ST must have (n-m) more (with algebraic multiplicity) 0 eigenvalues than TS,right? since ST has n eigenvalues, and TS has m eigenvalues and they must correspond, then the leftover (n-m) must be 0 – femto Mar 19 '24 at 08:04
  • The algebraic and geometric multiplicities are the same, I have edited the answer. Your conclusion about the algebraic multiplicity of $0$ is correct. – Martin Argerami Mar 19 '24 at 13:46
  • say n>m, if excluding extra (n-m) 0 eigenvalues, based on your result, (since the nonzero algebraic multiplicity is same, say they occupy total r algebraic multiplicity),the left over 0 eigenvalues' algebraic multiplicity (which is m-r) must be same, but does the left over 0's eigenvalues' geometric multiplicity preserved or not? – femto Mar 20 '24 at 07:08
  • No, it's not preserved. $$S=\begin{bmatrix}1&0\0&0\end{bmatrix},\qquad\qquad T= \begin{bmatrix}0&1\0&0\end{bmatrix}.$$ Then $$ST= \begin{bmatrix}0&1\0&0\end{bmatrix},\qquad\qquad TS=\begin{bmatrix}0&0\0&0\end{bmatrix}.$$ So $0$ has geometric multiplicity $1$ in $ST$ and $2$ in $TS$. – Martin Argerami Mar 20 '24 at 11:11