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If $X \in \mathbb{C}^{m \times n}$ and $Y \in \mathbb{C}^{n \times m}$ ($m \geq n$), how to prove the following?

$$\sigma (XY) = \sigma(YX) \cup \underbrace{\left \{ 0, ..., 0 \right \}}_{m-n}$$

Here, $\sigma$ denotes the set of eigenvalues/spectrum.

PinkyWay
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user79230
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  • What's $\lambda$ mean? – dezign May 02 '13 at 01:54
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    Presumably (in view of the title) $\lambda$ is supposed to mean "the multi-set of eigenvalues of". But this is not standard notation, so it should have been defined in the question. – Andreas Blass May 02 '13 at 01:56
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    See also http://math.stackexchange.com/questions/124888/are-the-eigenvalues-of-ab-equal-to-the-eigenvalues-of-ba-citation-needed?rq=1 (square case). – Jonas Meyer May 02 '13 at 02:06
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    @JonasMeyer Given ${0,\ldots,0}$, I believe it is about multiplicities. – Julien May 02 '13 at 02:14
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    @julien: Good point. See also http://math.stackexchange.com/questions/332674/do-the-non-zero-eigenvalues-of-ab-and-ba-have-the-same-algebraic-multiplicity-f. – Jonas Meyer May 02 '13 at 02:16
  • not ideal for me to post this comment here, but there's no other SU comment for me to do so. So, re the SU q you asked http://chat.stackexchange.com/rooms/29420/users-it-windows-7-messing-up – barlop Sep 22 '15 at 22:56

3 Answers3

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It is very easy to check that $XY$ and $YX$ have the same nonzero eigenvalues. Just apply $Y$ to the identity $XYv=\lambda v$ and note $Yv\neq 0$ if $\lambda v\neq 0$.

But if you mean $\lambda$ to be the set of eigenvalues counted with multiplicities, then what you are asking is equivalent to $$ \chi_{XY}(t)=t^{m-n}\chi_{YX}(t) $$ where $\chi_A(t)=\det(tI-A)$ is the charateristic polynomial.

Big hint: $$ \left(\matrix{I&X\\Y&tI}\right)\left(\matrix{tI&-X\\0&I}\right)=\left(\matrix{tI&0\\*&tI-YX}\right) $$ and $$ \left(\matrix{I&X\\Y&tI}\right)\left(\matrix{tI&0\\-Y&I}\right)=\left(\matrix{tI-XY&*\\0&tI}\right). $$

Julien
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  • @Sam Did you manage to deduce the formula for the characteristic polynomials from the hint? – Julien May 02 '13 at 05:22
  • The proof of Sylvester's determinant theorem (mentioned by Alex) can be modified to show that: $det(\lambda I_m - AB) = \lambda^{m-n} det(\lambda I_n - BA)$. The proof is based on decomposing a matrix M into LU and UL factorizations. Thanks to you and Alex for the hints. – user79230 May 02 '13 at 05:39
  • @Sam You're welcome. This is an algberaic trick one can not really invent. You might be interested in knowing that for square matrices, the formula can be proved easily when, say, $X$ is invertible (just by conjugation). And then extended to every matrix by density of the invertible matrices. – Julien May 02 '13 at 05:43
  • You are right Julien, the proof is very easy for square matrices as you mentioned. – user79230 May 02 '13 at 05:45
  • @Sam If you observe carefully, the proof of Sylvester to which Alex linked is precisely what I had written above. See the upper and lower triangular matrices. – Julien May 02 '13 at 05:46
  • Will the Sylvester's determinant theorem applies for any field?@Julien – cmi Nov 19 '18 at 06:47
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You could modify the proof of Sylvester's determinant theorem to show that $$\text{det} \left( \lambda I_m - XY \right) = \text{det} \left( \lambda I_n - YX \right)$$ for all $\lambda \neq 0$. This shows equivalence for all nonzero eigenvalues. For the zero eigenvalues, an application of the fundamental theorem of algebra is sufficient. Note that the characteristic polynomial of $XY$ (or $YX$) must have $m$ (or $n$) roots. Since we have examined all roots $\lambda\neq0$, the remaining roots must be zero.

Edit: This proof does not work, see the comments.

Alex L
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    The identity you wrote is false in the non-square case. – Julien May 02 '13 at 03:55
  • @julien is right; upon closer inspection, this identity does not work for $m\neq n$. You should pursue his hint instead. – Alex L May 02 '13 at 05:16
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    The proof of Sylvester's determinant theorem can be modified to show that: $det(\lambda I_m - AB) = \lambda^{m-n} det(\lambda I_n - BA)$. Thanks to Alex for this hint. – user79230 May 02 '13 at 05:35
  • Will the Sylvester's determinant theorem applies for any field?@AlexL – cmi Nov 19 '18 at 06:46
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Let $\lambda\in\mathbb C-\{0\}$. Partition matrix $M$ as follows: $$M=\left(\begin{array}\\I_m&A\\\lambda^{-1}B&I_n\end{array}\right).$$ Using determinant factorization property of Schur complements, we have $$\begin{split} |M|&=|I_m||M/I_m|\\ &=|I_n-\lambda^{-1}BI_m^{-1}A|\\ &=(-\lambda^{-1})^n|BA-\lambda I_n|\\ &=(-\lambda)^{-n}|BA-\lambda I_n|.\end{split}$$ Using the same factorization, we have $$\begin{split} |M|&=|I_n||M/I_n|\\ &=|I_m-AI_n^{-1}\lambda^{-1}B|\\ &=(-\lambda^{-1})^m|AB-\lambda I_m|\\ &=(-\lambda)^{-m}|AB-\lambda I_m|.\end{split}$$ As a result, we have $(-\lambda)^{-m}|AB-\lambda I_m|=(-\lambda)^{-n}|BA-\lambda I_n|$, which in turn implies $|AB-\lambda I_m|=(-\lambda)^{m-n}|BA-\lambda I_n|$. The last equation is called Weinstein–Aronszajn identity.

From Weinstein–Aronszajn identity, it is clear that $AB$ and $BA$ has the same set of non-zero eigenvalues. The remaining eigenvalues, if any, must be zeroes because otherwise it would already have been considered previously.

zzzhhh
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