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Suppose $X\in \mathbb R^{m\times n}$ $(m>n)$ is a left-invertible matrix, and its left-inverse is $X^+=(X^\top X)^{-1}X^\top$ (so that $X^+X=I$).

Now I have a matrix $A\in \mathbb R^{n\times n}$. Define $B\triangleq XAX^+$.

By computing some examples, I observe that the eigenvalues of $B$ are the same as the ones of $A$ apart from additional ($m-n$) zero eigenvalues. For example, $A=\begin{bmatrix} 1 & -2 \\ 3 & 4 \end{bmatrix}$, $X=\begin{bmatrix} 1 & 0 \\ 1 & 1 \\ 0 & 1 \end{bmatrix}$. Then $eig(A)=\{2.5000 + 1.9365i, 2.5000 - 1.9365i\}$, and $eig(B)=\{2.5000 + 1.9365i, 2.5000 - 1.9365i, 0\}$

Is it always true?

fibon
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    This follows directly from the fact that for any pair of (possibly rectangular) matrices $X$ and $Y$ of appropriate sizes, $XY$ and $YX$ share the same multi-set of nonzero eigenvalues. See this answer for a proof. This result was first discovered by J.J. Sylvester in mid-1800's. – user1551 May 20 '22 at 02:52
  • Amazing. Thanks for pointing it out. – fibon May 20 '22 at 03:24

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