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I am looking for a proof that two matrix share the same eigenvalues, even if they have different size. Similar matrices share the same eigenvalues and the similarity relationship is defined for $A,B,C \in K^{n,n}$ where $\operatorname{rank} (C)=n$ as:

$AC = CB \Leftrightarrow A = C^-1BC$

Now, if the matrices have different sizes, $A \in K^{(n,n)}$, $B \in K^{(m,m)} $ and $C \in K^{(m,n)}$ where $n>=m$ and $\operatorname{rank} (A)=m$. It can also be stated that

$AC=CB$

And that all eigenvalues for $B$ are eigenvalues for $A$.

I just can't come up with a way to map them to a common diagonal matrix the same way you would do if $C$ is invertible. I have tried using the pseudo inverse of $C$ and applying the SVD decomposition but I am not able to prove any relationship between eigenvalues.

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As it stands, what you want to prove is false (e.g. when $C=0$). The followings are true, though:

  1. If $AC=CB$ and $C$ is a "tall" matrix with full column rank, then all eigenvalues of $B$ (counting multiplicities) are eigenvalues of $A$. You may use the rank of $C$ and the equality $(A-\lambda I)^kC=C(B-\lambda I)^k$ to prove that every generalised eigenpair $(\lambda, v)$ of $B$ gives rise to a generalised eigenpair $(\lambda, Cv)$ of $A$.
  2. If $AC=CB$ and $C$ is nonzero, then $A$ and $B$ shares at least one eigenvalue over the algebraic closure $\overline{K}$ of $K$. More specifically, from $AC=CB$, we get $A\pmatrix{C&0}=\pmatrix{C&0}\pmatrix{B\\ &D}$ for every $(n-m)\times(n-m)$ matrix $D$ over $\overline{K}$. Now the conclusion follows from the analogous result for square matrices and the infinite cardinality of $\overline{K}$ (which implies the infinitude of choices of the spectrum of $D$).
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  • $\begingroup$ Thanks! Both the idea of extending it to square matrices and the reference really helped. $\endgroup$ – ebabio Nov 9 '16 at 21:24

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