Let $A\ \text{and}\ B$ be two $n\times n$ matrix. Where $a_1,a_2,...,a_n$ and $b_1,b_2,...,b_n$ be the eigenvalues of $A$ and $B$ respectively.
Now can we describe eigenvalues of AB in terms of eigenvalues of A and B?
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2No. See this question for instance. – user1551 Oct 12 '20 at 10:10
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I understood. Thank you for sharing. – Saikat Oct 12 '20 at 10:29
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In general no, because the matrices do not share their Eigenvectors. – Oct 12 '20 at 12:03
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No, in general you can say nothing. But there may be some specific circumstances where you can say something:
Case 1: all the eigenvalues of $B$ are nonzero.
In this case, $B$ is invertible. Let $D$ be a diagonal matrix with entries $r_1, \ldots, r_n$, and let $A = DB^{-1}$. Then $AB = D$, i.e., AB has the $r$s as eigenvalues, and those could be anything.
Case 2: $k$ of the eigenvalues of $B$ are zero.
In this case, at least $k$ of the eigenvalues of $AB$ are zero as well, for if $v$ is an eigenvector of $B$ for the eigenvalue $0$, i.e, if $Bv = 0$, then $$ (AB)v = A(Bv) = A0 = 0 $$ as well. The remaining $n-k$ eigenvalues can be prescribed just as in Case 1, although it takes a little bit more work to show this.
John Hughes
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