Putnam 2015/A6

Let $n$ be a positive integer. Suppose that $A,B,$ and $M$ are $n\times n$ matrices with real entries such that $AM=MB,$ and such that $A$ and $B$ have the same characteristic polynomial. Prove that $\det(A-MX)=\det(B-XM)$ for every $n\times n$ matrix $X$ with real entries.

Solution

Since the desired identity is continuous in the entries of $X$ and invertible matrices are dense, it suffices to prove the identity for invertible $X$. Since $\det (A - tI) = \det (B - tI)$ for all $t \in \mathbb{R}$, and $\det(A - tI)$ is a polynomial with finitely many roots, there exists a sequence $t_n \to 0$ such that defining $A_n = A - t_n I$ and $B_n$ likewise, we have $\det A_n = \det B_n \neq 0$ for all $n$, and $\lim \limits_{n\to \infty} A_n = A, \lim \limits_{n \to \infty } B_n = B$. Note also that we have $AM - t_nM = MB - t_nM$ so $A_n M = MB_n$.

It thus suffices to show that for all $n\in\mathbb{N}$ we have $\det (A_n - MX) = \det(B_n - XM)$. But
\[\det(A_n - MX) = \det (A_n - A_nMB_n^{-1} X) = \det A_n \det (I - MB_n^{-1}X) = \det A_n \det(X^{-1}B_n - M) \det X \det B_n^{-1},\]
and using that $\det A_n = \det B_n$, this is equal to
\[ \det X \det(X^{-1}B_n - M) = \det (B_n - XM).\]

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