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When matrix $A$ and $B$ have a common eigenvalue, is it true that the matrix $A - B$ will have the eigenvalue $0$?

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No. The matrices $$ A=\left(\begin{array}{cc}-1 & 0 \\ 0 & 0\end{array}\right),\quad B=\left(\begin{array}{cc}0 & 0 \\ 0 & -1\end{array}\right) $$ have common eigenvalue $0$. Yet the difference $A-B$ has eigenvalues $\pm 1$. Zero is not an eigenvalue of $A-B$.

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What if the common eigenvalue is NOT zero? –  Jeroen Dec 19 '12 at 14:02
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@JeroenfromBelgium: Consider $A=\left(\begin{array}{cc}-1 & 0 \\ 0 & 2\end{array}\right)$, $B=\left(\begin{array}{cc}2 & 0 \\ 0 & -1\end{array}\right)$. Then $A$ and $B$ have common eigenvalues $2$ and $-1$ and $A-B=\left(\begin{array}{cc}-3 & 0 \\ 0 & 3\end{array}\right)$ still does not have $0$ as an eigenvalue. –  Eckhard Dec 19 '12 at 14:40

A sufficient condition for $A-B$ to admit the eigenvalue $0$ is that the common eigenvalue $\lambda$ has non trivially intersecting $\lambda$-eigenspaces. Indeed, if $0\neq v$ is a $\lambda$-eigenvector for both $A$ and $B$, then $$ (A-B)v=Av-Bv=\lambda v-\lambda v=0. $$

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