Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question

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 $A-B$.

share|cite|improve this answer
What if the common eigenvalue is NOT zero? – Jeroen Dec 19 '12 at 14:02
@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 $$ (A-B)v=Av-Bv=\lambda v-\lambda v=0. $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.