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Eigenvalues of sum of anti-commuting matrices

I know two anti-commuting (nxn)-matrices A and B, n -even. I know also that +-a and +-b are real eigenvalues among all eigenvalues of A and B respectively. How to show that the matrix A+B has also at least two real eigenvalues of the form +-\sqrt{a^2+b^2} ? (I know also that (A+B)(A+B)^t=(A+B)^t(A+B)=(a^2+b^2)I, where I is (nxn)-identity matrix and AA^t=A^tA=a^2I, BB^t=B^tB=b^2I and ^t denotes transposition).

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Please could you edit your original question, rather than post a second almost identical one. – Robin Chapman Oct 11 '10 at 8:04
Note: I deleted the duplicate, so the banner on the top is spurious. – S. Carnahan Jun 14 '12 at 11:25
up vote 1 down vote accepted

Suppose for simplicity's sak that $A$ and $B$ are diagonalizable over $\mathbb{R}$ and are non-singular.

Let $V_a$ be the $a$-eigenspace of $A$. Then by anti-commutativity, we find $BV_a\subseteq V_{-a}$ etc. As $A$ and $B^2$ commute then there is an eigenvector $v\in V_a$ with $B^2v=b^2 v$ for some $v$. If we let $w=bv+Bv$ then $Bw=bw$ so $b$ is real (assuming $B$ has real eigenvectors). On the space $W$ spanned by $v$ and $w=b^{-1}Bv$ the linear transformation $A+B$ has matrix $$\left(\begin{array}{rr} a&b\\\\ b&-a\\\\ \end{array}\right)$$ which has an eigenvectors with eigenvalues $\pm\sqrt{a^2+b^2}$.

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Thanks a lot! But I don't understand how this: AV_a ⊆ V_a follows from anti-commutativity and how you got this: w=b^{−1}Bv. I also don't understand why B^2v=b^2v holds for v\in V_a. Sorry and thank you again. – Anna Oct 11 '10 at 9:00
corrected ! – Robin Chapman Oct 11 '10 at 18:35

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