
> 
> **Possible Duplicate:**  
> 
> [Eigenvalues of sum of anti-commuting matrices](https://mathoverflow.net/questions/41755/eigenvalues-of-sum-of-anti-commuting-matrices) 
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> 


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).



Eigenvalues of sum of two anti-commuting matrices
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 $\pm\sqrt{a^2+b^2}$.


