I have two matrices and I know the eigenvalues and eigenvectors of both, it also happens to be the case that they share the same set of eigenvalues. I now need to sum the two matrices and take the average, and finally, I need to compute the eigenvalues of the average matrix. Is there any shortcut I can take to perform the last step without digaonlization?


Unless one of the matrices happens to be scalar (i.e. only one eigenvalue, and eigenvectors form a basis), knowing the eigendecompositions of two matrices is hardly of any help.

Consider this example: let $A=\begin{pmatrix}1 & 0 \\ 0 & 2\end{pmatrix}$ with eigenvalues $1$ and $2$, and let $B$ be the same matrix, but in another basis, say rotated by $\varphi$: $B=\begin{pmatrix}1+\sin(\varphi)^2 & \sin(\varphi)\cos(\varphi) \\ \sin(\varphi)\cos(\varphi) & 1+\cos(\varphi)^2\end{pmatrix}$. You can check that $B$ has eigenvalues $1$ and $2$ corresponding to eigenvectors $\begin{pmatrix}\cos(\varphi) \\ -\sin(\varphi)\end{pmatrix}$ and $\begin{pmatrix}\sin(\varphi) \\ \cos(\varphi)\end{pmatrix}$.

Now, the characteristic polynomial of $\frac{A+B}{2}$ is $\lambda^2-3\lambda+2+\frac{\sin(\varphi)^2}{4}$, with discriminant $D=1-\sin(\varphi)^2$, so eigenvalues are $\frac{3\pm\sqrt{1-\sin(\varphi)^2}}{2}$, which can be anything between $1$ and $2$.

Note that for $\varphi=0$ we get $A=B$, while for $\varphi=\frac{\pi}{2}$ we get $B=\begin{pmatrix}2 & 0 \\ 0 & 1 \end{pmatrix}$ and $\frac{A+B}{2}=\frac{3}{2}I$.

In general, although there does exist rich theory concerning the structure of a single matrix/operator (everything about eigenvalues & eigenvectors), it is in general extremely difficult to get something similar for a pair of matrices (unless you know something else about them, e.g. that they commute). Problems that involve classification of pairs of matrices up to change of basis are even called wild.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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