# Eigenvalues of Matrix for small $h$

Let $$Q$$ be a square matrix, $$h > 0$$ small and $$I$$ the identity matrix. I want to calculate the eigenvalues or more specifically the biggest eigenvalue of $$M:=(I+hQ)^T(I+hQ)$$.

Then $$M = (I+hQ)^T(I+hQ) = I +h(Q^T+Q)+h^2Q^TQ$$.

Therefore I think the eigenvalues of M should be of the form $$1+h\lambda+ \mathcal{O}(h^2)$$, where $$\lambda$$ is an eigenvalue of $$Q^T+Q$$, if $$h$$ is small enough. Is there a way to prove this or is this statement wrong?

If $$Q^T+Q$$ and $$Q^TQ$$ are simultaneously diagonalizable this should be true since I can then diagonalizable the matrices $$M$$, $$I$$, $$Q^T+Q$$ and $$Q^TQ$$ simultaneously. So the diagonal matrix $$D_M$$ of $$M$$ is equal to $$D_M = I + hD_{Q^T+Q} + h^2 D_{Q^TQ}$$ and I get my result.

Can I also show the result if the matrices aren't simultaneously diagonalizable?

• since $M$ is SPSD using the operator 2 norm and associated triangle inequality should do it. $\Big\Vert M\Big\Vert_2 = \Big\Vert I +h(Q^T+Q)+h^2Q^TQ \Big\Vert_2 \leq \Big\Vert I +h(Q^T+Q) \Big\Vert_2 + h^2\Big\Vert Q^TQ \Big\Vert_2$ – user8675309 Nov 13 '20 at 19:04
• Thank you! That solves my question. – Vango Nov 13 '20 at 19:14
• note that the above only gives you the dominant eigenvalue, Typically when I see big-O I infer that people are interested in the dominant eigenvalue though this may not always be the case. – user8675309 Nov 13 '20 at 19:28