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?

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    $\begingroup$ 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 $ $\endgroup$ – user8675309 Nov 13 '20 at 19:04
  • $\begingroup$ Thank you! That solves my question. $\endgroup$ – Vango Nov 13 '20 at 19:14
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    $\begingroup$ 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. $\endgroup$ – user8675309 Nov 13 '20 at 19:28

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