Say we have some symmetric positive definite $n\times n$ matrix $M$ with $n$ distinct eigenvalues $\{\lambda_1,...,\lambda_n\}$. Is there a general formula for the maximum angle $\theta$ for which $M$ can rotate some vector, in terms of matrix invariants?

I worked out the $2\times 2$ case and the answer is $$\theta=\text{arccos}\Big(2\sqrt{\frac{\text{det}M}{(\text{tr}M)^2}}\Big)$$

In general, the answer is $$\theta=\text{arccos}\Bigg(\min_{v\neq0}\frac{v^TMv}{||v||\cdot||Mv||}\Bigg)$$

However I would like to find an answer analogous to the $2\times 2$ case for the general $n\times n$ case.

In trying to work out the $3\times 3$ case, the minimization procedure became extremely complicated. The best I could do was the case where two of the eigenvalues are equal $\lambda_1,\lambda_2,\lambda_2$. In that case, the angle turns out to be the same angle you would calculate for a $2\times 2$ matrix with the same eigenvalues $\lambda_1,\lambda_2$.

In fact, upon further analysis it should be true that for an $n\times n$ matrix with $k$ distinct eigenvalues, the formula for $\theta$ is equal to that which you would get from a $k\times k$ matrix with the corresponding eigenvalues.

Edit: Using the above fact, my guess for the $3\times 3$ case with $3$ distinct eigenvalues is $$\theta=\text{arccos}\Bigg(2\sqrt{\frac{6\text{det}M}{(\text{tr}M)^3-\text{tr}M^3}}\Bigg)$$

And my guess for the $4\times4$ case is $$\theta=\text{arccos}\Bigg(2\sqrt{\frac{48\text{det}M}{(\text{tr}M)^4-3(\text{tr}M^2)^2-4\text{tr}M\text{tr}M^3+6\text{tr}M^4}}\Bigg)$$

Technically there are $6$ possible solutions to the $3\times3$ case, however this is the only solution with rational coefficients on the traces. The $4\times4$ solution shown is also the only solution whose coefficients are rational.

In general, for an $n\times n$ matrix there are at most $\prod_{k=3}^{n}2\times p(k)$ possible solutions based on the statement just above the edit. Where $p(k)$ is the partition function.

These guesses are based on the assumption that you can write the solution as a ratio of linear combinations of power traces, for which each terms power sums to $n$. Based on Carlo Beenakker's answer this is an incorrect assumption.

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    $\begingroup$ You may wish to look into anti-eigenvalues and anti-eigenvectors. en.wikipedia.org/wiki/Antieigenvalue_theory $\endgroup$ – Benjamin Apr 22 at 12:49
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    $\begingroup$ @Benjamin I had no idea there was a whole theory based on this! $\endgroup$ – LucashWindowWasher Apr 22 at 16:23
  • $\begingroup$ Well, there is. And it's cool. And you should learn it. And then tell others. $\endgroup$ – Benjamin Apr 23 at 12:54
  • $\begingroup$ I used this to give intuition on the moment of inertia tensor in my class :) $\endgroup$ – LucashWindowWasher Apr 23 at 21:48

As explained in these notes, the maximum rotation angle $\theta$ of a symmetric positive definite matrix $M$ is related to the condition number $K=\mu_{\rm max}/\mu_{\rm min}$ of the matrix (the ratio of largest and smallest eigenvalue) by $$K=\frac{1+\sin\theta}{1-\sin\theta}\Leftrightarrow\cos\theta=\frac{2\sqrt{K}}{{1+K}}.$$ For $n=2$ this reduces to the first equation in the OP. The formulas for $\cos\theta$ in the OP for $n=3,4$ do not agree with the above.

I just noticed a similar answer at MSE.

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    $\begingroup$ Note that the OP's equation for the $n = 3$ case agrees with the equation from the notes in the case where $K$ has two identical eigenvalues. I don't think this is true of the OP's $n = 4$ case, though (in either the case where $K$ has two "double" eigenvalues or in the case where it has one "single" and one "triple" eigenvalue.) $\endgroup$ – Michael Seifert Apr 22 at 14:17
  • $\begingroup$ @MichaelSeifert Yes I am also noticing that my $4\times4$ case does not satisfy the condition I was trying to make hold. $\endgroup$ – LucashWindowWasher Apr 22 at 15:19
  • $\begingroup$ Do you know of a reference book/paper where this fact about the condition number is shown? $\endgroup$ – B Merlot Apr 25 at 17:55
  • $\begingroup$ no, but the MSE answer I linked to has a proof, you could just cite that if you would need to... $\endgroup$ – Carlo Beenakker Apr 25 at 20:11

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