# Maximum rotation made by a symmetric positive definite matrix?

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.

• You may wish to look into anti-eigenvalues and anti-eigenvectors. en.wikipedia.org/wiki/Antieigenvalue_theory – Benjamin Apr 22 '19 at 12:49
• @Benjamin I had no idea there was a whole theory based on this! – fewfew4 Apr 22 '19 at 16:23
• Well, there is. And it's cool. And you should learn it. And then tell others. – Benjamin Apr 23 '19 at 12:54
• I used this to give intuition on the moment of inertia tensor in my class :) – fewfew4 Apr 23 '19 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.
• 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.) – Michael Seifert Apr 22 '19 at 14:17
• @MichaelSeifert Yes I am also noticing that my $4\times4$ case does not satisfy the condition I was trying to make hold. – fewfew4 Apr 22 '19 at 15:19