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It's well known that for symmetric square matrices $A$, the inequality $\textrm{tr}{(A^2)}\leq (\textrm{tr}{A})^2$. I was wondering if there are any nice generalization for higher powers; for example is it true that $\sqrt[m]{\textrm{tr}{(A^m)}}\geq \sqrt[m+1]{\textrm{tr}{(A^{m+1})}}$?

What if the matrix is symmetric so that it has real eigenvalues? What happens when the matrix is nonnegative (so that the Perron Theorem states that the root with the largest magnitude is positive)?

What other nice trace inequalities regarding powers of matrices exist?

Thanks!

Anonymous
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    Why is $\textrm{tr}{(A^2)}\leq (\textrm{tr}{A})^2$ true? Any restrictions for $A$? – Nirvanacs Oct 29 '16 at 03:02
  • Your questions look as a topic for Richard Bellman, so you may try to search for answers in his books. For instance, in “Inequalities” by him and Edwin F. Beckenbach (Springer-Verlag, 1961) or “Introduction to matrix analysis” (McGraw-Hill, 1960). – Alex Ravsky Oct 30 '16 at 02:51

2 Answers2

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The originally stated inequality is in fact known to be false.

Consider:

$A = \begin{bmatrix} -2 & 0\\ 0 & 3 \end{bmatrix}$

$\mathrm{trace}\big(A\big) = 1$

$\mathrm{trace}\big(A^2\big) = 13$

$\mathrm{trace}\big(A^2\big) \gt \mathrm{trace}\big(A\big)^2$

If the matrices in question are Hermitian (or in reals, symmetric) and also positive semi definite matrices, then it is true. Proving that is a worthwhile exercise tied in with submultiplicativity of the Frobenius norm.

M.Mass
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MrD
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As pointed out in another answer, the stated inequality is false. A simple example is given by the diagonal matrix $A=\operatorname{diag}(1,-1)$, where $\operatorname{tr}(A^2)=2>\operatorname{tr}(A)^2=0$.

If $A$ is positive semidefinite, by diagonalising it, the inequality reduces to $$ \left(\sum\lambda_i^m\right)^{1/m}\ge\left(\sum\lambda_i^{m+1}\right)^{1/{m+1}}, $$ where $\lambda_1,\ldots,\lambda_n\ge0$ are the eigenvalues of $A$, which is indeed true.

Alternatively, if you use singular values instead, the following is true for every symmetric $A$: $$ \left(\sum\sigma_i^m\right)^{1/m}\ge\left(\sum\sigma_i^{m+1}\right)^{1/{m+1}}. $$ Both of these are special cases of the fact that $\|\cdot\|_p\ge\|\cdot\|_q$ when $0<p<q$. Proofs of this fact are abundant on this site. See, the following threads, for instances:

user1551
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