Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is there any general result characterizing real matrices $A$ such that $$[\mathrm{tr}(A)]^2\leq n\mathrm{tr}(A^2)?$$

I can see that the inequality holds if:

  1. all eigenvalues of $A$ are real (by the Cauchy-Schwarz inequality) or

  2. $A$ is a nonnegative matrix. To see this write $$n\mathrm{tr}(A^2)=n\sum_{i=1}^{n}(A_{ii})^{2}+n\sum_{i,j=1,i\neq j}^{n}A_{ij}A_{ji},$$ and note that, by the sum of squares inequality, $$n\sum_{i=1}^{n}(A_{ii})^{2}\geq\left( \sum_{i=1}^{n}A_{ii}\right)^{2}=\left[\mathrm{tr}(A)\right]^{2}.$$ If $A$ is nonnegative
    $$n\sum_{i,j=1,i\neq j}^{n}A_{ij}A_{ji}\geq 0,$$ and therefore the inequality holds.

But what about matrices not satisfying 1. or 2.? Are there more general conditions (or other specific ones) under which the inequality above holds?

share|improve this question
For $2 \times 2$ real matrices the precise condition is $$(A_{11} - A_{12})^2 + 4 A_{12} A_{21} \ge 0$$ –  Robert Israel Feb 4 '13 at 18:55
It is not the result you are looking for, but all real matrices satisfy $|\text{tr } A|^2 \leq n \text{tr } (A^TA)$. This is just Cauchy Schwartz using the inner product $\langle A, B \rangle = \text{tr } (A^TB)$ applied to $I,A$. –  copper.hat Feb 4 '13 at 19:19

2 Answers 2

The inequality is not true in general for a real diagonalizable $n\times n$ matrix $A=SDS^{-1}$ with complex eigenvalues, where $D$ is the diagonal matrix containing the eigenvalues of $A$; that is $D_{i,i}=e_{i}$, $i=1...n$.

The inequality:

$[\mathrm{tr}(A)]^2\leq n\mathrm{tr}(A^2)$


$[\mathrm{tr}(SDS^{-1})]^2\leq n\mathrm{tr}(SDS^{-1}SDS^{-1})$

and by the cyclic property of the trace:

$[\mathrm{tr}(D)]^2\leq n\mathrm{tr}(D^2)$,

because $D$ is diagonal this is equivalent to:

$\left(\sum_{i=1}^n e_{i}\right)^2\leq n\left(\sum_{i=1}^n e_{i}^2\right)$.

As $A$ and $A^2$ are real their traces are real and thus:

$\mathrm{tr}(D)=\sum_{i=1}^n e_{i}=\sum_{i=1}^n \Re(e_{i})$

$\mathrm{tr}(D^2)=\sum_{i=1}^n e_{i}^2=\sum_{i=1}^n \Re(e_{i}^2)=\sum_{i=1}^n \left(\Re(e_i)^2-\Im(e_i)^2\right)$

where we used:


The inequality then becomes:

$\left(\sum_{i=1}^n \Re(e_{i})\right)^2\leq n\sum_{i=1}^n \Re(e_i)^2-n\sum_{i=1}^n \Im(e_i)^2$

and as:

$0\leq\left(\sum_{i=1}^n \Re(e_{i})\right)^2$

the inequality fails to hold if (..but not iff):

$\sum_{i=1}^n \Re(e_i)^2< \sum_{i=1}^n \Im(e_i)^2$.

To demonstrate failure, assume the real diagonalizable matrix $A$ has complex eigenvalues such that:

$0< \sum_{i=1}^n \Im(e_i)^2$

then the real diagonalizable matrix:


has pure imaginary eigenvalues because:


For an example of a matrix built in such a way take:

$A=\left( \begin{array}{cc} 1 & 2 \\ -3 & 4 \\ \end{array} \right)$,

from which we get:

$B=\left( \begin{array}{cc} -3/2 & 2 \\ -3 & 3/2 \\ \end{array} \right)$, $e_{i}=\pm i/2\sqrt{15}$

$B^2=\left( \begin{array}{cc} -15/4 & 0 \\ 0 & -15/4 \\ \end{array} \right)$,



and thus we do not have:

$[\mathrm{tr}(B)]^2\leq n\mathrm{tr}(B^2)$.

share|improve this answer

The inequality in question can be rewritten as $$\renewcommand{\tr}{\operatorname{tr}}\tr(X^2)\ge0,$$ where $X=A-\frac{\tr(A)}{n}I$ is the traceless part of $A$. With this alternative formulation, I don't expect any nice characterisation of the feasible $A$s, but we immediately see that it is easier to work with this formulation:

  1. When all eigenvalues of $A$ are real, all eigenvalues of $X^2$ are nonnegative. Hence $\tr(X^2)\ge0$. Cauchy-Schwarz inequality is not needed.
  2. When $A$ has nonnegative off-diagonal entries, write $X=D+F$, where $F$ is the off-diagonal part of $X$ or $A$. Then $DF$ is a matrix with a zero diagonal and both $D^2$ and $F^2$ are nonnegative matrices. Hence $\tr(X^2)=\tr(D^2)+\tr(F^2)\ge0$. No tedious summation is needed here and we can even obtain a weaker sufficient condition than yours.
share|improve this answer
Nice spot with the$X$... –  Graham Hesketh Mar 25 at 20:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.