Trace inequality for real matrices

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?

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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

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.
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Nice spot with the$X$... – Graham Hesketh Mar 25 '15 at 20:11

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:

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

implies:

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

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:

$(\Re(e_i)+i\Im(e_i))^2=\Re(e_i)^2-\Im(e_i)^2+i2\Re(e_i)\Im(e_i)$.

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:

$B=A-S^{-1}\Re(D)S$

has pure imaginary eigenvalues because:

$SBS^{-1}=SAS^{-1}-\Re(D)=D-\Re(D)=i\Im(D)$.

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:

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

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

$[\mathrm{tr}(B)]^2=0$,

$2\mathrm{tr}(B^2)=-15$,

and thus we do not have:

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

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