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: 


*

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

*$A$ 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?
 A: 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:


*

*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.

*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.

A: 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:
$[\mathrm{tr}(B)]^2\leq n\mathrm{tr}(B^2)$.
A: $\renewcommand{\tr}{\operatorname{tr}}$There is a close result that addresses your inequality:

Let $A$ be a square matrix with real eigenvalues such that $\rho(A)=\rho(A^2)$ and $\tr(A^2)\ne0$. Then $$\rho(A)\ge\frac{(\tr(A))^2}{\tr(A^2)}$$
'$=$' holds iff the non-zero characteristic roots of $A$ are equal. $\quad[\rho(A)$ denotes the rank of $A]$

A short proof sketch:
Let $\lambda_1,\lambda_2,...,\lambda_r$ be the non-zero eigenvalues of $A$, so that $\rho(A)=r$.
By C/S inequality, $\left(\sum_{i=1}^r\lambda_i^2\right)\left(\sum_{i=1}^r1^2\right)\ge\left(\sum_{i=1}^r\lambda_i\right)^2$
$\qquad\qquad\qquad\qquad\Rightarrow r\tr(A^2)\ge(\tr A)^2,$
using the fact that if $\lambda$ is an eigenvalue of $A$, then $\lambda^2$ is an eigenvalue of $A^2$.
So if $A$ is a nonsingular matrix of order $n$, under the above conditions it follows that  $$(\tr(A))^2\le n\tr(A^2)$$

Reference: Linear Algebra - Rao/Bhimasankaram (2nd ed), Pg. 303
