Prove $\mathrm{tr}(A^2) \leq\mathrm{tr}(A^TA)$ 
Prove that $\mathrm{tr}(A^2) \leq \mathrm{tr}(A^TA)$.

I saw the below link before and think it is related to this question $A,B$ be Hermitian.Is this true that $tr[(AB)^2]\le tr(A^2B^2)$? but still can't solve it.
I also tried to use $\mathrm{tr}(A^TA)\geq0$.
 A: We begin by using the property $\mathrm{tr}(AB) = \mathrm{tr}(BA)$:
\begin{align} 
\mathrm{tr}(A^TA) - \mathrm{tr}(AA) &= \frac{1}{2}\mathrm{tr}(A^TA) - \mathrm{tr}(AA) + \frac{1}{2}\mathrm{tr}(AA^T) 
\end{align}
Then if we look at each individual entry that makes up the trace, we have:
\begin{align}
(A^TA)_{ii} &= \sum_j a^2_{ji}\\
(A^2)_{ii} &= \sum_j a_{ij} a_{ji}\\
(AA^T)_{ii} &= \sum_j a^2_{ij}
\end{align}
Thus we see that for all entries of the trace, we have:
\begin{align}
\frac{1}{2}(A^TA)_{ii} - (A^2)_{ii} + \frac{1}{2}(AA^T)_{ii} &= \frac{1}{2}\sum_j (a_{ij} - a_{ji} )^2 > 0
\end{align}
A: Presumably $A$ is a real square matrix. Let $H=\frac12(A+A^T)$ and $K=\frac12(A-A^T)$ be respectively the symmetric and skew-symmetric parts of $A$. Then
$$
A^TA-A^2=(A^T-A)A=K^T(H+K)=K^TH+K^TK.
$$
Thus, using the fact that symmetric matrices are orthogonal to skew-symmetric matrices with respect to the inner product $\langle X,Y\rangle=\operatorname{tr}(Y^TX)$, we obtain
$$
\operatorname{tr}(A^TA-A^2)=\langle H,K\rangle+\langle K,K\rangle=\langle K,K\rangle\ge0.
$$
A: In fact, you can show that $\langle A, B \rangle := \mathrm{tr}(A^T B) $ defines an inner product. An inner product must verify the following: $$ \langle v, v \rangle \geq \langle v, w \rangle$$ whenever $|w|=|v|$. Now since $ |A| = |A^T|$, one gets $$ \mathrm{tr}(A^TA) \geq \mathrm{tr}(A^2). $$
A: $\mathrm{tr}(A^2)=\sum_{i} {[A^2 ]}_{i,i}$
$=\sum_{i} \sum_j A_{i,j}\cdot A_{j,i}$
$=\sum_{i,j, j\neq i}2A_{i,j}\cdot A_{j,i}+\sum_{i,j, j= i}A_{i,j}\cdot A_{j,i}$
$=\sum_{i,j, j\neq i, i\gt j}2A_{i,j}\cdot A_{j,i}+\sum_{i}A_{i,i}^2$
$\mathrm{tr}(A^TA)=\sum_{i} {[A^TA]}_{i,i}$
$=\sum_{i} \sum_j A^T_{i,j}\cdot A_{j,i}$
$=\sum_{i} \sum_j A_{j,i}\cdot A_{j,i}$
$=\sum_{i,j}A_{j,i}^2$
$$\mathrm{tr}(A^TA)-tr(A^2)=\sum_{i,j,i\neq j,i\gt j}(A_{j,i}^2+A_{i,j}^2-2A_{i,j}\cdot A_{j,i})$$
$$=\sum_{i,j,i\neq j}(A_{j,i}-A_{i,j})^2 \ge 0$$
