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

Question

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 $\mathrm{tr}(A^TA)\geq0$.

Answer 1

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}

Answer 2

$\mathrm{tr}(A^2)=\sum_{i} {[A^2 ]}_{i,i}$

$=\sum_{i} \sum_j A_{i,j}\cdot A_{j,i}$

$=\sum_{i,j, j\neq i}2A_{i,j}\cdot A_{j,i}+\sum_{i,j, j= i}A_{i,j}\cdot A_{j,i}$

$=\sum_{i,j, j\neq i, i\gt j}2A_{i,j}\cdot A_{j,i}+\sum_{i}A_{i,i}^2$

$\mathrm{tr}(A^TA)=\sum_{i} {[A^TA]}_{i,i}$

$=\sum_{i} \sum_j A^T_{i,j}\cdot A_{j,i}$

$=\sum_{i} \sum_j A_{j,i}\cdot A_{j,i}$

$=\sum_{i,j}A_{j,i}^2$

$$\mathrm{tr}(A^TA)-tr(A^2)=\sum_{i,j,i\neq j,i\gt j}(A_{j,i}^2+A_{i,j}^2-2A_{i,j}\cdot A_{j,i})$$

$$=\sum_{i,j,i\neq j}(A_{j,i}-A_{i,j})^2 \ge 0$$

Answer 3

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 $$ \operatorname{tr}(A^TA-A^2)=\langle H,K\rangle+\langle K,K\rangle=\langle K,K\rangle\ge0. $$

Answer 4

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 $$ \mathrm{tr}(A^TA) \geq \mathrm{tr}(A^2). $$