An inequality for the trace of a matrix

Question

Let $A$ be a square matrix. Are there any simple conditions under which

$$\operatorname{Tr}(A^T A) \geq \operatorname{Tr}(A)^2 $$

or vice versa?

I do know the following: if $A$ is diagonalizable with eigenvalues $\lambda_j \geq 0$, then

$$\operatorname{Tr}(A^T A) = \sum \lambda_j^2 \leq (\sum \lambda_j)^2 = \operatorname{Tr}(A)^2 .$$

The same is true if $A$ has only negative eigenvalues. I am looking for more general circumstances under which we have inequality one way or the other.

Do you care about the underlying field? Are you working in $\mathbb{R}, \mathbb{C}$? — Calvin Lin, Jan 01 '13 at 04:08
This is over $\mathbb R$; over a different field (one which is not linearly ordered) the $\geq$ sign would not make sense. — user15464, Jan 01 '13 at 04:13
$\text{Tr}(A^2) = \sum \lambda_j^2$, but that's not the same in general as $\text{Tr}(A^T A)$. Of course it would be the same if $A$ was symmetric. — Robert Israel, Jan 01 '13 at 04:16
For example, try $A = \pmatrix{0 & 1\cr 0 & 0\cr}$: eigenvalues are $0$, $\text{Tr}(A^T A) = 1 > 0 = \text{Tr}(A)^2$. — Robert Israel, Jan 01 '13 at 04:22
No, $\text{Tr}(A^T A) = +2$ in that case. $\text{Tr}(A^T A) = \sum_i \sum_j a_{ij}^2 \ge 0$. — Robert Israel, Jan 01 '13 at 04:30

score 3 · Accepted Answer · edited Jan 20 '13 at 23:15

3

Presumably you're working over the real numbers. Your inequality is true for matrices of rank $1$. In fact if $A = u v^T$ for column vectors $u$ and $v$, $\text{Tr}(A^T A) = (u^T u)(v^T v)$ while $\text{Tr}(A)^2 = (u^T v)^2$, and the Cauchy-Schwarz inequality says $(u^T v)^2 \le (u^T u)(v^T v)$.

edited Jan 20 '13 at 23:15

answered Jan 01 '13 at 04:14

Robert Israel

448,999

What about a matrix of the form $A+B$, where $A$ is rank 1 and $B$ is positive definite symmetric? – user15464 Jan 01 '13 at 04:19
For symmetric positive definite $B$ the inequality goes the other way. – Robert Israel Jan 01 '13 at 04:25

An inequality for the trace of a matrix

1 Answers1