Let $A \in \mathbb{R}^{n \times n}$ such that for all $x \in \mathbb{R}^n, x^TAx \geq 0$. Prove that $\ker(A) = \ker(A^T)$

Question

Let $A \in \mathbb{R}^{n \times n}$ such that for all $x \in \mathbb{R}^n, x^TAx \geq 0$. Prove that $\ker(A) = \ker(A^T)$.

My idea: if we can prove $A$ is symmetric, then we can solve it?

We do not have enough information to deduce that $A$ is symmetric. For instance, consider $$ A = \pmatrix{1&-1\1&1}. $$ — Ben Grossmann, Oct 15 '20 at 18:15
You can write $A = P + S,$ where $P$ is symmetric positive semidefinite and $S$ is skew symmetric, that is $S^T = -S.$ If $v$ is a column vector and $Av = 0$ but $Pv \neq 0,$ what does that say? — Will Jagy, Oct 15 '20 at 18:26

Coriolanus · Answer 1 · 2020-10-16T16:06:51.660

3

Let $x \in \ker A$ or $x \in \ker A^T$. Then we have

$$x^T\frac{A + A^T}{2}x = x^T A x = 0 \tag 1$$

(See here for the first equality.)

This implies

$$\frac{A + A^T}{2}x = 0 \tag 2$$

so that

$$(A + A^T)x = 0 \tag 3$$

where $(2)$ follows since $\frac{A + A^T}{2}$ is symmetric positive semidefinite (see here).

Thus if $x \in \ker A$ $(3)$ gives $A^Tx = 0$ and vice versa so that $\ker A = \ker A^T$.

$\square$

edited Oct 16 '20 at 16:06

answered Oct 15 '20 at 22:41

Coriolanus

Yes. With (1) you easily prove the rest but you did not prove (1). You provided a link with a lemma. – Oliver Kayende Oct 16 '20 at 16:43

1 Answers1