Sum of symmetric, positive semidefinite matrices

Question

Let $A \in \mathbb{R}^{m \times n}, B \in \mathbb{R}^{p \times n}$. Show that $A^{T}A+ B^{T}B$ is invertible if and only if $\ker A \cap \ker B =\lbrace 0 \rbrace$.

I could show that if it's invertible, then $\ker A \cap \ker B= \lbrace 0 \rbrace$. Any help for the converse?

Answer 1

$\newcommand{\0}{\mathbf{0}}\newcommand{\x}{\mathbf{x}}$Hints: Recall or try and show that $\color{blue}{\x^T M^T M\x = 0\text{ iff }\x \in \ker M}$ for any matrix $M$, and recall that for any square matrix, it is invertible iff its kernel is $\{\0\}$. Then to show the converse, your goal is to show that if $\ker A \cap \ker B = \{ \0 \}$, then $A^T A + B^T B$ has kernel $\{ \0 \}$. To show this, suppose that $\x \in \ker\left(A^T A + B^T B\right)$ and try and deduce that $\x = \0$. Note also that $\x^T \left(A^T A + B^T B\right) \x = \x^T A^T A \x + \x^T B^T B\x$.

Answer 2

Here's an approach: suppose that $A^TA + B^TB$ is not invertible. Then, there exists a non-zero vector $x$ such that $(A^TA + B^TB)x = 0$. It follows that $$ 0 = x^T(A^TA + B^TB)x = (Ax)^T(Ax) + (Bx)^T(Bx) = \|Ax\|^2 + \|Bx\|^2 $$ conclude that $x \in \ker(A) \cap \ker(B)$.