# Sum of symmetric, positive semidefinite matrices

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?

• How did you show the forward direction? Possibly the converse can be shown similarly. – Minus One-Twelfth Mar 9 at 10:58

$$\newcommand{\0}{\mathbf{0}}\newcommand{\x}{\mathbf{x}}$$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 $$\x^T \left(A^T A + B^T B\right) \x = \x^T A^T A \x + \x^T B^T B\x$$.
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 $$x \in \ker(A) \cap \ker(B)$$.