Suppose that $A \in \mathbb{R}^{m \times n}$ and $B \in \mathbb{R}^{p \times n}$ such that $\ker A \cap \ker B = \{0\}$. Let $\alpha > 0$ and $b \in \mathbb{R}^m$.
I'm trying to show that $(A^T A + \alpha B^T B)x = A^T b$ has a unique solution.
If $(A^T A + \alpha B^T B)$ were invertible, I'd be done, but I'm not sure this is true. I've tried doing this by showing that $(A^T A + \alpha B^T B)x = 0$ implies $x = 0$, but I'm getting stuck in the case analysis. (Maybe it's not true.)
On
Suppose $(A^T A + \alpha B^T B)x = 0$. Then $x^T(A^T A + \alpha B^T B)x = x^T A^T A x + \alpha x^T B^T B x = 0$. That is $$ \underbrace{\|Ax \|_2^2}_{\geq 0} + \alpha \underbrace{\|Bx\|_2^2}_{\geq 0} = 0. $$ Since norms are positive definite, $Ax = 0$ and $Bx = 0$, i.e. $x \in \ker A \cap \ker B$. Our assumption that $\ker A \cap \ker B = \{0\}$ then implies $x = 0$. Hence $\ker (A^T A + \alpha B^T B) = \{0\}$.
Therefore, $(A^T A + \alpha B^T B)$ is invertible, and $(A^T A + \alpha B^T B)x = A^T b$ is uniquely solvable with the solution $x = (A^T A + \alpha B^T B)^{-1} A^T b$.
Hint: $(A^TA+\alpha B^TB)x=0$ implies $\langle Ax, Ax\rangle+\alpha\langle Bx, Bx\rangle=0$.