Consider some form of Tikhonov regularization, where we seek to minimize the objective described by:

$\min\_{x \in \mathbb{R}^n} ||Ax - b||^2\_2 + \lambda||Dx||^2\_2$

(with matrices $A, D$ and vectors $b, x$ of appropriate shape, and some scalar $\lambda >0$). I've read from a text that a unique solution exists for the above objective if and only if $ker(A) \cap ker(D) = \{0\}$, and I'm struggling with the "if" direction of the proof. In particular, I've derived the normal form of the solution, which much satisfy:

$(A^TA + \lambda D^TD)x = A^Tb$

and I've shown that if $z \in ker(A) \cup ker(D)$ for non-zero $z$, then if $x^\*$ is a solution to the above normal form, then so is $x^\* + z$. However, I'm not sure how to proceed with assuming that if $ker(A) \cap ker(D) = \{0\}$, then we must have a unique $x^\*$ as the solution.

EDIT: In fact, is there a cleaner solution that does not require proving both conditionals in the biconditional separately like I've formulated?