If $A \mathbf{x} = 0$, is $\mathbf{x}^T A^{\dagger} = 0$ true? ($A^{\dagger}$ is the pseudoinverse of $A$)

Question

Let $A$ be a $n \times n$ matrix, which can be non-singular. We know that $A \mathbf{x} = 0$.

Is it true that $\mathbf{x}^T A^{\dagger} = 0$ as well, where $A^{\dagger}$ is the left-pseudoinverse of $A$. If so, why?

Answer 1

Consider the more general case in which $A$ is an $m\times n$ matrix with rank $r$ and let $A = U\Sigma V^T$ be the SVD of $A$. Then $\operatorname{null}(A) = \operatorname{span}\{v_{r+1},\ldots,v_n\}$. So, if $Ax = 0$, then

$$ x^TA^\dagger = y^T{V_{r+1:n}}^T[V_{1:r}~V_{r+1:n}] \Sigma^\dagger U^T = y^T[0_{n\times r}~I_{n\times(n-r)}] \begin{bmatrix} \hat{\Sigma}^{-1} & 0\\ 0 & 0 \end{bmatrix}U^T = 0^T, $$ where $\hat{\Sigma}$ is the leading $r\times r$ principal submatrix of $\Sigma$.