How can I show $A^+B = 0$ if and only if $A^TB=0$?

Question

where $A^+$ is the pseudo inverse of $A$ and $A^T$ is the transpose of A.

Please do not assume that $A$ has full row rank or column rank.

Thanks in advance.

Answer 1

Let $A$ be a rank $r$ matrix with SVD $A = U\Sigma V^T$. Then

$$ A^+B = V \begin{bmatrix} \Sigma_r^{-1} & 0\\ 0 & 0 \end{bmatrix} U^TB = V_r\Sigma_r^{-1}U_r^TB = 0 \iff U_r^TB = 0 $$ and $$ A^TB = V \begin{bmatrix} \Sigma_r & 0\\ 0 & 0 \end{bmatrix} U^TB = V_r\Sigma_rU_r^TB = 0 \iff U_r^TB = 0 $$ where $V_r$ is the first $r$ columns of $V$, $U_r$ is the first $r$ columns of $U$, and $\Sigma_r$ is the $r\times r$ diagonal matrix of strictly positive singular values.

Answer 2

If $A=PQ$ is a full rank decomposition, that is, $P$ is left invertible and $Q$ is right invertible, both with the same rank as $A$, we have that $$ A^+=Q^T(QQ^T)^{-1}(P^TP)^{-1}P^T $$ See Moore-Penrose pseudoinverse on Wikipedia

The hypothesis $A^TB=0$ becomes $Q^TP^TB=0$. Since $Q$ is right invertible, $Q^T$ is left invertible, so we get $P^TB=0$. Therefore $A^+B=0$.

Similarly for the converse direction.

Alternatively, you can use the identities $$ A^T=A^TAA^+,\qquad A^+=A^+(A^T)^+A^T $$ that you can find at this Wikipedia page (the page uses the Hermitian transpose, but here we're assuming matrices are real).