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