For $A\in\mathbb{R}^{m\times n}$, prove that $\mathcal{R}(A^+) = \mathcal{R}(A^T)$.

For $A\in\mathbb{R}^{m\times n}$, prove that $\mathcal{R}(A^+) = \mathcal{R}(A^T)$.

Attempt: Let $x\in\mathcal{R}(A^+)$, then $A^+y = x$ for $y\in\mathbb R^m$. But since $\mathcal{R}(A^T) = \mathcal N(A)^\perp \implies x = A^+y \in\mathcal N(A)^\perp = \mathcal{R}(A^T)$.

Let $x\in\mathcal R(A^T) = \mathcal N(A)^\perp$. Then $A^Ty = x$ for $y\in \mathbb R^m$. I am not really sure how to proceed from here. Anything would be great!

Note that $\mathcal R(A^T) = \mathcal N(A)^\perp$ and $\mathcal R(A)^\perp = \mathcal N(A^T)$ and referencing this:

and

and given an SVD for $A = U\Sigma V^T = \left[\begin{matrix}U_1 & U_2\end{matrix}\right]\left[\begin{matrix}S & 0 \\ 0 & 0 \end{matrix}\right]\left[\begin{matrix}V_1^T \\ V_2^T\end{matrix}\right]$:

Solution 1:

Let $A$ be an $m \times n$ rank $r$ matrix. Let $U = [u_1, \ldots, u_m]$ be the left singular vectors (note they are pairwise orthonormal). Similarly, let $V = [v_1, \ldots, v_n]$ be the right singular vectors. Let $\sigma_i$ be the singular values. We can now write $A = \sum_{i=1}^r \sigma_i u_i v_i^T$.

Now for any vector $x$, there exists $c_1, \ldots, c_n$ such that $x = \sum_{i=1}^n c_i v_i$. Then we see that

$$ Ax = \left(\sum_{i=1}^r \sigma_i u_i v_i^T\right)\left(\sum_{i=1}^n c_i v_i\right) = \sum_{i=1}^rc_i\sigma_i u_i. $$

Thus $u_1, \ldots, u_r$ are a basis for the range of $A$. Rephrasing if $A$ is a rank $r$ matrix, then its first $r$ left singular values form a basis for the range.

Finally, we have that $A^T = V\Sigma^T U^T$ and $A^{+} = V \Sigma^{+} U^T$. Thus, they both have the same left singular vectors and have the same rank. Thus, their ranges are spanned by $v_1, \ldots, v_r$. Thus, their ranges are equal.

Solution 2:

Let $x\in\mathcal R(A^T)$. Then $\exists y$ s.t. $$x = A^Ty = A^T(A^T)^+A^Ty = (A^T(A^T)^+)^TA^Ty = A^+(AA^Ty).$$ Thus, $x\in\mathcal R(A^+)$.

Let $x\in\mathcal R(A^+)$. Then $\exists y$ s.t. $$x = A^+y = A^+AA^+y = (A^+A)^TA^+y = A^T\left[(A^+)^TA^+y\right].$$ Thus, $x\in\mathcal R(A^T)$.

Hence $\mathcal R(A^T) = \mathcal R(A^+)$.