TITLE: Does $Ran(A_+ + A_-) = Ran(A_+) + Ran(A_-)$ hold?
QUESTION [2 upvotes]: Consider a symmetric matrix $A$ and its eigenvalue decomposition as $A = V \Sigma V^T$ where $V$ is the matrix whose columns are eigenvectors and $\Sigma$ is the matrix whose diagonal elements correspond to the eigenvalues. Define $\Sigma_+$ as the diagonal matrix whose elements correspond to the absolute values of the positive eigenvalues of $A$, padded with zeros. Define $\Sigma_-$ in a similar fashion using the negative eigenvalues. With this, we can define $A_\pm$ as $A_\pm = V \Sigma_\pm V^T$ and have the relation that $A = A_+ - A_-$.
Is it true that $Ran(A_+ + A_-) = Ran(A_+) + Ran(A_-)$?
I believe this is true but can't seem to show it.
Some possibly helpful facts:
$\bullet$ $x \in Ran(A_\pm) \implies x \in Ker(A_\mp)$ and therefore $Ran(A_+) \perp Ran(A_-)$
$\bullet$ $A_+A_- = A_-A_+ = 0$
Here, $Ran(A)$ is the column space/range of $A$ and $Ker(A)$ is the kernel/null space of $A$.

REPLY [1 votes]: Yes, it is true. In particular, every element $Ax \in \operatorname{Ran}(A)$ can be written as
$$
Ax = (A_+ x) + (-A_-x),
$$
so that $\operatorname{Ran}(A) \subseteq \operatorname{Ran}(A_+)+\operatorname{Ran}(A_-)$. Conversely, we see that
$$
\operatorname{Ran}(A) \supseteq \operatorname{Ran}(A_\pm A) = \operatorname{Ran}(\pm A_\pm^2 \mp A_\pm A_\mp) = \operatorname{Ran}(A_\pm^2) = \operatorname{Ran}(A_\pm).
$$
$\operatorname{Ran}(A)$ contains both $\operatorname{Ran}(A_+)$ and $\operatorname{Ran}(A_-)$, so it contains their sum $\operatorname{Ran}(A_+) + \operatorname{Ran}(A_-)$.
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