We first observe that by spectral theorem of a selfadjoint matrix, any selfadjoint matrix $A\in M_n(\mathbb{C})$ can be written as (up to unitarily equivalence) $A=A_{+}-A_{-}$ where $A_{+}, A_{-}\geq 0$ with $A_{+}A_{-}=0$. Now I have a question which is following:
Question: Let $A\in M_n(\mathbb{C})$ be a selfadjoint matrix. Does there exist a sequence (or net) $\{A_{\lambda}\}$ where $A_{\lambda}=A_{+}(\lambda)-A_{-}(\lambda)$ with $A_{+}{(\lambda)}\geq 0, A_{-}(\lambda)>0$ and $A_{+}{(\lambda)}A_{-}(\lambda)=0$ s.t. $\lim A_{\lambda}=A$?
Any comment/hint is highly appreciated. Thanks in advance.
Misunderstood the question: I thought $>0$ meant $\geq 0$ and non-zero. It is actually meant as positive definite, which changes things.
This is not true generally. Indeed, suppose that $A = A^* \in M_p$ can be approximated by $A_n$ where $(A_n)_+(A_n)_- = 0$, where $(A_n)_- > 0$ for all $n$. Being positive definite, $(A_n)_-$ has strictly positive eigenvalues, and is therefore invertible. So since $(A_n)_+(A_n)_- = 0$, this implies that $(A_n)_+ = 0$, so $-(A_n)_- = (A_n)_+ - (A_n)_- = A_n \to A = A_+ - A_-$, from which its clear that $A_+$ must be 0, so the positive part of $A$ must be 0.
So if $A$ has a strictly positive eigenvalue, this cannot be done. For example, the 2x2 matrix below cannot be approximated in such a way.
Below is the original answer I put for which I interpreted $> 0$ as $\geq 0$ and non-zero.
Sequences will be good enough since $M_n$ is finite-dimensional. The obvious answer to your question if there are negative eigenvalues is to take $A_\lambda = A, A_+(\lambda) = A_+, A_-(\lambda) = A_-$.
The only difference now is that if you have no negative eigenvalues, can you approximate $A_-$ by strictly positive matrices? Yes!
All that's really happening here is that if $A$ is a self-adjoint matrix, its unitarily diagonalizable (actually any normal matrix is), so lets suppose that $A = \lambda_1 \oplus \cdots \oplus \lambda_p \in M_p$. All that your decomposition into $A_+$ and $A_-$ is doing is letting $A_+$ be the diagonal matrix with the positive $\lambda_i$'s, and $A_-$ will be the positive parts of the negative $\lambda_i$'s.
Edit: you can write this out explicitly: $$A_+ = \max(0,\lambda_1)\oplus \cdots \oplus\max(0,\lambda_p), A_- = \max(0,-\lambda_1)\oplus\cdots\oplus\max(0,-\lambda_p).$$
So if you wanted an approximation, just approximate the eigenvalues in $\mathbb{R}$! For example if $$A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, $$ we can appriximate it by $$A_n = \begin{pmatrix} 1 & 0 \\ 0 & -\frac{1}{n} \end{pmatrix}. $$ Then $(A_n)_-$ will just be $0 \oplus \frac{1}{n}$, which is a matrix $> 0$. You can do the same thing in general for any diagonal matrix.