Let's $A \in R^{n x n}$ singular PSD matrix, $B \in R^{n x n}$ full rank indefinite matrix and $C \in R^{n x n}$ full rank PSD matrix, I want to show that in order to fulfill the next equation: $C=A+B$, $B$ must be PSD matrix. I have an intuition behind that, but I can't prove it is correct (which I'm not sure 100% it's correct).
My intuition is based on: Let us assume that B hold a negative eigenvalue, and w.l.o.g let's arrange the rows of the matrices by the size of the eigenvalues: $\lambda_1\le\lambda_2\le..\lambda_{n-1}\le\lambda_{n}$ hence we will have rows in C that will be negetive dominated or with a negative value on the diagonal, and C will not be a PSD matrix anymore.
Any Ideas? thoughts? if I'm in the right direction ? how to make it formal?
Thanks for your help!!!
No. this is false. Let $A$ be the identity. Then take $B$ to be $\epsilon$ times an arbitrary diagonal matrix with diagonal entries $\pm 1.$ Then $A+B$ will be $PSD.$
EDIT You had asked for a singular $A,$ but the same example works - let $A$ be identity with one entry on the diagonal zeroed, and make sure that $B$ has a $1$ in that position.