Let A be a symmetric $m$ x $m$ matrix of rank r, and B a symmetric $m$ x $m$ matrix of rank $m - r$, such that $AB = 0$. Show that $A^+A+B^+B=I$, where $A^+$ is the Moore-Penrose pseudoinverse of $A$.
My professor said to try using the spectral decomposition of A to show this, and I know that if we let $A=XLX^T$ then $A^+=XL^+X^T$ where the non-zero entries of $L^+$ are the reciprocals of the non-zero entries of $L$. I'm stuck from here. Thanks for your help.
Since $A$ and $B$ are symmetric, that means they have an orthogonal eigendecomposition. Thus $$A = Q^TDQ$$ $$B = Z^TRZ$$ where $D,R$ are diagonal and $Q, Z$ are orthogonal. As you noted, $$A^+=Q^TD^+Q$$ where $D^+$ has the reciprocal of the non-zero entities. Thus by orthogonality $$A^+A = (Q^TD^+Q)(Q^TDQ) = I_r$$ where $I_r$ is the partial identity matrix with ones on the diagonal up to row/column $r$ and zeros everywhere else. If we do a similar method for $B$ we get $I_{m-r}$ which only has ones on the last $m-r$ elements of the diagonal (zeros everywhere else). Thus $A^+A + B^+B = I_r + I_{m-r} = I.$