The equivalence of $Y=XX^T$ - by Schur complement and rank constraint

Question

I am confused about the following lemma which is useful to convex optimization problem: ( From http://ieeexplore.ieee.org/document/599549/ )

I know the left one is by Schur complement (https://en.wikipedia.org/wiki/Schur_complement): $$Z\succeq 0 \Longleftrightarrow Y\succeq 0, Y-XX^T\succeq 0 $$

However, the above only shows $Y\succeq XX^T$. The rank constraint (right hand side) tells me what?

I have no idea how to use the right constraint to show the equality.

Answer 1

Use elementary operations on the right part of

$$ \text{rank} \begin{bmatrix} Y & X \\ X^T & I_q \\ \end{bmatrix} $$ to subtract it multiplied by $X^T$ from the left part. Get the following $$ \text{rank} \begin{bmatrix} Y-XX^T & X \\ 0 & I_q \\ \end{bmatrix} $$

For the resulted block matrix it is known that $$q = \text{rank} \begin{bmatrix} Y-XX^T & X \\ 0 & I_q \\ \end{bmatrix} \ge \text{rank}(Y-XX^T)+\text{rank}(I_q)\ge \text{rank}(Y-XX^T)+q $$

Then $\text{rank}(Y-XX^T) = 0$, and the matrix $Y-XX^T$ contains only zero eigenvalues. So $$Y-XX^T = 0.$$

Edit

Elementary operation on matrices do not change their ranks. So

$$ \text{rank} \begin{bmatrix} Y & X \\ X^T & I_q \\ \end{bmatrix} =\text{rank} \begin{bmatrix} Y-XX^T & X \\ 0 & I_q \\ \end{bmatrix} $$