If $T=\begin{bmatrix}x&Y^*\\Y &Z\end{bmatrix}$, $Z\ge 0$ $n\times n$ matrix, $x>0$ and $Y$ is $n\times 1$ vector, $T\ge 0$ iff $Y\in\text{ran }(Z)$

Question

This is what I have to prove:

If $T=\begin{bmatrix} x & Y^* \\ Y & Z\end{bmatrix}$ where $Z$ is positive semidefinite $n\times n$ matrix, $x>0$ and $Y$ is $n\times 1$ column vector then $T$ is positive semidefinite iff $Y\in \text{ran} (Z)$.

Here's what I tried: By Schur's complement theorem for positivity we have that $T\ge 0$ iff $Z-x^{-1}Y Y^* \ge 0$. So it suffices to prove that $xZ-YY^{*}\ge 0$ iff $ Y \in \text{ran} (Z)$. I tried a lot but I am unable to show this . Any hint in the right direction will be appreciated.

Answer 1

The statement you wish to prove is not correct. Look at $\begin{bmatrix} 1 & 3\\ 3 & 1 \end{bmatrix}$.

What you can prove is that if $T$ is positive semi-definite, then $Y$ is in the range of $Z$. The reverse implication is false, as I have just told you.

Let $U$ be an $n$-by-$n$ orthonormal matrix such that $U^* Z U$ is diagonal.

Now look at $\begin{bmatrix} 1 & 0\\ 0 & U^* \end{bmatrix} T \begin{bmatrix} 1 & 0\\ 0 & U \end{bmatrix}= \begin{bmatrix} x & (U^*Y)^*\\ U^*Y & U^*Z U \end{bmatrix}$.

Of course, $Y$ is in the range of $Z$ if and only $U^*Y$ is in the range of $U^*ZU$.

Therefore, we can simplify the problem.

If $D$ is a diagonal matrix with non-negative entries, prove that (1) implies (2), where

1. that $M=\begin{bmatrix} x & W^*\\ W & D \end{bmatrix}$ is positive semi-definite;
2. that for each $k$ such that $2\leq k\leq n+1$, if the $k^\text{th}$ diagonal entry of $D$ is zero, then the $k^\text{th}$ entry of $W$ is also zero.

The argument is simple. If condition (2) isn't satisfied, then for some $k$ between $2$ and $n+1$, the $k^\text{th}$ diagonal entry of $D$ is zero but the corresponding entry of $W$ is $w_k\neq0$, then referring to the $k^\text{th}$ unit column vector as $e_k$, we have $(e_1+re_k)^*M(e_1+e_k)=x+2rw_k$, which can be made negative for certain values of $r$.