$A$ is Hermitian, $B$ is leading principal submatrix of $A$, $rank B = rank A$. Why does $A$ is positive semidefinite?

Question

Let $n ≥ 2$, and $A ∈ M_n$ be Hermitian, and let ${\rm{B }} \in {\rm{ }}{{\rm{M}}_{n - 1}}$ be a leading principal submatrix of A.

If $B$ is positive semidefinite and $rank B = rank A$, why does $A$ is positive semidefinite?

Answer 1

The hermitian form $a$ associated with $A$ has a diagonalizing basis $\mathbf{b}$, so the matrix which rapresents the form can be considered diagonal. The rank is $n-1$ so we can assume (if needed, we can interchange the basis vectors order $\mathbf{b}_{i}$ to achieve this) the last row of the matrix as zero vector $\mathbf{0}$ (one row must be of zeros coordinates becouse there are $n-1$ independent rows in the matrix by hypothesis, and the matrix is diagonal, so one line must be linear combination of the remaining $n-1$).

$\qquad \qquad \qquad \qquad \qquad A=$$\begin{bmatrix} b_{11} & 0 & \cdots & 0 & 0 \\0 & b_{22} & \cdots & 0 & 0\\ \vdots & \vdots & \vdots &\vdots & \vdots \\ 0 & 0 & \cdots & b_{n-1 n-1} &0 \\ 0 & 0 & \cdots & 0 & 0 \end{bmatrix}$

The matrix $B$ corresponding to the form $b$ is the diagonal submatrix of $A$ with the $b_{ii}$ as diagonal elements. The hermitian form $a(\mathbf{v},\mathbf{w})$ can be expressed as

$\qquad \qquad \sum_{i,j} x_i \bar y_ja(\mathbf{b}_i,\mathbf{b}_j) = \ ^\text{t}\mathbf{x}A \mathbf {\bar y}$ ($\mathbf{x}$ and $\mathbf{y}$ are respectively the $\mathbf{v}$ and $\mathbf{w}$ coordinate vectors)

So we have to test that $\sum_{i,j} x_i \bar x_ja(\mathbf{b}_i,\mathbf{b}_j) = \ ^\text{t}\mathbf{x}A \mathbf {\bar x}\ge 0$. This is an immediate consequence of the hypothesis that $B$ is positive semidefinite.

(To fix ideas I assumed rank $= n-1$, but the matter obviously remains substantially unchanged by assuming any rank - some $b_{ii}$ will be now $0$.)

Answer 2

The size of $B$ doesn't matter (you don't need $B\in M_{n-1}$; $B$ can be smaller sized). As long as it is a positive semidefinite leading principal submatrix of a Hermitian matrix $A$ of the same rank, $A$ must be positive semidefinite.

Let $B$ be $r\times r$ and $A=\pmatrix{B&X\\ X^\ast&Y}$. Since $A$ has the same rank as $B$, its last $n-r$ columns must be linear combinations of its first $r$ columns. That is, $\pmatrix{X\\ Y}=\pmatrix{B\\ X^\ast}C$ for some matrix $C$. It follows that $X=BC,\ Y=C^\ast BC$ and $$ A=\pmatrix{B&BC\\ C^\ast B&C^\ast BC} =\pmatrix{B^{1/2}\\ C^\ast B^{1/2}} \pmatrix{B^{1/2}&B^{1/2}C}, $$ which is positive semidefinite because it is a Gram matrix.