how to show that K is positive-definite,knowing that A is positive-definite Given a positive-definite and symmetric matrix $A$,which can be written as:
$A=\begin{bmatrix}
d & u^{T}\\ 
u & H 
\end{bmatrix}=\begin{bmatrix}
\sqrt d & 0\\ 
\frac{u}{\sqrt d} & I_{n-1} 
\end{bmatrix}\begin{bmatrix}
1 & 0\\ 
0 & K 
\end{bmatrix}\begin{bmatrix}
\sqrt d & \frac{u^{T}}{\sqrt{d}}\\ 
0 & I_{n-1} 
\end{bmatrix}$
where $K=H-\frac{1}{d}uu^{T}$
how can I show that the matrix $K\in \mathbb{R}^{n-1,n-1}$ is also positive-definite?

Note that the matrix
$$P = \begin{bmatrix}\sqrt{d} & \frac{u^T}{\sqrt{d}}\\ 0 & I_{n-1}\end{bmatrix}$$
is invertible, with inverse
$$P^{-1} = \begin{bmatrix}\frac{1}{\sqrt{d}} & -\frac{u^T}{d}\\ 0 & I_{n-1} \end{bmatrix},$$
thus from
$$A = P^T \begin{bmatrix}1 & 0\\ 0 & K\end{bmatrix} P$$
you obtain
$$\begin{bmatrix}1 & 0\\ 0 & K\end{bmatrix} = (P^T)^{-1} A P^{-1},$$
and the positive definiteness of $K$ follows from that of $A$ considering vectors with first component $0$.
