Let K be a $n\times n$ symmetric positive definite matrix. Consider $K=LU$ the $LU$ decomposition of $K$. Let $u_{1}^{T}$ be the first row of U without $u_{1,1}$ (a $1\times (n-1)$ vector), and Let $\ell_1$ be the first column of $L$ without $l_{1,1}$ (a $(n-1)\times 1$ vector).
Prove: $B=K_{2}-\ell_{1}u_{1}^{T}$ is a symmetric positive definite matrix. ($K_2$ is the sub matrix of $K$ without the first row and column.
My way:
I have managed to prove that $B=B^T$ and therefore $B$ is symmetric. Now I need to show that $B$ is positive definite:
$\forall x\ne 0$: $x^TBx>0$ but I am stuck here.
Let $K = (\sigma_{ij})_{n \times n}$. Equating the first row and first column of $K$ with $LU$ we can easily see that $$\ell_{1} = \frac{1}{\sigma_{11}}\begin{pmatrix} \sigma_{21} \\ \sigma_{31} \\ \dots \\ \sigma_{n1} \end{pmatrix}$$ and $u_1 = \sigma_{11} \ell_1.$
Now you can verify the following identity $$ B_1KB_1^{\top} = \begin{pmatrix} \sigma_{11} & \bf{0}_{1 \times (n-1)} \\ \bf{0}_{(n-1) \times 1} & B \end{pmatrix} $$ where $B_1$ is the invertible matrix
$$ B_1 = \begin{pmatrix} 1 & \bf{0}_{1 \times (n-1)} \\ -\ell_1 & I_{(n-1) \times (n-1)} \end{pmatrix}.$$
Since $K$ is positive definite, so is $B_1KB_1^T$ and $B$ being a principal submatrix of a positive definite matrix is also positive definite.