Given a real $n \times n$ symmetric matrix $A$ and $a_{11}$ is non-zero, if you use Gaussian Elimination to reduce it to $$ \begin{pmatrix} a_{11} & a_1^T \\ 0 & A_2 \end{pmatrix} , $$ $A_2$ will be an $(n-1) \times (n-1)$ symmetric matrix.
How do you prove this?
I played around with $LA = U \to(LA)^T = U^T \rightarrow A^TL^T = U^T \to AL^T = U^T $ and so on, but I can't draw any conclusions about $A_2$.
After the first step of the LU factorization, we have $$ L_1^{-1}A:= \begin{bmatrix} 1 & 0 \\ -l_{21} & I \end{bmatrix} \begin{bmatrix} a_{11} & a_{12}^T \\ a_{12} & A_{22} \end{bmatrix} = \begin{bmatrix} a_{11} & a_{12}^T \\ 0 & -l_{21}a_{12}^T+A_{22} \end{bmatrix} $$ with $l_{21}:=a_{12}/a_{11}$, which is clearly symmetric because both $A_{22}$ and $l_{21}a_{12}^T=a_{12}a_{12}^T/a_{11}$ are symmetric.