Suppose A $\in \Bbb{R}^{n\mathrm{x}n}$ is symmetric positive definite and we apply Gaussian Elimination without pivoting. After $k$ steps (each step places zeros in the $k$ column) the reduced form will be:
$A^{(k)} = \begin{pmatrix}A_{11}^{(k)} & A_{12}^{(k)}\\\ 0 & A_{22}^{(k)}\end{pmatrix}$
where $A_{22}^{(k)} \in \Bbb{R}^{(n-k)\mathrm{x}(n-k)}$
The question is how to prove by induction that $A_{22}^{(k)}$ is positive definite.
The base case is trivial as $A^{(0)} = A_{22}^{(0)}$ and $A^{(0)} = A$ which is symmetric definite positive. However, I am striving to prove the inductive case. $A_{22}^{(k-1)}$ is positive definite $\Rightarrow$ $A_{22}^{(k)}$ is positive definite.
Now, my intuition made me do the following:
$ \begin{pmatrix}I & & 0\\\ 0 & a & I\end{pmatrix} A^{(k)} = A^{(k-1)}$ where $a \in \Bbb{R}^{(n-k)\mathrm{x}1}$ the column vector with the coefficients from the Gaussian elimination and $I$ is the identity matrix of the proper dimension. I tried to make some form of block multiplication with no luck.
Thanks in advance.