If Hermitian $A$ is positive definite, is this block matrix also positive definite?

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Let $A$ be an Hermitian matrix, and suppose $A$ is positive definite, i.e., $(Ax, x)>0$ for all $x\in \mathbb C^n$.

If I let $A=\begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & \\ \vdots &\ &\ \ \tilde{\large A} \\ a_{n1} & \end{pmatrix}$,

then is $\tilde A$ also positive definite ?


If Hermitian $A$ is positive definite, I have $(Ax,x)=\displaystyle \sum_{k=1}^n \sum_{j=1}^n a_{kj}x_j \overline x_k> 0$ for all $x=(x_j)_{1\leqq j\leqq n} \in \mathbb C^n$

And I want to check whether $(\tilde Ay, y)> 0$ for all $y=(y_j)_{2\leqq j\leqq n}\in \mathbb C^{n-1}$.

I have $\displaystyle (\tilde Ay, y)=\sum_{k=2}^n\sum_{j=2}^n a_{kj}x_j \overline{x_k}$.

Can I show this is positive, using $\sum_{k=1}^n \sum_{j=1}^n a_{kj}x_j \overline x_k> 0$ for all $x=(x_j)_{1\leqq j\leqq n} \in \mathbb C^n$ ? I have no idea for checking this.

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Let $y\in\mathbb C^{n-1}$ with $y\neq 0$. Then create $x=(0, y^T)^T$.

By definition, $(Ax, x) > 0$

Now notice that $(Ax, x) = (\tilde A y, y)$