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I have a question that asks me to use the following symmetric positive definite matrix of order $n + 1$

$$B = \begin{bmatrix} \alpha & a^T \\ a & A \end{bmatrix} $$

With this matrix, I have to show that the scalar $\alpha$ must be positive and the $n \times n$ matrix $A$ must be positive definite.

My attempt:

Given that the matrix $B$ is positive definite, we know that

$$ \begin{bmatrix} x_1 & \vec{x_2} \end{bmatrix} \cdot \begin{bmatrix} \alpha & a^T \\ a & A \end{bmatrix} \cdot \begin{bmatrix} x_1 \\ \vec{x_2} \end{bmatrix} > 0 $$

and therefore

$$\alpha x_1^2 + 2ax_1\vec{x_2} + A\vec{x_2}^2 > 0$$ $$\alpha x_1^2 + 2ax_1\vec{x_2} + \vec{x_2}^TA\vec{x_2} > 0$$

We know that $\alpha$ must be positive because if $\alpha$ was negative then there would exist some vector $x$ such that $\alpha x_1^2 + 2ax_1\vec{x_2} + \vec{x_2}^TA\vec{x_2} \leq 0$. We also know that $A\vec{x_2}^2 > 0$ by the same reasoning. As a result, matrix $A$ must be positive definite since we know that

$$ A\vec{x_2}^2 = \vec{x_2}^TA\vec{x_2} > 0 $$

which is the definition of a positive definite matrix.

Would this proof work for showing that $\alpha$ is positive and that $A$ is a positive definite matrix?

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    $\begingroup$ I think it's important to point out exactly choices of $x_1$ and $\vec{x_2}$ that make it work instead of just saying they exist. Otherwise it looks fine to me. See also Cauchy's Interlacing Theorem. $\endgroup$ – Git Gud Feb 18 '15 at 23:46
  • $\begingroup$ Thanks for the response! I'll make sure to add that in and also look into Cauchy's Interlacing Theorem. $\endgroup$ – Josh Black Feb 19 '15 at 0:18

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