Prove that matrix with parameter is positive definite

Question

I want to prove that the following matrix is positive definite for $a \in (0.5,1)$. \begin{align} A = \begin{bmatrix} 1 & a & a \\ a & 1 & a \\ a & a & 1 \end{bmatrix} \end{align}

Let $x,y,z \in \mathbb{R}$:

\begin{align} \begin{bmatrix} x & y & z \end{bmatrix} \cdot A \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix} = x^2+y^2+z^2+2a(yx+zx+zy) \end{align} Obviously, $x^2+y^2+z^2 \geq 0$, $\forall x,y,z \in \mathbb{R}$. What's left to prove is that

\begin{align}2a(yx+zx+zy) > 0\end{align}

Any ideas? Thank you!

Answer 1

You can't actually prove that $2a(yx + zx + zy)> 0$ because that isn't always the case. However, you can say that in the event that $(yx + zx + zy) < 0,$ $$ \begin{align} x^2 + y^2 + z^2 + 2a(yx + zx + zy) &> x^2 + y^2 + z^2 + 2(yx + zx + zy) \\ &= (x+y+z)^2 \geq 0 \end{align} $$ Which leads to the desired conclusion.

Alternatively, it's enough to simply check that all leading principal sub-determinants of $A$ are positive (see the fourth explanation on this list).

Answer 2

The matrix $A$ is positive definite if and only if the principal determinants are positive hence if and only if: $$1>0\; \text{and}\; \det \begin{bmatrix} 1 & a \\ a & 1 \\ \end{bmatrix}=1-a^2>0\; \text{and}\; \det A =(2a+1)(a-1)^2>0\iff a\in(-0.5,1) $$