Prove matrix is positive definite for any alpha

Question

My question is:

Prove that the matrix

$A=\begin{bmatrix} 1 & \alpha \\ 0& 1 \end{bmatrix}$

is positive definite in $\Re ^{2}$ for any |$\alpha$|<2.

Here is my reasoning:

Apparently there are several definitions out there regarding if a matrix is positive definite or not. One of the definitions that I read was that it must be a symmetric matrix and have positive eigen values.

There is another definition out there which says that the matrix must be a linear operator on a finite dimensional inner product space. Where the operator is self adjoint and $\left \langle T(x),x \right \rangle>0$.

Now I understand that a matrix is a linear operator, but I do not understand the adjoint aspect.

Also when I attempt to get the eigen values of the matrix, it seems to me that alpha doesn't even matter as the zero element in the matrix will always make alpha zero. And according to the one definition I stated I thought that the original matrix must be a symmetric matrix in the first place an order to be positive definite.

Any help with this?

Answer 1

HINT The common definition is indeed that $<T(x), x> \ge 0 \ \forall x \in \mathbb{R}^n$. So for you, this means $(x,y) A (x,y)^T \ge 0$ for all $x,y \in \mathbb{R}$. Note that $$ (x,y) A (x,y)^T = (x,y) \cdot (x+\alpha y, y) = x(x+ \alpha y) + y^2 = x^2+ y^2 + \alpha xy $$ Can you take it from here?

UPDATE

Then, assuming $x\ne 0$ and $y \ne 0$ (in which case it is true for all $\alpha$), we have $$ 0 \le x^2+ y^2 + \alpha xy \iff -\alpha xy \le x^2 + y^2 \iff \alpha \ge \frac{x^2+y^2}{xy}. $$

Can you take if from here?