How to prove using the definition that matrix $A$ is negative definite?

Question

I'm trying to prove that

$$A=\begin{bmatrix} -2 & 2\\ 2 & -5\\ \end{bmatrix}$$

is negative definite. I can do it using the principal minors, but I am trying to understand how to do it using only the definition of a negative definite matrix, i.e., using the quadratic form associated with it.

Answer 1

Definition. A real symmetric matrix $A$ is called positive definite if $x^\intercal A x > 0$ for every real nonzero vector $x$ of compatible size.

Definition. A matrix $A$ is called negative definite if the matrix $-A$ is positive definite.

In your case, $$ x^\intercal (-A) x = \cdots = 2(x_1 - x_2)^2 + 3x_2^2, $$ as desired. You should work through the missing steps to verify the above.

Remark. Your matrix $-A$ happens to be a weakly chained diagonally dominant L-matrix. By a well-known result, this means that it is an M-matrix. A symmetric M-matrix is also called a Stieltjes matrix. Stieltjes matrices are positive definite.

Answer 2

Lemma $:$ A matrix $A$ is negative definite $\implies$ all the eigen values of $A$ are strictly negative.

Proof $:$ Let $\lambda$ be an eigen value of $A$ corresponding to the eigen vector $X (\neq 0).$ Then by the given hypothesis $$\begin{align} X^t A X & < 0. \\ \implies \lambda X^t X & < 0. \\ \implies \lambda & < 0. \end{align}$$ (since $X^t X > 0$ for $X \neq 0$).

In your case the eigen values of the given matrix $A$ are $-1$ and $-6$. So both the eigen values of the given matrix $A$ are negative. Hence by the above lemma the given matrix $A$ is negative definite.