Show that the matrix $A$ is positive definite

Question

I'm trying to solve the following problem:

Let

$A = \begin{bmatrix} 1 & 2 & 3 & -2\\ 2 & 5 & 4 & -5\\ 3 & 4 & 14 & -2 \\ -2 & -5 & -2 & 10 \\ \end{bmatrix}$

Show that the matrix $A$ is positive definite (that is $x^T A x >0$ for each $x \neq 0$)

I have noticed that $A$ is a symmetric matrix and hence it induces a quadratic form $q_A(x)=x^T A x$. I also know that a quadratic form $q_A$ is positive definite (which is $x^T A x >0$ for each $x \neq 0$) if and only if all the eigen values of $A$ are positive, but ¿Is there an easier way to solve the problem, sice it is a $4 \times 4$ matrix and it will take some time to compute $det(A-\lambda I)$? and It also might be hard to find the roots of the characteristic polynomial of $A$

Answer 1

You can use Sylvester's criterion. Essentially matrix is positive-definite if and only if all the leading principle minors determinants are positive-definite. We can see $$|1|>0$$ $$A = \begin{vmatrix} 1 & 2 \\ 2 & 5\\ \end{vmatrix}=1>0$$ $$A = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 5 & 4 \\ 3 & 4 & 14 \\ \end{vmatrix}=1>0$$ $$A = \begin{vmatrix} 1 & 2 & 3 & -2\\ 2 & 5 & 4 & -5\\ 3 & 4 & 14 & -2 \\ -2 & -5 & -2 & 10 \\ \end{vmatrix}=1>0$$ Thus by Sylvester's criterion we can conclude that $A$ is positive-definite matrix.

Answer 2

$A$ is positive definite because it has a Cholesky decomposition $A=LL^T$ where $$ L=\pmatrix{1&0&0&0\\ 2&1&0&0\\ 3&-2&1&0\\ -2&-1&2&1} $$ is invertible.