I'm learning linear algebra, specifically quadratic forms, and need help with the following exercise :
Let $f(x, y, z) = 5x^2 + 12xy + 9y^2 + 6xz - 10yz$.
$(1)$ Find the matrix representation $\mathbf x^T A \mathbf x$ of $f$.
$(2)$ Find the signs of the eigenvalues of $A$.
Here's my work so far :
$(1)$ The matrix representation of $f$ is given by $$\begin{bmatrix}x&y&z\\ \end{bmatrix} \begin{bmatrix}5&6&3\\6&9&-5\\3&-5&0\\ \end{bmatrix} \begin{bmatrix}x\\y\\z\\ \end{bmatrix} = \mathbf x^T A \mathbf x$$
where $A$ is a symmetric matrix, i.e. $A$ has real eigenvalues.
$(2)$ This is the part where I need your help. I'm still going to share my thoughts and what I have found so far.
First we note that $\det(A) = -386 < 0.$ Therefore, by Sylvester's criterion, the matrix $A$ is not positive definite since all principal minors determinants are not all positive. Secondly, by the relation between the eigenvalues and the determinant of $A$ we know that $\det(A) = \lambda_1\lambda_2\lambda_3 < 0$. We therefore have two options : either all eigenvalues are negative or either two of them are positive and the other one is negative.
It is unclear to me how I should proceed from here to decide which of the two options is the correct one.
Recall that a symmetric matrix $A$ is called negative definite if and only if all of its eigenvalues are negative. Moreover, a symmetric matrix $A$ is called negative definite if and only if all its leading principal minors are negative.
Note that $\det(5)=5>0$. Hence, $A$ is not negative definite and as you noted $\det(A) = \lambda_1\lambda_2\lambda_3 < 0$. Hence, it must be that two of the eigenvalues are positive one of them is negative.