Proving positive-definiteness using the characteristic equation How does one prove that $q(x,y,z) = 2x^2+5y^2+2z^2+2xz$ is positive definite by solving its characteristic equation?
 A: I assume that the characteristic equation of the quadratic form $q(x, y, z)$ is in fact the characteristic equation $p_A(\lambda) = 0$ of the coefficient matrix $A$ of $q$, as explained below.
Suppose we set $\mathbf r = (x, y, z)^T$ and 
$A = \begin{bmatrix} 2 & 0 & 1 \\ 0 & 5 & 0 \\ 1 & 0 & 2 \end{bmatrix}; \tag{1}$
then it is easy to see (just do the indicated matrix operations) that $q(x, y, z)$ is given by
$q(x, y, z) = \mathbf r^T A \mathbf r = 2x^2 + 5y^2 + 2z^2 + 2xz. \tag{2}$
We next observe that $A$ is a symmetric matrix and its characteristic polynomial $p_A(\lambda)$ is given by
$p_A(\lambda) = \det(A - \lambda I) = \det(\begin{bmatrix} 2 - \lambda & 0 & 1 \\ 0 & 5 - \lambda & 0 \\ 1 & 0 & 2 - \lambda \end{bmatrix})$
$= (2 - \lambda)^2(5 - \lambda) - (5 - \lambda) = (5 - \lambda)(\lambda^2 -4\lambda + 3). \tag{3};$
the roots of $p_A(\lambda)$ are evidently $\lambda = 1, 3, 5$, all positive.  Furthermore, since $A = A^T$ is symmetric, there exists an orthogonal matrix $O$ which diagonalizes $A$, thus:
$O^TAO = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5 \end{bmatrix} = D. \tag{4}$
If we now set
$\mathbf s = \begin{pmatrix} s_x \\ s_y \\ s_z \end{pmatrix} = O^T\mathbf r \tag{5}$
for any $\mathbf r = (x, y, z)^T \in \Bbb R^3$, we see, since (5) implies $\mathbf s^T = \mathbf r^T O$, that
$q(x, y, z) = \mathbf r^T A \mathbf r = \mathbf r^T O O^T A O O^T \mathbf r = \mathbf s^T D \mathbf s = s_x^2 + 3s_y^2 + 5s_z^2 > 0 \tag{6}$
as long as $\mathbf s \ne 0$, which, since $\mathbf s = O^T\mathbf r$, is equivalent to $\mathbf r \ne 0$ ($O$ is of course nonsingular, as is $O^T$).  We have used the defining property of an orthogonal matrix, $O^TO = OO^T = I$, in the derivation (6).  QED.
And this is how one proves it!
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
