I have a problem finding a matrix of coefficients for $$Q(x,y,z)=3x^2+2y^2+3z^2-2xy-2yz$$ original question is "Find the eigenvalues and the eigenvectors of the matrix of its coefficients"
I am able to find eigenvalues and eigenvectors, but I get stuck at the first step.
Assume $$\begin{bmatrix}x & y & z \end{bmatrix} \begin{bmatrix}3 & a & b \\ c & 2 & d \\ e & f & 3 \end{bmatrix}\begin{bmatrix}x \\ y \\ z \end{bmatrix}=3x^2+2y^2+3z^2-2xy-2yz$$ then find $\ a,b,c,d,e,f \ $ by equality, like below $$\begin{bmatrix}x & y & z \end{bmatrix}\begin{bmatrix}3x+ay+bz \\ cy+2y+dz \\ ex+fy+3z \end{bmatrix}\\ x(3x+ay+bz)+y(cx+2y+dz)+z(ex+fy+3z)$$ Is there an easy way to find the coefficients in the matrix with respect to $ -2xy-2yz \ $ in $Q(x,y,z)$?
As you can see the entries $a$ and $c$ will both give the terms in $(a+c)xy$. The entries $b$ and $e$ give the terms in $(b+e)xz$. The entries $d$ and $f$ will give the terms in $(d+f)yz$.
The simples way is to suppose that the matrix is symmetric (i.e. $a = c$, $b = e$, $d = f$). In this way the non diagonal entries of our matrix our simply half the coefficient of the respective term in $Q$.
More explicitly the non diagonal entry $a_{ij}$ in the matrix will be half the coefficient of $x_{i}x_j$ in $Q(x,y,z)$ where ($x_1=x , x_2 = y, x_3 = z)$.
In your example you find the term $-2xy$ in $Q$ so take $a$ the half of $-2$. That is, $a = -1$. The term $-2yz$ in $Q$ gives $d = -1$. There is no term in $xz$ so the coefficient of $xz$ is $0 \implies b = 0$.
The matrix you're looking for is the following
$$ \begin{pmatrix} 3 & -1 & 0 \\ -1 & 2 & -1 \\0 & -1 & 3\end{pmatrix}.$$