Find symmetric matrix of polynomial equation

Question

Let f be a polynomial such that $x, y, z, t$ belong to $\mathbb{R}^4$.

$$f(x) = 4x^2+4y^2+4z^2+4t^2+8xy+6xt+6yz+8zt$$

Find the symmetric matrix and determine whether $A$ is a positive definite or not.

I understand how to find a symmetric matrix and check whether or not it is a positive matrix. I'm having trouble starting. Would I change this equation to a matrix?

Any guidance on how to start this problem would be appreciated.

Answer 1

$\begin{eqnarray*}f(x) &=& 4x^2+4y^2+4z^2+4t^2+8xy+6xt+6yz+8zt\\ &=&\begin{bmatrix} x&y&z&t \end{bmatrix} \begin{bmatrix} 4&4&0&3\\4&4&3&0\\0&3&4&4\\3&0&4&4 \end{bmatrix} \begin{bmatrix}x\\y\\z\\t\end{bmatrix} \end{eqnarray*} $

It's eigenvalues are -3, 3, 5, 11,so it's not positive definite.

Answer 2

To build the matrix, assign each row and each column of the matrix to a variable, in the same order, for example $(x,y,z,t)$. Then put in the main diagonal the coefficients of $x^2$, $y^2$, etc., and put in the other entries the half of the corresponding monomials. As example, the first row of the matrix for the mentioned order of the variables would be $$(4,4,0,3)$$

Answer 3

You need to find a $ 4 \times 4 $ symmetric matrix $A$ such that $$ 4x^2+4y^2+4z^2+4t^2+8xy+6xt+6yz+8zt = \pmatrix { x & y & z & t } A \pmatrix { x \\ y \\ z \\ t } $$

To determine whether $A$ is positive definite, compute $f(1,-1,0,0)$ and $f(1,-2,0,0)$ for instance.