Answer to Question #340436 in Linear Algebra for Micky

Question

Determine whether the polynomial x²+2y²+4xy+2yz+6xz is a quadratic form and if so write it in the form X^T AX, where A is a symmetric matrix.

Answer 1

By definition, quadratic form is a polynomial with terms all of degree two. Thus, $p=x^2+2y^2+4xy+2yz+6xz$ is a quadratic form. Denote:$X=\left(\begin{array}{ll}x\\y\\z \end{array}\right)$. The aim is to find the matrix $A=\left(\begin{array}{lll}a_{11}&a_{12}&a_{13}\\a_{12}&a_{22}&a_{23}\\a_{13}&a_{23}&a_{33}\end{array}\right)$ satisfying $p=(x\,\,y\,\,z)\left(\begin{array}{lll}a_{11}&a_{12}&a_{13}\\a_{12}&a_{22}&a_{23}\\a_{13}&a_{23}&a_{33}\end{array}\right)\left(\begin{array}{ll}x\\y\\z \end{array}\right)$.

The right side of the latter equality is: $2xya_{12}+2xza_{13}+2yza_{23}+a_{11}x^2+a_{22}y^2+a_{33}z^2$.

By comparing the coefficients of the latter expression and $p$, we get: $a_{11}=1$, $a_{12}=2$, $a_{13}=3$, $a_{22}=2$, $a_{23}=1$, $a_{33}=0.$ Thus, $A=\left(\begin{array}{lll}1&2&3\\2&2&1\\3&1&0\end{array}\right)$.

Answer: $p=x^2+2y^2+4xy+2yz+6xz$ can be presented as: $p=X^{\top}AX$, where $X=\left(\begin{array}{ll}x\\y\\z \end{array}\right)$, $A=\left(\begin{array}{lll}1&2&3\\2&2&1\\3&1&0\end{array}\right)$.