Quadratic Forms

If $Q(x)$ is a function, then,

$$Q(\vec{x})={\vec{x}}^{T}A\vec{x}$$

$A$ is symmetric

Examples

Find $Q$

$Q(\vec{x})={\vec{x}}^{T}A\vec{x}$, $A=\left[\begin{array}{cc}4& 1\\ 1& -3\end{array}\right]$

\b{x & y} \b{4 &1 \\ 1 &-3} \b{x \\ y} = 4x^2+2xy-3y^2

The $2xy$ term is called a cross-product due to it having both variables.

Finding $A$

${\mathbb{R}}^2$

$Q=x^2-6xy+9y^2$ We will use our eyes, the main diagonal will be the coefficients for second order terms. The other diagonal will be $\frac{1}{2}$ of the coefficient of the cross-product. So, A = \b{1 &-3 \\ -3 & 9}. We could compute ${\vec{x}}^{T}A\vec{x}$ to verify this result.

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${\mathbb{R}}^3$

$Q=5x_1^2-x_2^2+3x_3^3+6x_1{x}_3-12x_2{x}_3$ $Q=5x_1^2-x_2^2+3x_3^3+6x_1{x}_3-12x_2{x}_3+0x_1{x}_2$ We will once again use our eye (consider resting them after this). Like last time the main diagonal will be the coefficients for second order terms. The other terms will be $\frac{1}{2}$ the coefficient of the cross-products. We will look at the variables being crossed take $x_1{x}_3$ for example. The location $1,3$ and $3,1$ in the matrix $A$ will be $\frac{1}{2}$ the coefficient of $x_1{x}_3$.

A= \b{5&0&3\\0&-1&-6\\3&-6&3}

We can once again compute ${\vec{x}}^{T}A\vec{x}$ to verify this result.