7.2 Quadratic Forms

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Section 7.2 Quadratic Forms

Handout 7.2 Quadratic Forms

Objectives: Topics

Quadratic forms

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Change of variables

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Principal axes theorem

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Classification of quadratic forms

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Objectives: Goals

Characterize and classify quadratic forms using eigenvalues and eigenvectors.

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Express quadratic forms in the form $Q(\vec x)=\vec x^{T}A\vec x\text{.}$

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Apply the principal axes theorem to remove cross-product terms.

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Motivating Question

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Does the inequality below hold for all $x,y\text{?}$

\begin{equation*} x_1^2-6x_1x_2+9x_2^2 \geq 0. \end{equation*}

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Definition 7.23. Quadratic Form.

A quadratic form is a function $Q:\mathbb{R}^n\to\mathbb{R}$ defined by

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\begin{equation*} Q(\vec x)=\vec x^{T}A\vec x = \begin{bmatrix} x_1 \amp x_2 \amp \cdots \amp x_n \end{bmatrix} \begin{bmatrix} a_{1,1} \amp a_{1,2} \amp \cdots \amp a_{1,n} \\ a_{1,2} \amp a_{2,2} \amp \cdots \amp a_{2,n} \\ \vdots \amp \vdots \amp \ddots \amp \vdots \\ a_{1,n} \amp a_{2,n} \amp \cdots \amp a_{n,n} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}, \end{equation*}

where $A$ is an $n\times n$ symmetric matrix.

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Example 7.24.

Compute $Q_A(\vec x) = \vec x^{T}A\vec x$ and $Q_B(\vec x) = \vec x^{T}B\vec x$ for the following matrices.

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(a)

$A=\begin{bmatrix}4 \amp 0\\ 0 \amp 3\end{bmatrix}$

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(b)

$B=\begin{bmatrix}4 \amp 1\\ 1 \amp -3\end{bmatrix}$

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The graphs of the quadratic forms from Example 7.24 are given below. Students are not expected to sketch quadratic surfaces.

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$Graphs of the two quadratic curves Q_A and Q_B.$

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As we will see shortly, properties of the symmetric matrix $A$ used to generate the quadratic form $Q_A(\vec x)$ will allow us to determine characterisitcs about the quadratic form, such as whether it is always nonnegative. Therefore, if first presented with the quadratic form, it is useful to find the symmetric matrix $A$ which generates the quadratic form.

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Example 7.25.

Write the quadratic form below in the form $\vec x^{T}A\vec x$ for $\vec x\in\mathbb{R}^3\text{.}$

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\begin{equation*} Q(\vec x)=5x_1^2+x_2^2+3x_3^2+6x_1x_3-12x_2x_3. \end{equation*}

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Note in the above example it is difficult to determine whether $Q(\vec x)$ is always nonnegative. This is due to the cross-product terms, terms of the form $x_ix_j\text{,}$ since these terms could be positive or negative depending on the input.

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To help us better determine properties of $Q(\vec x)\text{,}$ we would like to perform a change of variables to obtain an equivalent quadratic form $Q(\vec y)$ which no longer has any cross-product terms. Recall that we can find a symmetric matrix $A$ so that $Q(\vec x) = \vec x^T A \vec x\text{.}$ Since $A$ is symmetric, we know by Theorem 7.18 that $A = PDP^T$ for a diagonal matrix $D$ and orthogonal matrix $P\text{.}$

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What happens when we make the change of variables $\vec x = P\vec y\text{?}$

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Theorem 7.26. Principal Axes Theorem.

If $A$ is symmetric, then there exists an orthogonal change of variables $\vec x=P\vec y$ such that the quadratic form $\vec x^{T}A\vec x$ becomes $\vec y^{T}D\vec y\text{,}$ with no cross-product terms.

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Example 7.27.

Use the orthogonal diagonalization of $A$ to remove cross-product terms from $Q=\vec x^{T}A\vec x\text{.}$

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\begin{equation*} A=\begin{bmatrix}3 \amp 2\\ 2 \amp 6\end{bmatrix} =PDP^{T} = \frac{1}{\sqrt5}\begin{bmatrix}2 \amp 1\\ -1 \amp 2\end{bmatrix}\begin{bmatrix}2 \amp 0\\ 0 \amp 7\end{bmatrix}\frac{1}{\sqrt5}\begin{bmatrix}2 \amp -1\\ 1 \amp 2\end{bmatrix} \end{equation*}

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