 Quadratic Forms - Maple Help

Main Concept

Let A be an  symmetric matrix with real entries , and let $\stackrel{\mathbf{\to }}{\mathbit{x}}$ be an $n×1$ column vector of the form . Therefore,   is said to be the quadratic form of A. The expansion of

A quadratic form, Q, and its corresponding symmetric matrix, A, can be classified as follows:

 • Positive definite if  for all
 • Positive semi-definite if  for all $\stackrel{\to }{\mathbit{x}}$ and  for some
 • Negative definite if  for all $\stackrel{\to }{\mathbit{x}}\ne 0.$
 • Negative semi-definite if $Q\le 0$ for all $\stackrel{\to }{\mathbit{x}}$ and $Q=0$ for some $\stackrel{\to }{\mathbit{x}}\ne 0.$
 • Indefinite if $Q>0$ for some $\stackrel{\to }{\mathbit{x}}$ and $Q<0$ for some other $\stackrel{\to }{\mathbit{x}}$. Graphical Representation

If $\stackrel{\mathbf{\to }}{\mathbit{x}}$ has only two elements, , then we can graphically represent the quadratic form, $Q\left(x,y\right)$, as a function .  This is shown in the plot below.

This also allows us to visually determine the classification of the $2×2$ symmetric matrix A as:

 • Positive definite if $Q\left(x,y\right)$ is bounded below by $z=0$ and intersects this plane at only a single point, $\left[0,0,0\right].$
 • Positive semi-definite if $Q\left(x,y\right)$ is bounded below by  and intersects this plane along a straight line.
 • Negative definite if $Q\left(x,y\right)$ is bounded above by $z=0$ and intersects this plane at only a single point, $\left[0,0,0\right].$
 • Negative semi-definite if $Q\left(x,y\right)$ is bounded above by  and intersects this plane along a straight line.
 • Indefinite if $Q\left(x,y\right)$ lies above $z=0$ for some values of $\stackrel{\to }{\mathbit{x}}$ and below $z=0$ for other values of $\stackrel{\to }{\mathbit{x}}$, thereby intersecting this plane along a curve which is not a straight line. Application in Multivariable Calculus

Using quadratic forms to classify matrices as definite, semi-definite, or indefinite can be useful in performing the multivariable second derivative test.

Let  have continuous second partial derivatives in some neighborhood of a critical point $\left(a,b\right)$ and let  be the Hessian matrix of $f$ evaluated at $\left(a,b\right)$.

 • If $H\left(a,b\right)$ is positive definite, then $\left(a,b\right)$ is a local minimum.
 • If $H\left(a,b\right)$ is negative definite, then $\left(a,b\right)$ is a local maximum.
 • If $H\left(a,b\right)$ is indefinite, then $\left(a,b\right)$ is a saddle point.
 • If $H\left(a,b\right)$is positive semi-definite or negative semi-definite, then the second derivative test is inconclusive as to the nature of the point $\left(a,b\right).$

Change the values in the symmetric matrix, A, and observe how the plot and formula of its quadratic form, $Q\left(x\mathit{,}y\right)$, change in response. The 3-D plot below can be rotated for visual representation.

Try to find a 2 × 2 symmetric matrix of each type: positive definite, positive semi-definite, negative definite, negative semi-definite, and indefinite.

 ${\stackrel{\mathbit{\to }}{\mathbit{x}}}^{\mathbit{T}}$ $\mathbit{A}$ $\stackrel{\mathbf{\to }}{\mathbit{x}}$ $\mathbf{=}$ ${\stackrel{\mathbf{\to }}{\mathbit{x}}}^{T}\cdot \mathbit{A}\mathbf{\cdot }\stackrel{\mathbf{\to }}{\mathbit{x}}$  $\left[\begin{array}{cc}{x}_{}& y\end{array}\right]$ $\left[\begin{array}{c}x\\ y\end{array}\right]$ $=$ More MathApps