Home page for accesible maths 3 Inner products 3.A Bilinear forms 3.C The Cauchy-Schwarz inequality

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3.B Positive definiteness

In the previous section, we generalized the idea of the scalar product, to that of bilinear forms, and described those forms on ${\mathbb{R}}^{n}$ using $n\times n$ matrices. But the scalar product has further properties which ensure we can define a notion of length, distance, and even angle between two non-zero vectors. To generalize these notions to bilinear forms, we will insist on a few additional “natural” properties. For example, the distance from $\vec{x}$ to $\vec{y}$ should be the same as the distance from $\vec{y}$ to $\vec{x}$ . So we define the following property of a function $\langle\cdot,\cdot\rangle:V\times V\to{\mathbb{R}}$ , for any real vector space $V$ over ${\mathbb{R}}$ :

\langle\vec{x},\vec{y}\rangle=\langle\vec{y},\vec{x}\rangle,

for all $\vec{x},\vec{y}\in V$ . If a bilinear form satisfies this property, then we call it a symmetric bilinear form. It is natural to ask: Which bilinear forms are symmetric? Here is the answer for ${\mathbb{R}}^{n}$ :

Theorem 3.11.

The bilinear form $\langle\cdot,\cdot\rangle_{A}$ on ${\mathbb{R}}^{n}$ is symmetric if and only if the matrix $A$ is a symmetric matrix (i.e. $A=A^{T}$ ).

The proof of this Theorem is given as Exercise 3.47.

But there are further hurdles in using the formulas 3.1 to define lengths and angles. For example, if $\langle\vec{x},\vec{x}\rangle<0$ , the square root will not be real (and then there is no consistent way of choosing between the two square roots). Even if $\langle\vec{x},\vec{x}\rangle=0$ , then we could define the length as $||\vec{x}||=0$ , but that would prevent us from using the formula for the angle. To avoid these problems, it is better to consider forms with the following property, called positive difiniteness:

\langle\vec{x},\vec{x}\rangle>0

for any non-zero vector $\vec{0}\neq\vec{x}\in V$ . The standard scalar product on ${\mathbb{R}}^{n}$ obeys this property. We will call a symmetric bilinear form obeying the positive definiteness property an inner product of $V$ . Another way of thinking about this condition is this: “The distance between two distinct vectors is always positive.” Recall that distance was defined as $||\vec{x}-\vec{y}||$ .

Example 3.12.

$A:=\begin{bmatrix}1&2&2\\ 2&1&2\\ 2&2&1\end{bmatrix}$ is symmetric. Is $\langle\cdot,\cdot\rangle_{A}$ positive definite?

Solution: Consider the vector $\vec{x}=\begin{bmatrix}1&-1&0\end{bmatrix}^{T}$ . Then

\langle\vec{x},\vec{x}\rangle_{A}=\vec{x}^{T}A\vec{x}=\begin{bmatrix}1&-1&0% \end{bmatrix}\begin{bmatrix}1&2&2\\ 2&1&2\\ 2&2&1\end{bmatrix}\begin{bmatrix}1\\ -1\\ 0\end{bmatrix}=-2.

Therefore the bilinear form corresponding to $A$ is not positive definite.

In the above example, I first tried a few other vectors ( $\begin{bmatrix}1&0&0\end{bmatrix}^{T}$ and $\begin{bmatrix}1&1&0\end{bmatrix}^{T}$ ), and found they had $\langle\vec{x},\vec{x}\rangle_{A}>0$ . But to prove positive definiteness, you need to prove $\langle\vec{x},\vec{x}\rangle_{A}>0$ for all non-zero vectors. Later, Theorem 5.15 will give us an easier method.

Exercise 3.13:

Define a function ${\mathbb{R}}^{2}\times{\mathbb{R}}^{2}\to{\mathbb{R}}$ by $\langle\begin{bmatrix}x_{1}\\ x_{2}\end{bmatrix},\begin{bmatrix}y_{1}\\ y_{2}\end{bmatrix}\rangle=x_{1}x_{2}+y_{1}y_{2}$ . Determine which of the following properties are satisfied by $\langle\cdot,\cdot\rangle$ :

i.

Bilinear,
ii.

Symmetric,
iii.

Positive definite.

[End of Exercise]

Exercise 3.14:

Let $A=\begin{bmatrix}3&1\\ 1&1\end{bmatrix}$ . Prove that $\langle,\rangle_{A}$ is an inner product.

[End of Exercise]