Dot Product vs Inner Product

Question

eStoY confUSo

Why didn't he use the dot product?

I'm hoping that $\mathbb P_2$ is a polynomial space (I don't even know what that means), probably representable as a vector in $\mathbb R^3$.

So now that I'm already so confused I want to make a LITTLE bit of sense here. Two vectors are orthogonal in this vector space if their inner product is zero? Because what he did accomplishes that.

Answer 1

You're right, the inner product of two polynomials is defined as $$\langle\, p,q\, \rangle=\int_0^1p(t)q(t)\,\mathrm dt, $$ and two polynomials are orthogonal if and only if $\;\langle\, p,q\,\rangle=0$.

These definitions, b.t.w. are valid in the (infinite-dimensional) vector space $\mathbf R[t]$ of all polynomials, and not only in the vector space of degree at most $2$.

He didn't use the dot product of the polynomial coefficients (if I understand well what you mean) simply because the ordinary dot product isn't interesting here, and he chooses another, more useful, inner product.

Answer 2

Polynomials as vectors.

The set of polynomials of degree 2 are 3 dimensional vectors. You have an addition property, and a scalar multiplication property. That is what makes a vector space.

You can even represent them with with the same notation. $(a,b,c) \equiv ax^2+bx+c$

The set of function defined at $x_1,x_2,\cdots x_n$ is also a vector space. And and we could represent $f(x)$ as $(f(x_1), f(x_2),\cdots, f(x_n))$

The inner product might be: $\langle f(x),g(x)\rangle = \frac 1n (f(x_1), f(x_2),\cdots, f(x_n))\cdot (g(x_1), g(x_2),\cdots, g(x_n)) = \frac 1n \sum_\limits{i=1}^n f(x_i) g(x_i)$

This is just multiplying the traditional dot product by $\frac 1n.$ It is also the definition of co-variance in statistics.

And we extend concept this to continuous functions over the interval $[0,1]$

As $n$ goes to infinity $\frac 1n (f(x_1), f(x_2),\cdots, f(x_n))\cdot (g(x_1), g(x_2),\cdots, g(x_n))$ approaches $\int_0^1 f(x)g(x) dx$

It is a well defined inner product, and it is not too far removed from the Euclidean definition you are more comfortable with.

Next you define "orthogonal" as meaning having inner product equal to 0.

Unfortunately this definition of orthogonal no longer ties out with perpendicular in the geometric sense. It does tie out with "independent" as used in prob / statistics.

Answer 3

The dot product is defined on $\Bbb R^n$, or more generally $\Bbb F^n$ by the formula $$ u\cdot v = \sum u_iv_i, $$ but for vector spaces like $\Bbb P_2$, which is, concretely, $\{a_0 +a_1x+a_2x^2:a_i\in\Bbb R\}$, the vector space of polynomials of degree no more than $2$ with coefficients from $\Bbb R$, you can define different inner products which generalize the dot product for Euclidean spaces.

Formally, an inner product is a map $V\times V\to \Bbb R$ that satisfies some extra properties. (You can also have complex inner products.) For instance, the inner product is linear in the first component, and it should be positive-definite, meaning that $\langle v,v\rangle \ge 0$ and $\langle v,v\rangle=0$ if and only if $v=0$. Inner products should also be symmetric, meaning $\langle v,w\rangle = \langle w,v\rangle$. (If the inner product takes values in $\Bbb C$, then symmetry is replaced with conjugate symmetry $\langle v,w\rangle = \overline{\langle w,v\rangle}$.)

You should check that the dot product is an example of an inner product.

Vector spaces equipped with an inner product are called inner product spaces and if $V$ is an inner product space, then two vectors are declared to be orthogonal if $\langle v,w\rangle = 0$.

Finally, check that $\langle p,q\rangle = \displaystyle \int_0^1 p(t)q(t)\,dt$ is an inner product on $\Bbb P_2$.