Linear Algebra: coordinates and basis

Question

I'm asked to show that if $f,g \in \mathrm{Sig}_3$ (sig stands for signal) have the coordinates $x,y \in \mathbb{R^4}$ with respect to the basis $(v_0, v_1, v_2, v_3)$, then $$\left=\pi(x_1y_1+x_2y_2+x_3y_3+x_4y_4)$$ where $$v_0=\frac{1}{\sqrt2}, v_1=\cos x, v_2=\cos(2x), v_3=\cos(3x)$$ and $$C^0([0;2\pi])$$

Any idea how to show this? I'm completely stuck. All of the examples online use vectors with numbers, but I'm just given functions.

Answer 1

Assume $f=[a_1,a_2,a_3,a_4]$ and $g=[b_1,b_2,b_3,b_4]$, you'll have $$f(x)=a_1\frac1{\sqrt2}+a_2\cos x+a_3\cos (2x)+a_4\cos (3x)\\g(x)=b_1\frac1{\sqrt2}+b_2\cos x+b_3\cos (2x)+b_4\cos (3x)$$ The inner product $\left$ would be $$\left=\int_0^{2\pi}f(x)g(x)\mathrm{d}x$$ Before we begin integrating it's worth noting that for every distinct $m,n \in\Bbb{Z}$ and $m,n\geq0$ $$ \begin{align} \int_0^{2\pi}\cos(mx)\cos(nx)\mathrm{d}x&=\frac12\int_0^{2\pi}\cos((m+n)x)+\frac12\int_0^{2\pi}\cos((m-n)x)\mathrm{d}x\\ &=0+0\\&=0 \end{align} $$ (Note that the RHS integrals are on a whole period of $\cos({kx})$ where $k\in\Bbb{N}$, so they equal zero.)

Going back to the inner product, now we know that the integrals $\int_0^{2\pi}\cos x\mathrm{d}x$, $\int_0^{2\pi}\cos(2x)\mathrm{d}x$, $\int_0^{2\pi}\cos(3x)\mathrm{d}x$, $\int_0^{2\pi}\cos x\cos(2x)\mathrm{d}x$, $\int_0^{2\pi}\cos x\cos(3x)\mathrm{d}x$, and $\int_0^{2\pi}\cos(2x)\cos(3x)\mathrm{d}x$ are zero, so we can omit them from the multiplication result. What remains is $$ \begin{align} \left&=\int_0^{2\pi}f(x)g(x)\mathrm{d}x\\ &=\int_0^{2\pi}\left(a_1b_1\frac12+a_2b_2\cos^2x+a_3b_3\cos^2(2x)+a_4b_4\cos^2(3x)\right)\mathrm{d}x \end{align} $$

but since for $k\in\Bbb{Z}$ $$ \begin{align} \int_0^{2\pi}\cos^2(kx)\mathrm{d}x&=\int_0^{2\pi}\left(\frac12+\frac12 \cos(2kx)\right)\mathrm{d}x\\ &=\pi+0\\&=\pi \end{align} $$

we'll have $$ \begin{align} \left&=a_1b_1\pi+a_2b_2\pi+a_3b_3\pi+a_4b_4\pi\\ &=\pi(a_1b_1+a_2b_2+a_3b_3+a_4b_4) \end{align} $$ which is what you want.

Answer 2

The question is a bit hard to understand because it's written in a weird order. Usually the question goes (environment) (specifics) (statement to be proved). If I understand your problem well, I think it would be better written this way:

Let $f$ and $g$ be elements of $\mathrm{Sig}_3$, which has inner product $$ \left = \int_0^{2\pi} f(x)g(x)\,dx $$ [I don't actually know what $\mathrm{Sig}_3$ is, even with the information that Sig stands for signal. But I'm trying to work from the context.] Let \begin{align*} v_0 &= \frac{1}{\sqrt{2}} \\ v_1 &= \cos(x) \\ v_2 &= \cos(2x) \\ v_3 &= \cos(3x) \end{align*} Suppose that with respect to the basis $(v_0, v_1, v_2, v_3)$, $f$ and $g$ have coordinates $x$ and $y$ in $\mathbb{R}^4$. Show that $$ \left = \pi(x_1 y_1 + x_2 y_2 + x_3 y_3 + x_4 y_4) $$

Please correct me if that's not what your problem is. But notice how it starts with the environment of the vector space $\mathrm{Sig}_3$, along the definition of the inner product. Then comes the specific information: for these two elements, we have these two coordinate vectors. Finally, the statement we are to prove.

So I think you will be home free if you understand the meaning of this sentence:

Suppose that with respect to the basis $(v_0, v_1, v_2, v_3)$, $f$ and $g$ have coordinates $x$ and $y$ in $\mathbb{R}^4$.

What does this tell you about the formulas for $f(x)$ and $g(x)$? Once you know this, you should be able to compute $\left$ directly.