Orthonormal basis proof.

Question

Let $\beta=(v_1,\ldots,v_n)$ be an orthonormal basis for $V$. Show that for any $x,y\in V$,

$$\langle x,y\rangle=\sum_{i=1}^n \langle x,v_i\rangle \overline{\langle y,v_i\rangle}$$

How would you go about this one? I'm a little confused how being orthonormal affects the summation part.

Answer 1

Since $(v_1,\ldots,v_n)$ is a basis, we have

$$x=x_1v_1+\cdots+x_nv_n$$

for some scalars $x_1,\ldots,x_n$. Using the bilinearity of the inner product and orthonormality of the basis, show that

$$\langle x,v_i\rangle=x_i$$

Then let $y=y_1v_1+\cdots+y_nv_n$ for scalars $y_1,\ldots,y_n$. Using the bilinearity of the inner product and orthonormality of the basis, show that

$$\langle x,y\rangle=x_1y_1+\cdots+x_ny_n$$

Convert $\langle x,v_i\rangle=x_i$, and $\langle y,v_i\rangle=y_i$, and the theorem will come out easily.

(I left out the overline: I assume this is the complex conjugate? You should be able to put it in easily, depending on its meaning.)

Answer 2

Let's see how it works in the special case in which $n=3$: \begin{align} \langle x,y\rangle = {} & \langle x_1 v_1+x_2v_2+x_3v_3,\ y_1 v_1 + y_2 v_2 + y_3 v_3 \rangle \\[8pt] = {} & \phantom{{}+{}} \langle x_1 v_1,\ y_1 v_1+y_2 v_2+y_3 v_3\rangle \\[2pt] & {} + \langle x_2 v_2,\ y_1 v_1+y_2 v_2+y_3 v_3\rangle \\[2pt] & {} + \langle x_3 v_3,\ y_1 v_1+y_2 v_2+y_3 v_3\rangle \\[8pt] = {} & \phantom{{}+{}} \langle x_1 v_1,\ y_1 v_1\rangle + \langle x_1 v_1,\ y_2 v_2\rangle + \langle x_1 v_1,\ y_3 v_3\rangle \\[2pt] & {} + \langle x_2 v_2,\ y_1 v_1\rangle + \langle x_2 v_2,\ y_2 v_2\rangle + \langle x_2 v_2,\ y_3 v_3\rangle \\[2pt] & {} + \langle x_3 v_3,\ y_1 v_1\rangle + \langle x_3 v_3,\ y_2 v_2\rangle + \langle x_3 v_3,\ y_3 v_3\rangle \\[8pt] = {} & \phantom{{}+{}} x_1 \overline y_1\langle v_1,v_1\rangle + x_1 \overline y_2\langle v_1,v_2\rangle + x_1 \overline y_3\langle v_1,v_3\rangle \\[2pt] & {} + x_2 \overline y_1\langle v_2,v_1\rangle + x_2 \overline y_2\langle v_2, v_2\rangle + x_2 \overline y_3 \langle v_2,v_3\rangle \\[2pt] & {} + x_3 \overline y_1\langle v_3,v_1\rangle + x_3 \overline y_2\langle v_3, v_2\rangle + x_3 \overline y_3 \langle v_3,v_3\rangle \\[8pt] = {} & \phantom{{}+{}} x_1 \overline y_1\cdot 1 + x_1 \overline y_2 \cdot 0 + x_1 \overline y_3 \cdot 0 \\[2pt] & {} + x_2 \overline y_1\cdot 0 + x_2 \overline y_2 \cdot 1 + x_2 \overline y_3 \cdot 0 \\[2pt] & {} + x_3 \overline y_1\cdot 0 + x_3 \overline y_2 \cdot 0 + x_3 \overline y_3 \cdot 1 \\[8pt] = {} & x_1 \overline y_1+x_2 \overline y_2+x_3 \overline y_3. \end{align}

Is $x_1$ equal to $\langle x,v_1\rangle$? Let's see: \begin{align} \langle x,v_1\rangle & = \langle x_1 v_1+x_2v_2+x_3v_3,v_1\rangle \\[8pt] & = x_1\langle v_1,v_1\rangle + x_2\langle v_2,v_1\rangle + x_3\langle v_3,v_1\rangle \\[8pt] & = x_1\cdot 1 + x_2\cdot0 + x_3\cdot 0 \\[8pt] & = x_1. \end{align} And the same applies to $x_2,x_3,y_1,y_2,y_3$.