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.

  • $\begingroup$ If these are standard inner products, I would try just expanding both of these things by their definitions and seeing if it works out. $\endgroup$ – Alfred Yerger May 5 '15 at 23:16
  • $\begingroup$ Rory Daulton's answer is a standard way of doing things. Mine may perhaps also make some claim to some degree of standardness, but some people feel that his is the adult way, and certainly it is something one must learn to reach mathematical adulthood. Here's my private (crank?) hypothesis: You don't really understand things like his answer until you understand that they're a way of expressing the idea of my answer, but without singling out a concrete example, as I did when I wrote $n=3$. ${}\qquad{}$ $\endgroup$ – Michael Hardy May 6 '15 at 0:29

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


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.)


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 $x_2,x_3,y_1,y_2,y_3$.


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