# Orthonormal basis and inner product

Let $$\langle x, y\rangle$$ be an inner product on a vector space $$V$$, and let $$e_1, e_2, \cdots, e_n$$ be an orthonormal basis for $$V$$.

Question.

Prove: $$\langle a_1e_1 + a_2e_2 + \cdots + a_ne_n, b_1e_1 + b_2e_2 + \cdots + b_ne_n\rangle = a_1b_1 + a_2b_2 + \cdots + a_nb_n$$, where $$a_i,b_i\in F$$.

Since $$\{e_1, e_2, \cdots, e_n\}$$ is an orthonormal set, $$\langle e_i, e_j\rangle = 0$$ when $$i \neq j$$ and $$\langle e_i, e_j\rangle = 1$$ when $$i = j$$. So:

$$\langle a_1e_1 + a_2e_2 + \cdots + a_ne_n, b_1e_1 + b_2e_2 + \cdots + b_ne_n\rangle = \langle a_1e_1, b_1e_1\rangle + \langle a_2e_2, b_2e_2\rangle + \cdots + \langle a_ne_n, b_ne_n\rangle = a_1\overline{b_1} + a_2\overline{b_2} + \cdots + a_n\overline{b_n}$$

Associated question

Prove: $$\langle x, y\rangle = \langle x, e_1\rangle\langle y, e_1\rangle + \langle x, e_2\rangle\langle y, e_2\rangle + \cdots + \langle x, e_n\rangle\langle y, e_n\rangle$$

Similarly, I get:

$$\langle x, y\rangle = \langle x, e_1\rangle\overline{\langle y, e_1\rangle} + \langle x, e_2\rangle\overline{\langle y, e_2\rangle} + \cdots + \langle x, e_n\rangle\overline{\langle y, e_n\rangle}$$

...am I missing something? If yes, I'd really appreciate it if someone would point out what.

• Where did the conjugate symbols come from?
– user403337
Jan 17, 2020 at 4:30

The second follows immediately from the first, using $$x=a_1e_1+\dots+a_ne_n$$ and $$y=b_1e_1+\dots+b_ne_n$$.

There shouldn't be any complex conjugate symbols. You said inner product, not Hermitian inner product.

• Thank you! That's what I was missing.
– ben
Jan 17, 2020 at 5:25

The first answer is right, also, note that \begin{align} \left\langle \sum_{i=1}^n a_ie_i, \sum_{j=1}^n b_je_j \right\rangle &= \sum_{i=1}^n a_i \left\langle e_i, \sum_{j=1}^n b_je_j \right\rangle \\ &= \sum_{i=1}^n \sum_{j=1}^n a_i\overline{b_j} \langle e_i, e_j \rangle \\ &= \sum_{i=1}^n \sum_{j=1}^n a_i\overline{b_j} \delta_{ij} \\ &= \sum_{i=1}^n a_i\overline{b_i} \end{align} where $$\delta_{ij}$$ is the Kronecker's delta: $$\delta_{ij} = 1$$ if $$i=j$$ and $$\delta_{ij} = 0$$ if $$i\neq j$$.

For the second part, you need some more elaboration. Since $$e_1,\dots,e_n$$ is orthonormal, every vector $$x$$ can be written as $$x = \sum_{i=1}^n \langle x,e_i \rangle e_i$$ and similarly, $$y = \sum_{j=1}^n \langle y,e_j \rangle e_j.$$ So, applying the previous result with $$a_i = \langle x,e_i \rangle$$ and $$b_j = \langle y,e_j \rangle$$ we get the desired result.

• My issue concerned the complex conjugate symbols on the terms involving bi and y. Sorry, I should have made that more clear. But my working more-or-less resembles yours, which is reassuring. So thanks!
– ben
Jan 17, 2020 at 5:23