summing inner product of orthonomal basis

Question

I need some help with some very basic linear algebra when doing calculations in inner product space.

Here is a line I got lost when reading a book... \begin{align*} (x,e_m)=\left(\sum_{n=1}^k\lambda_n e_n,e_m\right)=\sum_{n=1}^k\lambda_n \left(e_n,e_m\right)=\lambda_m,\quad m=1,\cdots, k. \end{align*}

Could anyone show me explicitly how $\sum_{n=1}^k\lambda_n \left(e_n,e_m\right)=\lambda_m?$

I got $\sum_m\lambda_m$ here.

Answer 1

If $e_1,\cdots,e_k$ is an orthonormal basis then $\langle e_i,e_j\rangle$ is $1$ when $i=j$ and $0$ when $i\ne j$.

Consider the case with $k=4$ and $m=3$. We have

$$\lambda_1\langle e_1,e_3\rangle+\lambda_2\langle e_2,e_3\rangle+\lambda_3\langle e_3,e_3\rangle+\lambda_4\langle e_4,e_3\rangle $$

$$=\lambda_1\cdot0+\lambda_2\cdot0+\lambda_3\cdot1+\lambda_4\cdot0=\lambda_3. $$

Answer 2

For an orthonormal basis, $(e_n,e_m) = \delta_{nm}$. $\delta_{nm} = 1$ when $n=m$ and $0$ when $n\neq m$. What happens when you plug this into your sum?