Expressing an inner product in terms of an orthonormal basis

199 Views Asked by At

Let $\{ v_1, v_2, v_3, \dotsc, v_n \}$ be an orthonormal basis of $V$. Show that for any vectors $w$ and $z$ of $V$: $$ \langle w,z \rangle = \sum_{k=1}^{n} \langle w,v_k \rangle \langle v_k,z \rangle. $$

2

There are 2 best solutions below

2
On BEST ANSWER

We have that

  • $w=\sum_{i=1}^{n} a_i v_i$
  • $z=\sum_{j=1}^{n} b_j v_j$

then

$$\langle w,z \rangle=\langle\sum_{i=1}^{n} a_i v_i,\sum_{j=1}^{n} b_j v_j \rangle=\sum_{k=1}^{n} a_kb_k=\sum_{k=1}^{n} a_kv_k^T\cdot v_kb_k= \sum_{k=1}^{n} \langle w,v_k\rangle\langle v_k,z\rangle$$

0
On

Since $\{v_1, \ldots, v_n\}$ is an orthonormal basis, we have $\langle v_k, v_j\rangle = \delta_{kj}$ and $x = \sum_{i=1}^n \langle x, v_i\rangle v_i$ for all $x \in V$.

Therefore $$\langle w,z \rangle= \left\langle \sum_{k=1}^n \langle w, v_k\rangle v_k, \sum_{j=1}^n \langle z, v_j\rangle v_k\right\rangle = \sum_{k=1}^n\sum_{j=1}^n \langle w,v_k\rangle\langle v_j,z\rangle \underbrace{\langle v_k, v_j\rangle}_{\delta_{kj}} = \sum_{k=1}^{n} \langle w,v_k\rangle\langle v_k,z\rangle$$