orthonormal basis for $\mathbb{R}^n$

Question

Let $\{u_1,u_2,\dots, u_n\}$ be an orthonormal basis for $\mathbb{R}^n$.

Show that

$$ \Vert v \Vert^2 = (v\cdot u_1)^2+(v\cdot u_2)^2+\dots+(v\cdot u_n)^2.$$

By the properties of a basis, $v$ can be written as $v=c_1u_1+c_2u_2+\dots+c_nu_n$.

I tried this

Answer 1

Since $\{u_1,\cdots, u_n\}$ is a basis for $\mathbb{R}^n$ then these vectors span $\mathbb{R}^n$ and we can write $$v=c_1 u_1+\cdots c_n u_n \qquad (1)$$ for some $c_i \in \mathbb{R}$.

Take the inner-product (the scalar product) of both sides of (1) by $u_i$ to get that

$$v\cdot u_i=c_1 u_1\cdot u_i+\cdots c_i u_i\cdot u_i+\cdots c_n u_n\cdot u_i= c_i $$ becaue $u_i\cdot u_i=1$ and $u_j\cdot u_i=0$ if $j\neq i$.

Therefore, by (1), we have the representation $$v= (u_1\cdot v) u_1+\cdots (u_n\cdot v) u_n.$$

Now, $$\|v\|^2= v\cdot v= \left((u_1\cdot v) u_1+\cdots (u_n\cdot v) u_n\right)\cdot\left( (u_1\cdot v) u_1+\cdots (u_n\cdot v) u_n\right)$$ $$=\sum_{i=1}^{n}\sum_{j=1}^{n} (u_i\cdot v)(u_j\cdot v) u_i \cdot u_j$$ $$=\sum_{i=1}^{n} (u_i\cdot v)(u_i\cdot v) u_i \cdot u_i= \sum_{i=1}^{n} (u_i\cdot v)^2.$$

Answer 2

For each $i\in\{1,\ldots,n\}$, let $u^i\colon \mathbf{R}^n\to\mathbf{R}$ be the unique linear function satisfying $$u^i(u_j)=\delta_{ij}$$ for all $i,j\in\{1,\ldots,n\}$. Then $$v=\sum_{i=1}^n u^i(v) u_i$$ for all $v\in\mathbf{R}^n$. This is true for any basis. Since we are considering an orthonormal basis, it follows from our definition of $u^i$ that $u^i(v)=\langle u_i, v\rangle$. Thus, \begin{align} \Vert v \Vert^2 =\langle v,v\rangle=\langle\sum_{i=1}^n \langle u_i, v\rangle u_i,\sum_{j=1}^n \langle u_j, v\rangle u_j\rangle=\sum_{i=1}^n\sum_{j=1}^n \langle u_i, v\rangle \langle u_j, v\rangle\langle u_i, u_j\rangle\\=\sum_{i=1}^n\sum_{j=1}^n \langle u_i, v\rangle \langle u_j, v\rangle\delta_{ij}= \sum_{i=1}^n\langle u_i,v\rangle^2 \end{align}