Coordinate vectors relative to an orthonormal basis

Question

Let's say that we have $B$ = $(u$_$1$$, ..., u$_$n$$)$, which is an ordered orthonormal basis for $\mathbb{R}^n$. We then have an $n \times n$ matrix $U$ = $(u$_$1$$, ..., u$_$n$$)$. Then, for any $v$ in $\mathbb{R}^n$, would its coordinate vector relative to $B$ be $U^Tv$? (basically, would $v$_$B$ = $U^Tv$?)

Answer 1

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If $\vec{v} \in \mathbb{R}^n$,

let $\vec{v}= c_1 \cdot \vec{u_1} +\cdots +c_n \cdot \vec{u_n} = U\begin{bmatrix}c_1 \\ \vdots \\ c_n\end{bmatrix} = U\cdot\vec{v_B}$

$\vec{v_B} = U^{-1}\cdot \vec{v}$

but $B$ is an orthonormal basis, $U^{-1} = U^T$, so yes.

You can see that $U^{-1} =U^T$ by

$$U^T \cdot U=\begin{bmatrix}\vec{u_1} \\ \vdots \\ \vec{u_n} \end{bmatrix} \begin{bmatrix}\vec{u_1} |& \vec{u_2}|& \cdots & |\vec{u_n}\end{bmatrix} = \begin{bmatrix} (\vec{u_1}\cdot\vec{u_1})& (\vec{u_1}\cdot\vec{u_2})&\cdots &( \vec{u_1}\cdot\vec{u_n} )\\(\vec{u_2}\cdot\vec{u_1})& (\vec{u_2}\cdot\vec{u_2})&\cdots &( \vec{u_2}\cdot\vec{u_n}) \\ \vdots& \vdots&\ddots & \vdots \\ (\vec{u_n}\cdot\vec{u_1})& (\vec{u_n}\cdot\vec{u_2})&\cdots & (\vec{u_n}\cdot\vec{u_n})\end{bmatrix}_{n\times n}$$ $$= I_{n\times n}$$

because by definition $\vec{u_i}\cdot\vec{u_j} =1$ if $i=j$, and $\vec{u_i}\cdot\vec{u_j} =0$ if $i\not=j$.