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Let $B$ = {$\vec v_1,\ldots,\vec v_n$} be an orthonormal basis for $R^n$ and let $P$ = [$\vec v_1 \dots \vec v_n$]. Show that for any $\vec x \in \mathbb R^n$ we have $[\vec x]_B = P^T \vec x$

Should I begin by rewriting the right hand side as a inner product?

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Yes, you should rewrite the RHS an inner product. Since $\{v_1, \ldots, v_n\}$ is a basis, how can you relate $x$ to that basis? –  Christopher A. Wong Oct 7 '12 at 22:07
    
@ChristopherA.Wong: I know that $[\vec{x}]_B$ = [$c_1 ... c_n$] where $\vec{x} = c_1v_1 + ... + c_nv_n$ oh, and $\vec{x} = c_1v_1 + ... c_nv_n$ since {$v_1,...,v_n$} is a basis. I think i might have it now. –  cdk Oct 7 '12 at 22:11
    
Do I know anything useful about $<P,v_i>$? –  cdk Oct 7 '12 at 22:19
    
inner product? $P$ is a matrix. Syntax error –  Berci Oct 7 '12 at 22:21
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1 Answer 1

up vote 1 down vote accepted

This requires 2 things:

  1. For any basis $B=(v_1,\dots,v_n)$ and for their matrix $P$, one has $$P^{-1}x = [x]_B$$
  2. If $B$ is orthonormal, then $P^T=P^{-1}$. For this, inner product is hidden in row-column products when you calculate $P^TP$.

This 1. can be shown by writing up $P\cdot\begin{bmatrix} \alpha_1\\ \alpha_2\\ \vdots \\ \alpha_n \end{bmatrix} = \alpha_1v_1+\alpha_2v_2+\dots+\alpha_nv_n$, so, by definition of $[x]_B$, it yields $P\cdot [x]_B = x$.

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I did not know (1), and the proof follows immediately after. Could you prove that result? –  cdk Oct 7 '12 at 22:39
    
Chris, does writing it as $x=P[x]_B$ help? That's practically the definition of the notation $[x]_B$. –  Gerry Myerson Oct 7 '12 at 22:49
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