Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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 $P\cdot [x]_B = x$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.