# standard matrix using a orthonormal bases

I need a small clarification. I was trying to solve the following question. If $u_1,u_2,....,u_n$ and $v_1,v_2,...,v_n$ are orthonormal bases for $\mathbb{R}^n$. construct the matrix A that transforms each $v_j$ into $u_j$ to give $Av_1=u_1,.... Av_n=u_n.$

So clearly A is an $n*n$ matrix, and $A[v_1 v_2 .. v_n] = [u_1 u_2 ... u_n]$. I don't know how to proceed after that..I think , each $v_j$ is a combination of $T(u_j)'s$ then using the dot product to find the coefficients. Any input is much appreciated.

• Hint: $A\left[ v_1\;v_2\;\dots\;v_n\right]=\left[ u_1\;u_2\;\dots\;u_n\right]$ – karakusc Dec 18 '14 at 11:15
• So A is a diagonal matrix with diagonal entries $u_1/v_1, u_2/v_2,... u_n/v_n$ Am I right? – user142807 Dec 18 '14 at 11:18
• $u_i/v_i$ is not even defined, note that these are vectors. Given this matrix equation, you can use the fact that for a matrix $V$ with orthonormal columns, $VV^T=V^TV=I$, where $I$ is the identity matrix. – karakusc Dec 18 '14 at 11:20
• You should first specify the base(s) of $\mathbb{R}^n$ relevant to set up the matrix. – Ivan Dec 18 '14 at 11:21
• Then, If both orthonormal bases are the standard bases, then A is the identity matrix. – user142807 Dec 18 '14 at 11:24

Convince yourself that if $\;\{w_1,..,w_n\}\;$ is any orthonormal basis, then for any $\;v\in V\;$ we have the easy expression
$$v=\sum_{k=1}^n \langle\,v\,,\,w_k\,\rangle\,w_k$$
$$\forall\;1\le i\le n\;,\;\;v_i=\sum_{k=1}^n\langle\,v_i\,,\,u_k\,\rangle\,u_k\implies$$
$$A=\left(\langle\,v_i\,,\,u_j\rangle\right)_{1\le i,j\le n}=\begin{pmatrix}\langle v_1,u_1\rangle&\langle v_1,u_2\rangle&\ldots&\langle v_1,u_n\rangle\\\ldots&\ldots&\ldots&\ldots\\\langle v_n,u_1\rangle&\langle v_n,u_2\rangle&\ldots&\langle v_n,u_n\rangle\end{pmatrix}$$
• one clarification.. suppose ${(v_1,v_2...v_n)}$, is an orthonormal bases for the input space V, then each $v_i$ can be expressed as a linear combination of the bases vectors. and all the coefficients are zero because of linear independence property. Am I right? – user142807 Dec 18 '14 at 11:47
• Of course, but not all coefficients: for any $\;i\;$ , $\;v_i=0\cdot v_1+\ldots+\color{red}1\cdot v_i+\ldots+0\cdot v_n\;$ – Timbuc Dec 18 '14 at 11:52