Prove : Matrix is orthogonal after transformation

Question

I need your help to prove the statement below, i tried several ways with out success. I apologize for the bad translation, I hope it is clear :

Let B be NxN matrix which transforms orthonormal base u1,u2,......un to other orthonormal base v1,v2,.....vn.

(It means that for all 1 ≤ i < n , exists Bui = vi)

According to this statement I want to prove that B is orthogonal Matrix.

Answer 1

Being an orthonormal base $\{v_i\}_{i=1}^n$ means spanning $\mathbb{R}^n$ and satisfying $\langle v_i, v_j \rangle=\delta_{ij}$. We want to show that $B$ is orthogonal, that means that for any $x,y \in \mathbb{R}^n$ we want to show that $\langle Bx, By\rangle=\langle x,y \rangle$. First of all write $x,y$ in terms of the orthonormal basis $\{u_i\}_{i=1}^n$: $$x=\sum_{i=1}^nc_iu_i, \, y=\sum_{j=1}^n d_j u_j $$ where $c_i,d_i \in \mathbb{R}$ for all $i,j=1,\dotso,n$. Now observe:

$$\langle Bx,By\rangle=\langle B(\sum_{i=1}^nc_iu_i),B(\sum_{j=1}^nd_ju_j)\rangle=\sum_{i=1}^n\sum_{j=1}^nc_id_j\langle Bu_i,Bu_j\rangle=\sum_{i=1}^n\sum_{j=1}^nc_id_j\langle v_i,v_j\rangle= \\\sum_{i=1}^n\sum_{j=1}^nc_id_j\langle u_i,u_j\rangle=\langle\sum_{i=1}^nc_iu_i,\sum_{j=1}^nd_ju_j\rangle=\langle x,y\rangle $$

Answer 2

"$Bu_i=v_i$ for all $i$" can be written under the compact form:

$$\tag{0}BU=V$$

where $U$ (resp. $V$) is the matrix whose columns are the $u_i$s (resp. the $v_i$s).

A matrix $M$ whose columns constitute an orthonormal basis is orthogonal (see explanation in the remark below)

Thus, $U$ and $V$ are orthogonal, and, as the set of orthogonal matrices forms a group (the orthogonal group), $(1)$, written under the form:

$$B=VU^{-1}$$

is meaningful and proves that $B$ is orthogonal.

Remark: the fact that an orthogonal matrix is characterized by the fact that its columns constitute an orthonormal basis is equivalent to the defining property of an orthonormal matrix, i.e.,

$$\tag{1}M^TM=I.$$

Proof: Let us consider the LHS of $(1)$; its generic element $a_{ij}$ is the product of line $L_i$ of $M^T$ with column $C_j$ of $M$, otherwise said, as the dot product of column $C_i$ with column $C_j$ of $M$. This dot product is $0$ if $i \neq j$, and $1$ in the case $i=j$ ; we have found back, in this way matrix $I$, ther RHS of $(1).$