Let $v,w \in R^n$ with $||u||=||w||=1$. Prove that there exists orthogonal matrix $A$ such that $Av=w$.

Question

Let $v,w \in \mathbb R^n$ with $\|u\|=\|w\|=1$. Prove that there exists orthogonal matrix $A$ such that $Av=w$. Also prove that $A$ can be chosen such that $\det(A)=1$

Here, $u\neq 0$ and $w\neq 0$ so I can extend these to get

$B_1 = \{u,x2,x3\dots x_n\}$ and $B_2=\{w,y2\dots y_n\}$ both orthonormal basis of $R^n$

and define $T:R^n \to R^n $ as $T(u)=w$ and $T(x_i)=y_i \;\; \forall i\leq i \leq n$

here will $A=[T]_{B_1}^{B2}$ satisfy the given conditions?

I can't think of another way to solve this. please help

Answer 1

You need no other way! That works. The matrix of a linear map $f$ with respect to two orthonormal bases is orthogonal if and only if, if $\{v_1,\ldots,v_n\}$ is the first basis, then $\{f(v_1),\ldots,f(v_n)\}$ is an orthonormal basis too. In order to have determinant $1$, you may have to change the sign of the image of one of the vectors.

Answer 2

What you have done is correct. To get determinant $1$ just note that if you replace $y_n$ by $-y_n$ (keeping everything else fixed) then determinant of the matrix becomes the negative of the old one. Hence we can get $A$ with $det(A)=1$.

The determinant of a real orthogonal matrix is $+1$ or $-1$.