An orthogonal matrix which sends a vector to other vector with same length.

Question

I know that For every to vectors $u,v\in \mathbb R^n$ where $|u|=|v|$ there exists an orthogonal matrix $A$ such that, $Au=v$. I have a problem so construct this matrix. is there any method to construct this matrix?

Answer 1

Do you know Gram-Schmidt? If so, complete $u$ to an orthogonal basis $\mathcal{B}_1$ and do the same to $v$ (let's call its o.b. $\mathcal{B}_2$), and take the linear map which sends one to the other.

In more detail, note that if $\Vert u\Vert=0$ the result is trivial. If not, complete $u$ to any basis $\{u,e_2,e_3,\cdots,e_n\}$ of $\mathbb{R}^n$, and do the same for $v$: $\{v,f_2,f_3,\cdots,f_n\}$. Apply Gram-Schmidt to both, yielding orthonormal bases $$\mathcal{B}_1=\{\widetilde{u},\widetilde{e_2},\widetilde{e_3},\cdots, \widetilde{e_n}\}$$ and $$\mathcal{B}_2=\{\widetilde{v},\widetilde{f_2},\widetilde{f_3},\cdots, \widetilde{f_n}\},$$ where $\widetilde{u}=\frac{u}{\Vert u\Vert}$ and $\widetilde{v}=\frac{v}{\Vert v\Vert}$. Consider the linear transformation $A$ which sends the elements of $\mathcal{B}_1$ to $\mathcal{B}_2$ "orderly" (i.e., $\widetilde{u} \mapsto \widetilde{v}$ and $\widetilde{e_i} \mapsto \widetilde{f_i}$ for all $i$). This is by construction an orthogonal linear map, and thus will have an orthogonal matricial representation in the canonical basis*. Note also that since $A\widetilde{u}=\widetilde{v}$ and $A$ is linear, we have that $$A(u)=A(\Vert u\Vert \cdot\widetilde{u})=\Vert u \Vert \cdot \widetilde{v}=\Vert v \Vert \cdot \widetilde{v}=v.$$

As a sidenote, note that such a matrix will depend on the initial bases you take.

*Just to be extremely clear, since you explicitly say in the comments that you are dealing with the definition that $A^TA=AA^T=I$, note that this is equivalent to asking that the collumn vectors of the matrix are orthonormal. The collumn vectors in the canonical basis are precisely $A\cdot E_i$, where $\{E_i\}$ is the canonical basis. Therefore, we must show that $$\langle A E_i,A E_j \rangle=\delta_{i,j}.$$ We will prove more generally that $\langle Av, Aw \rangle=\langle v,w\rangle$ (which is usually the definition of an orthogonal map). To see this, note that (letting $\widetilde{e_1}:=\widetilde{u}$ and $\widetilde{v_1}:=\widetilde{v}$). \begin{align*} \langle Av,Aw \rangle&= \langle A\left(\sum c_i\widetilde{e_i}\right),A\left(\sum d_j\widetilde{e_j}\right) \rangle \\ &=\langle \sum c_iA\widetilde{e_i},\sum d_jA\widetilde{e_j} \rangle \\ &=\langle \sum c_i\widetilde{f_i},\sum d_j\widetilde{f_j} \rangle \\ &=\sum c_id_i \\ &=\langle \sum c_i\widetilde{e_i},\sum d_j\widetilde{e_j} \rangle \\ &=\langle v, w \rangle. \end{align*} Another argument could be to use the above fact that $\langle Av,Aw\rangle=\langle v, w \rangle$ for all $v,w$ to conclude that $\langle A^TA v,w \rangle=\langle v, w\rangle$ for all $v,w$, and thus $\langle (A^TA-I)v,w \rangle=0$ for all $v,w$. Fixing any $v$ and taking $w:=(A^TA-I)v$, we have that $\Vert (A^TA-I)v\Vert^2=0$. Thus, $A^TA-I=0$, and then $A^TA=I$ (analogously, you have that $AA^T=I$).

Answer 2

Assuming that $u\ne v$, a reflection in the hyperplane that bisects $u$ and $v$ will do the trick. This hyperplane is normal to $u-v$ and passes through the origin. Using a well-known formula for this reflection, we get $$A = I - 2{(u-v)(u-v)^T \over (u-v)^T(u-v)}.$$ It’s easy to verify that $A=A^T$ and that $A^TA = A^2 = I$, so this matrix is indeed orthogonal. With a bit more work you can also verify that $Au=v$.