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- Feb 5, 2012

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Here's one of the questions that I encountered recently along with my answer. Let me know if you see any mistakes. I would really appreciate any comments, shorter methods etc.

Problem:

Let \(u,\,v\) be two vectors in a Euclidean space \(V\) such that \(|u|=|v|\). Prove that there is an orthogonal transformation \(f:\, V\rightarrow V\) such that \(v=f(u)\).

Solution:

We assume that \(u\) and \(v\) are non zero. Otherwise the result holds trivially.

Let \(B\) denote the associated symmetric bilinear function of the Euclidean space. Let us define the linear transformation \(f\) as,

\[f(x)=\begin{cases}x&\mbox{if}&x\neq u\\v&\mbox{if}&x=u\end{cases}\]

It's clear that, \(B(f(x),\,f(y))=B(x,\,y)\) whenever \(x,\,y\neq u\). Also \(B(f(u),\,f(u))=B(v,\,v)\) and since \(|v|=|u|\Rightarrow B(v,\,v)=B(u,\,u)\) we have \(B(f(u),\,f(u))=B(u,\,u)\).

It remains to show that, \(B(f(x),\,f(u))=B(x,\,u)\) for \(x\neq u\).

\[B(f(v+u),\,f(v+u))=B(f(v),\,f(v))+2B(f(v),\,f(u))+B(f(u),\,f(u))\]

Also since \(v+u\neq u\),

\[B(f(v+u),\,f(v+u))=B(v+u,\,v+u)=B(v,\,v)+2B(v,\,u)+B(u,\,u)\]

Using the above two results and the fact that \(B(u,\,u)=B(v,\,v)\) we get,

\[B(f(v),\,f(u))=B(v,\,u)\]

Now consider \(B(f(x+v),\,f(x+u))\).

**Case I: \(x+v\neq u\)**

\[B(f(x+v),\,f(x+u))=B(f(x),\,f(x))+B(f(x),\,f(u))+B(f(v),\,f(x))+B(f(v),\,f(u))\]

Also,

\[B(f(x+v),\,f(x+u))=B(x+v,\,x+u)=B(x,\,x)+B(x,\,u)+B(x,\,v)+B(v,\,u)\]

Using the above two results and the fact that \(B(f(v),\,f(u))=B(v,\,u)\) we get,

\[B(f(x),\,f(u))=B(x,\,u)\]

**Case II: \(x+v=u\)**

\[B(x,\,v)=B(u-v,\,v)=B(u,\,v)-B(v,\,v)\]

\[B(x,\,u)=B(u-v,\,u)=B(u,\,u)-B(v,\,u)\]

Therefore, \[B(x,\,u)=-B(x,\,v)~~~~~~(1)\]

\[B(f(x),\,f(u))=B(f(u-v),\,f(u))=B(f(u),\,f(u))-B(f(v),\,f(u))=B(v,\,v)-B(v,\,v)=0\]

Then since \(B(f(x),\,f(u))=B(x,\,v)=0\) by (1) we get \(B(x,\,u)=0\)

\[\therefore B(f(x),\,f(u))=B(x,\,u)\]