If $T:\mathbb{R^2}\to \mathbb{R^2}$ is a linear transformation such that $\langle x,y \rangle =0 \implies \langle T(x),T(y) \rangle =0 $ for each $x,y \in \mathbb{R^2} $, show that $T=aS$ ,where $S:\mathbb{R^2}\to \mathbb{R^2}$,is an orthogonal transformation.

My attempt.

Instead of showing $T$ orthogonal I have decided to show $S=\frac{1}{a}T$ orthogonal.

For this we need to show that $\langle S(x),S(x) \rangle = \langle x,x \rangle $ .So I let $x=x_1e_1+x_2e_2$.But I am unable to get rid of $a$.Is my approach right?

  • $\begingroup$ Your approach assumes that $a\neq0$, but $T$ might be the zero operator. $\endgroup$ – symplectomorphic Aug 28 '16 at 8:10
  • $\begingroup$ @ thanasissdr We don't know about orthogonality of $S$. $\endgroup$ – user114873 Aug 28 '16 at 8:21
  • $\begingroup$ @user114873 Note that $T$ is not orthogonal, in general (you have written so in your question) $\endgroup$ – 57Jimmy Aug 28 '16 at 8:43
  • $\begingroup$ @user114873 Your approach does not work because you need a specific $a$, not any one (see answer below). $\endgroup$ – 57Jimmy Aug 28 '16 at 8:46
  • $\begingroup$ This is also true for $\mathbb{R}^n$ (or $\mathbb{C}^n$ and unitary operators) - see math.stackexchange.com/questions/1790611/…. $\endgroup$ – levap Aug 28 '16 at 9:08

First of all, notice that if $x_1y_1 + x_2y_2 =0$, then by assumption $x_1y_1\langle T(e_1),T(e_1) \rangle + x_2y_2\langle T(e_2),T(e_2) \rangle =0.$ In particular, for $x= \left( \begin{array} \ 1 \\ -1 \end{array} \right) $ and $y= \left( \begin{array} \ 1 \\ 1 \end{array} \right) $ we get $\langle T(e_1),T(e_1) \rangle = \langle T(e_2),T(e_2) \rangle$. Now set $a:= \sqrt{\langle T(e_1),T(e_1) \rangle} = ||T(e_1)||$ and verify that $\frac{1}{a}T$ is indeed orthogonal.

  • $\begingroup$ Is it okay to take $x=(1,-1)$ and $y=(1,1)$? Won't it be a particular case? $\endgroup$ – user114873 Aug 28 '16 at 9:02
  • $\begingroup$ No, it is a true fact that the equality must hold for all values that satisfy $x_1y_1 + x_2y_2$, which is the case for this two special values. It is simply "providential" that with these special values we get exactly the equality we wanted, but the equality itself holds in general. $\endgroup$ – 57Jimmy Aug 28 '16 at 14:57
  • $\begingroup$ Note that this kind of proceding is quite common in linear algebra: to show that $ (\forall x,y: x^TAy=x^TBy) \Rightarrow A=B$ you can use for example the particular vectors $x=e_i$, $y=e_j$, which allow you to deduce that for all $i,j$: $A_{ij}=B_{ij}$, and thus $A=B$. You have used specific values, but the equality $A=B$ holds true in general. It is not a statement about those particular values, they were simply useful to prove it. $\endgroup$ – 57Jimmy Aug 28 '16 at 15:01
  • $\begingroup$ Thank you very much for this details.But I had to take $a^2:= \langle T(e_1),T(e_1) \rangle$ $\endgroup$ – user114873 Aug 28 '16 at 16:04
  • $\begingroup$ Sure, that's true, I'm sorry. I've edited it. $\endgroup$ – 57Jimmy Aug 28 '16 at 19:30

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