This is a problem from C.W. Curtis Linear Algebra. It goes as follows:

"Let V a vector space over R and let T a linear transformation, $T:V\mapsto V$ that preserves orthogonality, that is $(Tv,Tw)=0$ whenever $(v,w)=0$. Show that T is a scalar multiple of an orthogonal transformation."

My approach was to see the effect of $T$ to an orthonormal basis. So I started with the question if it was possible that could exists such a $T$ that could map an element of the orthonormal basis to the zero vector. But I have not been able to produce a contradiction. Because if seems to me that could be a case, and if right there could not be an orthogonal transformation $S$, such that $T=\lambda S$, for some $\lambda \in \mathbb{R}$.

Hope you can help.

  • $\begingroup$ do you assume $V$ is finite dimension? $\endgroup$
    – Red shoes
    Jul 12, 2017 at 0:52
  • $\begingroup$ Yes, I forgot that $\endgroup$
    – Rogelio
    Jul 12, 2017 at 0:52
  • $\begingroup$ If $V$ is finite dimensional, then just pick an orthogonal basis and use it to represent arbitrary $v,w\in V$ and the claim will immediately follow. $\endgroup$
    – dezdichado
    Jul 12, 2017 at 0:54
  • $\begingroup$ You're right -- $T$ could map one of the basis vectors to the zero vector -- when $T = \lambda I$, and $\lambda = 0$, for example. So that's one of the cases you have to handle. :) $\endgroup$ Jul 12, 2017 at 1:01

1 Answer 1


(Im assuming finite dimension too)

Let's have an orthonormal basis $(e_1,e_2,...,e_n)$.

We can verify that for any $i,j$ we have : $(e_i+e_j)\perp (e_i-e_j)$

$\langle e_i+e_j,e_i-e_j\rangle=||e_i||^2-\langle e_i,e_j\rangle+\langle e_j,e_i\rangle-||e_j||^2=1-0+0-1=0$

So when we apply $T$ to these vectors we get:

$\langle T(e_i+e_j),T(e_i-e_j)\rangle=||Te_i||^2-\underbrace{\langle Te_i,Te_j\rangle}_0+\underbrace{\langle Te_j,Te_i\rangle}_0-||Te_j||^2=0$

So $||Te_i||=\lambda$ is constant.

Now for a random vector $x=\sum\limits_{i=1}^n x_ie_i\quad$ it is easy to verifies that

$||Tx||^2=\sum\limits_{i=1}^n x_i^2\underbrace{||Te_i||^2}_{\lambda^2}+2\sum\limits_{i=1}^n\sum\limits_{j>i}^n x_ix_j\underbrace{\langle Te_i,Te_j\rangle}_0=\lambda^2||x||^2$

So $||Tx||=\lambda||x||$ since we deal with all positive quantities here.

  • If $\lambda=0$ the OP statement is trivially verified since $T=0\,I$.

  • Else $S=\frac 1\lambda T\ $ is an isometry and since it respects orthogonality, then $S$ is orthogonal.


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