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That is we have a linear transformation, i.e. an $ n\times n $ matrix $A$, such that for every pair of vectors $ v $ and $ w $ we have

$$ \langle v,w\rangle=0 \ \ \ \implies \ \ \ \ \ \langle Av,Aw \rangle = 0 $$

What are all such matrices?

I've got this solution: [The idea is to first prove that such a matrix is necessarily angle preserving, from which then by Angle preserving linear maps we get the answer.] If $A$ is not invertible, then the question is invalid, technically, so we deal with invertible maps only. Take any two vectors, and use Gram-Schmidt to orthoganalize them $$ z:=v- \frac{\langle v,w\rangle}{\langle w,w\rangle} \ w \implies \ \langle z,w\rangle=0.$$ -- which can be confirmed by directly writing out their dot product.

So, by assumption $$ \langle Az,Aw\rangle = 0 $$ Writing out: $$ 0= \langle Az,Aw\rangle \implies \langle Av,Aw\rangle=\frac{\langle v,w\rangle}{\langle w,w\rangle} \langle Aw,Aw\rangle $$

But this holds for any two vectors, in particular when you exchange roles of $v$ and $w$: So for any two $v, w $ $$ \langle Av,Aw\rangle=\frac{\langle v,w\rangle}{\langle w,w\rangle} \langle Aw,Aw\rangle =\frac{\langle w,v\rangle}{\langle v,v\rangle} \langle Av,Av\rangle \tag{$*$}$$

In the middle equality, divide (we will see that at the end the equality we want holds, even when $\langle v,w\rangle=0$, so this is safe!) to get $$ \frac{\langle Aw,Aw\rangle}{\langle w,w\rangle} = \frac{\langle Av,Av\rangle}{\langle v,v\rangle} $$ From which we get $$ \frac{\langle Aw,Aw\rangle}{\langle w,w\rangle}=\frac{\sqrt{\langle Aw,Aw\rangle}}{\sqrt{\langle w,w\rangle}}\frac{\sqrt{\langle Av,Av\rangle}}{\sqrt{\langle v,v\rangle}} .$$

Now replace in the middle equality of $(*)$ and rearrange to get $$ \frac{\langle Av,Aw\rangle}{\sqrt{\langle Aw,Aw\rangle}\sqrt{\langle Av,Av\rangle}}=\frac{\langle v,w\rangle}{\sqrt{\langle w,w\rangle}\sqrt{\langle v,v\rangle}} .$$

But see the denominators:

$$ \frac{\langle Av,Aw\rangle}{\|Aw\|\|Av\|}=\frac{\langle v,w\rangle}{{\|w\|\|v\|}} \tag{$**$}$$

Notice that the division above was safe, since for orthogonal pairs this already holds from the hypothesis.

But $(**)$ just means that the map $A$ is angle preserving (not just right angle preserving!)

Now see the beautiful proof in the link above that all and the only angle preserving maps are (nonzero) multiples of an orthogonal matrix.

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Behnam Esmayli
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    Why would $A$ not invertible make the question "invalid"? The matrix $A=0$ has the property you ask about... – David C. Ullrich May 03 '16 at 00:46
  • You're completely right. One reason is that I was working on Riemannian metrics, and this descended from my attempt to show something about conformal metrics, where metrics are positive definite, hence invertible. Here, we may work on, say the orthogonal complement of the null space, and prove that on this subspace we will have $A$ in this form, and zero on null-space of course. Though, I have not checked if details will work out nicely. But thanks for the question. – Behnam Esmayli May 03 '16 at 02:17

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Actually! I got another proof based on an answer I had previously on a related question, without directly using the angle preserving property.

Prove that there is a positive real number $\lambda$, so that $A =\lambda B$, for two positive definite square matrices

In that question, I proved the following claim. For positive definite matrices $A,B$, if $∀v, w ∈ R^n : v · Aw = 0 ⇔ v · Bw = 0.$, then there is a positive scalar $\lambda$ making $A=\lambda B$.

We can apply that conclusion directly to this proof. Since $∀v, w ∈ R^n : v^T w = 0 ⇔ v^TA^T Aw = 0.$. As you suggested, we only consider invertible $A$, then $A^TA$ is a positive definite matrix.

Using the result from the previous post, we know there exist scalar $\lambda$ that makes $A^TA = \lambda I$. Then we know $A$ can only be a positive multiple of an orthogonal matrix.

Binxu Wang 王彬旭
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