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Writing the ortogonality condition for any element of O(n), I've arrived to:

If we take n=2, we know that $\Lambda\Lambda^{T}=\mathbb{I}$, so:

$$\begin{pmatrix} x & y \\ z & t \end{pmatrix} \begin{pmatrix} x & z \\ y & t \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $$

This yields:

$$\begin{cases} x^{2} + y^{2} = 1\\ xz + yt = 0 \\ z^{2} + t^{2}= 1 \end{cases}$$

Visually, these equations mean that we can find two orthonormal vectors.

We could generalise our reasoning to arbitrary dimension easily.

How could I prove rigorously, without plugging in any numbers nor functions, that this system of equations has solution (infinite, in fact)?

Is there any method to prove the existence of solution of non non linear equations (or systems)?

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  • $\begingroup$ How can you prove rigorously, without plugging in any numbers or functions, that the equation $x+y=3$ has solutions (infinitely many, in fact)? $\endgroup$
    – Gerry Myerson
    May 24, 2013 at 10:13
  • $\begingroup$ @GerryMyerson I think that in your case you can isolate a term and see that: $x=3-y$. But for a nonlinear system there isn't usually a way to isolate the variables. $\endgroup$
    – jinawee
    May 24, 2013 at 10:23
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    $\begingroup$ Well, let's see: you can do $x=-yt/z$, so $(yt/z)^2+y^2=1$, which is $y^2(t^2+z^2)=z^2$, which becomes $y^2=z^2$, so $y=\pm z$, and then you get $x=\pm t$, shouldn't be too hard to isolate a term from there. $\endgroup$
    – Gerry Myerson
    May 24, 2013 at 10:31
  • $\begingroup$ In general, the matrix $G=M^TM$ is the Gramian matrix of the columns of $M$. So the latter form an orthonormal set iff $G$ is the identity, i.e. $M$ is orthogonal. $\endgroup$
    – Julien
    May 24, 2013 at 22:00

2 Answers 2

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The first and third equations states that points (x, y) and (z, t) are on a unit circle. Let's write that as: $x = \cos\phi$

$y = \sin\phi$

$z = \cos\psi$

$t = \sin\psi$

Now we may write the second equation as:

$\cos\phi\cos\psi + \sin\phi\sin\psi = 0$

This may be rewritten as:

$\cos (\phi-\psi) = 0$

Which has a solution:

$\phi - \psi = \frac{\pi}{2} + k\pi$

We should restrict the angles to $(-\pi, \pi)$:

$\phi = \psi \pm \frac{\pi}{2}$

Which means that for each (x,y) on a unit circle there are two possible (z, t).

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Your system for arbitrary dimension $n$ is equivalent to finding an orthonormal basis in $\mathbb R^n$ (put its elements as solumns in your matrix), which, in its turn is trivial.

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