Let $T:\Bbb R^3\to \Bbb R^3$ be an orthogonal transformation such that $\det T=1$ and $T$ is not the identity transformation .Let $S\subset \Bbb R^3$ be the unit sphere .

Show that $T$ fixes exactly $2$ points on $S$.

My effort: In order to show that $T$ fixes two points we have to show that there exists two eigen vectors corresponding to the eigen value $1$.

Since $T$ is an orthogonal transformation on $\Bbb R^3$ all the eigen values are of unit modulus and one of them must be real.

But how can I show that $1$ will be an eigen value of $T$? I am feeling confused.


2 Answers 2


First, we need to show that $1$ is an eigenvalue of $T$. If this was not the case, then $-1$ would be an eigenvalue of $T$. There are two cases:

  1. All eigenvalues are real. In this case, $-1$ is the only eigenvalue.

  2. At least one eigenvalue, $\lambda$, is not real. Then $\overline{\lambda}$ is the other eigenvalue.

In any case, we'd obtain $\det T=-1$, a contradiction.

If $v$ is an eigenvector of norm $1$, then $-v$ is also an eigenvector of norm $1$. This gives us at least two fixed points.

Now suppose there was another fixed point, say $w$. Then $w$ and $v$ are LI, which means that the space $V$ generated by $v$ and $w$ has dimension $2$, and all points of $V$ are fixed by $T$.

Let $z\in V^\perp$. Then $Tz\in T(V)^\perp=V^\perp$. Since $V^\perp$ has dimension $1$ and $\Vert Tz\Vert=\Vert z\Vert$, we have either $Tz=z$, in which case we'd obtain $T=Id$, a contradiction, or $Tz=-z$, in whihc case we'd obtain $\det T=-1$ (look at the matrix of $T$ in the basis $\{v,w,z\}$), another contradiction.

Therefore, $v$ and $-v$ are the only fixed points of $T$ in the sphere.

  • $\begingroup$ I am having some problems understanding a few facts:1.Why will be $\{w,v\}$ be linearly independent? 2.Since $V^\perp $ has dimension $1$ so $Tz=kz$ but why will $||Tz||=||z||$ be? $\endgroup$
    – Learnmore
    Aug 9, 2016 at 2:30
  • $\begingroup$ rest of the facts are quite barring the above $\endgroup$
    – Learnmore
    Aug 9, 2016 at 2:39
  • $\begingroup$ @S.Bandopadhaya Two elements $v,w$ of the unit sphere are linearly dependent if and only if $w=\pm v$ (If they are linearly dependent, then $v=\lambda w$ for some $\lambda\in\mathbb{R}$, so $|\lambda|=\Vert v\Vert/\Vert w\Vert=1$), which we are assuming is not the case. 2. Since $T$ is orthogonal, it preserves the norm: $\Vert Tz\Vert^2=(Tz)^T(Tz)=z^T(T^TT)z=z^Tz=\Vert z\Vert^2$. $\endgroup$ Aug 9, 2016 at 3:08
  • $\begingroup$ Its quite clear now;Thank you $\endgroup$
    – Learnmore
    Aug 9, 2016 at 3:22

Hint: $\det(T - \lambda I)$ is a polynomial on $\lambda$ with an odd degree, so it must have a real root. Why must one of these roots be $1$ (as opposed to $-1$)? Consider the determinant. Of course, if $v$ is a unit eigenvector, then so is $-v$.

  • $\begingroup$ Could you rephrase slightly? (the root mustn't be $1$, but one of the roots must be). $\endgroup$
    – Steve D
    Aug 8, 2016 at 17:47
  • $\begingroup$ Good catch, Steve. $\endgroup$ Aug 8, 2016 at 17:50

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