1

The book "Riemannian Geoemetry" by Peter Petersen says the following on pg 171:

Now recall that linear isometris $L:\Bbb{R}^k\to \Bbb{R}^k$ with $\text{det }L=(-1)^{k+1}$ has $1$ as an eigenvalue.

I've never heard of this theorem. Is this an elementary linear algebra proposition that I'm somehow unable to recall?

2 Answers2

2

Yes, this is correct. A linear isometry is also called an orthogonal transformation, preserving orthogonality and length. If $v$ is an eigenvector with eigenvalue $\lambda$ then $\|v\|=\|Lv\|=|\lambda|\|v\| $ and $|\lambda|=1$.The determinant is the product of the eigenvalues.In $R^k$ you have $k$ eigenvalues (some of them may be pairs of complex conjugate ones). A product like that is equal to $(-1)^{k+1}$ only if at least one of the eigenvalues is 1. Say, in $R^3$ an orthogonal transformation with determinant $=1$ has to be a rotation around an axis or the identity (well, this is also a rotation).

GReyes
  • 16,446
  • 11
  • 16
  • Consider $R^2$. Then $(-1)^{k+1}=-1$. In that case, if both the eigenvalues are $i$, then we are satisfying the given conditions without making any of the eiigenvalues $1$. So I don't follow the last part of your argument –  Feb 08 '20 at 01:19
  • Well, they cannot be both $i$. They have to be conjugate $\pm I$ (they are roots of a polynomial with real coefficients). You need the dimension of the space to be at least $3$ for the argument to work. – GReyes Feb 08 '20 at 01:21
  • Ah true. But what if in $R^4$, the eigenvalues are $i,i,-i,-i$? –  Feb 08 '20 at 01:29
  • Sorry I meant $R^3$ –  Feb 08 '20 at 01:35
  • That combination is impossible in $R^3$. You can have $1,1,1$ or $1,\cos\phi+i\sin\phi,\cos\phi-i\sin\phi$ The first one is the identity and the second is a rotation of angle $\phi$ around the first eigenvector. – GReyes Feb 08 '20 at 02:56
  • In order to have determinant $=-1$ in $R^4$ the possibilities are $1,1,1,-1$ or $1,-1,-1,-1$ or you can have $1,-1$ and one $2\times 2$ rotation block . Two rotation blocks are impossible since the determinant would be $1$. – GReyes Feb 08 '20 at 03:01
0

Remember, determinant measures the distortion a linear transformation imparts on a space. If the determinant has absolute value one then volumes are preserved.

Nick Castillo
  • 2,033