Let $V$ be a real inner product space of odd dimension and $S∈L(V,V)$ an orthogonal transformation. Prove that there is a vector $v$ such that $S^2(v)=v$.
proof about orthogonal transformation in an inner product space
1
$\begingroup$
linear-algebra
-
1Hint: Since $V$ has odd dimension, you know $S$ must have at least one real eigenvalue ... – 2012-12-02
-
0You might also consider posting what work you have done so far and a bit of motivation for recent spree of linear algebra question. – 2012-12-02
-
1Neal: and then? – 2012-12-02
-
0"Then"? Then you're done! What are the possible eigenvalues of an orthonormal transformation? Ho are the possible eigenvalues of a power of *any* transformation related to the eignevalues of the transformation? – 2012-12-02
-
0DonAntonio: I need more explanation here. What are the possible eigenvalues of an orthogonal transformation? – 2012-12-02
-
0Actually I don't understand why $S$ must have at least one real eigenvalue ... – 2012-12-02
1 Answers
0
Hint: Orthogonal transformation in inner product spaces satisfies the following relation
$$ =
and have the property $S=S^T$. Note that, $=u^T \, v$.
-
0Why $
=$, and why $S=S^T$? – 2012-12-02 -
0@i_a_n: Check the inner product operations. – 2012-12-02
-
0Sorry I don't think I get it. Can you explain them more clearly? – 2012-12-02