5
$\begingroup$

Let $O(n)$ be the standard orthogonal group of real matrices. I am trying to prove the following:

$N = \bigcup_{g\in GL_n(\mathbb{R})}g\cdot O(n)\cdot g^{-1}$ is not a subgroup of $GL_n(\mathbb{R})$.

I know that if it was a subgroup then it was equal to the normal closure of $O(n)$ but I do not know what that is...

Motivation:

It is proved here that a linear automorphism $T:V \rightarrow V$ preserves some inner product on $V$ if and only if the matrix of $T$ w.r.t an arbitrary basis is similar to an orthogonal matrix. I want to prove a composition of two transformation of this type is not necessarily also of that type. (Which amounts to proving $N$ is not a subgroup, since closure under taking inverses clearly holds).

$\endgroup$
4
$\begingroup$

Note that $N$ is the set of real diagonalizable matrices that only have (complex) eigenvalues of magnitude $1$.

In the case of $n =2$: let $$ A = \pmatrix{0&2\\-1/2&0} $$ Note that both $A$ and $A^T$ are elements of $N$. However, the matrix $AA^T$ has eigenvalues $4,1/4$, which are not of magnitude $1$.

You can generalize this counterexample by considering the block-matrix $$ \pmatrix{A & 0\\0 & I_{n-2}} $$ Note that this counterexample still works if you restrict to the special orthogonal group.

$\endgroup$
1
  • $\begingroup$ In fact: for $A \in N$, $AA^T$ (or $A^TA$) are elements of $N$ if and only if $A$ is normal (i.e. $AA^T = A^TA$). $\endgroup$ – Ben Grossmann Aug 12 '15 at 13:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.