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...


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).


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.

  • $\begingroup$ 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

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