For every real matrix $X$ with $\det X = 1$, there exists a real symmetric traceless matrix $Y$ such that
$$ X^TX = e^Y $$
Is this true? If so, why?
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As $X^TX$ is a symmetric and positive matrix, you have a representation of $X^TX$ : $$ X^TX = O^T D^2 O $$with $O$ orthogonal and $D$ diagonal with $>0$ diagonal elements (because $\det X\neq 1$).
Hen you can write $D^2 = \exp U$ with $U$ a diagonal real matrix;
Then check that$$ X^TX = O^T \exp (U) O = \exp (O^T U O) $$with the series definition.
mookid
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It would probably help the poster if you explained why positive definite matrices can be decomposed as you describe. – DanielV Mar 24 '14 at 21:41
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Is there a name for the $ X^TX = O^T D^2 O $ decomposition so that I can search more about it? – Gabriel Mar 24 '14 at 21:42
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This is a diagonalization. – mookid Mar 24 '14 at 21:43
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@mookid But diagonalization is when you write $ M = P^{-1} D P $. I don't know how to make the jump that $ P $ must be orthogonal in the case $ M = X^T X $. – Gabriel Mar 24 '14 at 21:45
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@Gabriel: You are right :) I tried to add some explanations. – mookid Mar 24 '14 at 21:45
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It is always true that ${\rm det}(e^{M}) = e^{\rm{trace}(M)}$ for any square complex matrix $M.$ This is because there is a unitary matrix $U$ such that $U^{-1}MU$ is upper triangular, so it suffices to consider the case that $M$ is upper triangular. In that case, if the diagonal entries of $M$ are $\{\lambda_{1},\ldots , \lambda_{n} \},$ then the diagonal entries of $e^{M}$ are $e^{\lambda_{1}},\ldots, e^{\lambda_{n}}$ and ${\rm det}(e^{M}) = e^{\lambda_{1} + \ldots + \lambda_{n}} = e^{{\rm trace}(M)}.$ Hence if ${\rm det}(X) = 1$ and $XX^{T} = e^{Y}$ for a real matrix $Y,$ it is necessarily the case that ${\rm trace}(Y) = 0.$ As mookid has explained, $XX^{T}$ may be diagonalized by a real symmetric matrix, so it suffices to consider the case that $XX^{T}$ is real diagonal with positive entries, in which case it may be written as the exponential of a real diagonal matrix.
Geoff Robinson
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