# Showing that a matrix cannot be obtained as an element of a one parameter subgroup of SL(2;R)SL(2; R)

I have to show that the matrix
[−λ00−1λ]\begin{bmatrix}
-\lambda & 0 \\
0 & -\frac{1}{\lambda}
\end{bmatrix}

cannot be obtained as an element of a one parameter subgroup of SL(2;R)SL(2; R) except when λ=1\lambda = 1. Additionally, show that it can be obtained as a combination of paths eAeBe^Ae^B and find AA and BB.

I don’t even know how do I start proving this result. Given a one parameter subgroup, I can verify versus some element or list all the elements but I need help with this.

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3

It isn’t even in SL(2;R)SL(2;R): the determinant is −1-1.
– Robert Israel
2 days ago

@RobertIsrael Sorry, typo! The first element is −λ-\lambda
– Cheeku
2 days ago

1

One parameter subgroups arise from taking the matrix exponential of some element in your lie algebra, right? Then if you had some two by two matrix that mapped to your given matrix you could compute the coefficients. Maybe those coefficients lead to a contradiction?
– Nikolas Wojtalewicz
2 days ago

@NikolasWojtalewicz Right! I see that. So, the parameter won’t satisfy A(s+t)=A(s)A(t)A(s+t) = A(s) A(t),
– Cheeku
2 days ago

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1

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Hint: eigenvalues. What are the eigenvalues of exp(tA)\exp(tA) in terms of the eigenvalues of AA?

So, eta1=−λe^{t a_1} = -\lambda and eta2=−1λe^{t a_2} = -\frac{1}{\lambda}, and a1a_1, a2a_2 are complex eigenvalues of AA. I really don’t see how this is a contradiction
– Cheeku
2 days ago

BTW, nobody said λ>0\lambda > 0. If λ\lambda is allowed to be negative, you can indeed have your matrix in a one-parameter subgroup.
– Robert Israel
2 days ago

If λ>0\lambda > 0 and λ≠1\lambda \ne 1, a1a_1 and a2a_2 are non-real, but are not complex conjugates.
– Robert Israel
2 days ago