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Problem of the Week #93 - March 10th, 2014

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Chris L T521

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Jan 26, 2012
995
Here's this week's problem!

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Problem
: Show that the one-parameter groups in $SL_n$ are the homomorphisms $t\to e^{tA}$, where $A$ is a real $n\times n$ matrix whose trace is zero.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

Well-known member
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Jan 26, 2012
995
No one answered this week's problem. You can find the solution below.

We begin with a lemma.

Lemma: For any square matrix $A$, $\exp(\mathrm{tr}\,A)=\det \exp(A)$

Proof of Lemma: If $\mathbf{x}$ is an eigenvector of $A$ with eigenvalue $\lambda$, then it is also an eigenvector of $\exp(A)$ with eigenvalue $\exp(\lambda)$. So, if $\lambda_1,\ldots,\lambda_n$ are eigenvalues of $A$, then the eigenvalues of $\exp(A)$ are $\exp(\lambda_i)$. The trace of $A$ is the sum $\lambda_1+\ldots+\lambda_n$, and the determinant of $\exp(A)$ is the product $\exp(\lambda_1)\cdot\ldots\cdot\exp(\lambda_n)$. Therefore, $\exp(\mathrm{tr}\,A) = \exp(\lambda_1+\ldots+\lambda_n) = \exp(\lambda_1)\cdot\ldots\cdot\exp(\lambda_n) = \det \exp(A)$.$\hspace{0.25in}\blacksquare$

We now go ahead and prove the main result.

Proof: The lemma shows that if $\mathrm{tr}\,A=0$, then $\det\exp(tA) = \exp(t\,\mathrm{tr}\,A) = \exp(0) = 1$ for all $t$, so $\exp(tA)$ is a one-parameter group in $SL_n$. Conversely, if $\det\exp(t A)=1$ for all $t$, the derivative of $\exp(t\,\mathrm{tr}\,A)$, evaluated at $t=0$, is zero. But $\left.\dfrac{d}{dt}\right|_{t=0}\exp(t\,\mathrm{tr}\,A)=\mathrm{tr}\, A$ and thus we have $\mathrm{tr}\,A = 0$. $\hspace{.25in}\blacksquare$
 
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