Difference between revisions of "Talk:Aufgaben:Problem 4"
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[[User:Carl|Carl]] ([[User talk:Carl|talk]]) 11:32, 22 June 2015 (CEST) | [[User:Carl|Carl]] ([[User talk:Carl|talk]]) 11:32, 22 June 2015 (CEST) | ||
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| + | Also I don't really understand what you are doing. You say you want to show that \(T_e\mathrm{SL}(n,\mathbb{R})\cong\{ A\in \mathbb{R}^{n\times n}\ |\ \mathrm{tr}A =0\} \) | ||
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| + | where I assume \(T_e\mathrm{SL}\) is the lie algebra of \(\mathrm{SL}\) (we don't want an isomorphism but equity). And then you show that the map \(\gamma_t: T_e\mathrm{SL}(n,\mathbb{R})\rightarrow \mathrm{SL}(n,\mathbb{R})\) is bijective, by looking at the dimension of \(\mathrm{SL}\) and \(\{ A\in \mathbb{R}^{n\times n}\ |\ \mathrm{tr}A =0\}\)? (Notice that you also forgot to show that it is homomorphic (which it isn't) if you want to show that something is isomorphic). And now what about your first isomorphism? Maybe you could clarify. | ||
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| + | [[User:Carl|Carl]] ([[User talk:Carl|talk]]) 13:16, 22 June 2015 (CEST) | ||
Revision as of 11:16, 22 June 2015
Did we have \(det(e^{tA}) = e^{tr(A)}\) in the lecture? Carl (talk) 23:51, 19 June 2015 (CEST)
Well actually it's \(det(e^{tA}) = e^{tr(tA)} = e^{t\cdot tr(A)}\).
The first equality is a property of the matrix exponential, see Wikipedia: [[1]] and Trubo used that property too in the lectures.
The second equality follows from the linearity of the trace. So I don't see any problems on using that.
Djanine (talk) 11:10, 22 June 2015 (CEST)
Great, thanks.
Not sure about the injectivity in the proposed shorter solution. What is your reasoning for \(e^{tA} =e^{tB} \Rightarrow A = B\)
Carl (talk) 11:32, 22 June 2015 (CEST)
Also I don't really understand what you are doing. You say you want to show that \(T_e\mathrm{SL}(n,\mathbb{R})\cong\{ A\in \mathbb{R}^{n\times n}\ |\ \mathrm{tr}A =0\} \)
where I assume \(T_e\mathrm{SL}\) is the lie algebra of \(\mathrm{SL}\) (we don't want an isomorphism but equity). And then you show that the map \(\gamma_t: T_e\mathrm{SL}(n,\mathbb{R})\rightarrow \mathrm{SL}(n,\mathbb{R})\) is bijective, by looking at the dimension of \(\mathrm{SL}\) and \(\{ A\in \mathbb{R}^{n\times n}\ |\ \mathrm{tr}A =0\}\)? (Notice that you also forgot to show that it is homomorphic (which it isn't) if you want to show that something is isomorphic). And now what about your first isomorphism? Maybe you could clarify.
