Talk:Aufgaben:Problem 4
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.
Carl (talk) 13:16, 22 June 2015 (CEST)
I moved your proposal here, so we can discuss it:
Proposal of a little bit shorter solution:
z.z.: \(T_e\mathrm{SL}(n,\mathbb{R})\cong\{ A\in \mathbb{R}^{n\times n}\ |\ \mathrm{tr}A =0\} \)
proof: Let \(A\in \mathbb{R}^{n\times n}\) be such that \(\mathrm{tr}A=0\). Let \(\gamma:\mathbb{R}\rightarrow\mathbb{R}^{n\times n}\) be a curve defined by \(\gamma(t)=exp(tA)\). This curve has the following properties:
- \(\gamma(0)=I_n\)
- \(\dot{\gamma}(0)=A\)
- \(\mathrm{det}(\gamma(t))=\mathrm{det}(\mathrm{exp}(tA))=\mathrm{exp}(\mathrm{tr}(tA))=1\)
Therefore we know, that \( \gamma(t) \) is a curve in \(\mathrm{SL}(n,\mathbb{R})\) and \(A\in T_e\mathrm{SL}(n,\mathbb{R})\).
For every \(M\in\mathrm{SL}(n,\mathbb{R})\) there is (for a fixed \(t\)) at most one \(A\in T_e\mathrm{SL}(n,\mathbb{R})\) with \(\mathrm{exp}(tA)=M\). Therefore the function
$$\gamma_t: T_e\mathrm{SL}(n,\mathbb{R})\rightarrow \mathrm{SL}(n,\mathbb{R})$$
is in injective. Because a linear mapping of two vector spaces with same dimension is injective if and only if it is surjective, we can show the surjectiviy of \(\gamma_t\) by showing that \(T_e\mathrm{SL}(n,\mathbb{R})\) and \(\{ A\in \mathbb{R}^{n\times n}\ |\ \mathrm{tr}A =0\} \) have the same dimension:\\
$$\mathrm{dim}(T_e\mathrm{SL}(n,\mathbb{R}) = \mathrm{dim}(\mathrm{SL}(n,\mathbb{R}))=n^2-1$$ $$\mathrm{dim}(\{ A\in \mathbb{R}^{n\times n}\ |\ \mathrm{tr}A =0\} =n^2-1$$
q.e.d
$$ e^{tA} = \sum^\infty_{k=0} \frac{A^k}{k!} = I_n + tA + \frac{1}{2}t^2A^2 + O(t^3) $$ $$ e^{tB} = \sum^\infty_{k=0} \frac{B^k}{k!} = I_n + tB + \frac{1}{2}t^2B^2 + O(t^3) $$
We want these two to be equal and by equating the coefficients we get \(A = B\)
