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 \(A = B\)
Djanine (talk) 11:33, 24 June 2015 (CEST)
That would make sense if t was a variable, but as far is I understand t is fixed for the map \(\gamma_t\)
Carl (talk) 13:12, 24 June 2015 (CEST)
Well I've found another problem of this "shorter solution". It says "a linear mapping of two vector spaces with same dimension is injective if and only if it is surjective,..." but \(\gamma_t\) isn't linear.
Anyway, I'd rather use the other solution.
Djanine (talk) 13:24, 24 June 2015 (CEST)
And for one of your other questions: In wikipedia it says that two vectorspaces are isomirph when there's a bijective, (again!) linear map, such that an inverse exists.
But I also don't know what to do with the isomoprhism if we want to show equality...
Djanine (talk) 13:39, 24 June 2015 (CEST)
Yes, I think all together we can agree that this solution is not correct.
Carl (talk) 13:48, 24 June 2015 (CEST)
I just corrected \(tr(tA)=t^2 * tr(A) \) to \(tr(tA)=t * tr(A)\). Note: The trace is linear but \(det(tA)=t^{dim(A)}* det(A) \). For unbelievers of \(det(e^{tA}) = e^{tr(tA)} = e^{t\cdot tr(A)}\), consider that the determinant is the product of the eigenvalues whereas the trace is the sum of the eigenvalues (as det(A) and tr(A) are invariant under conjugacy). With the conjugacy property of the matrix exponential, the eigenvalues of \(e^{A}\) are the exponential of the eigenvalues of A. --Brynerm (talk) 19:01, 29 June 2015 (CEST)
don't know what I was thinking :) Carl (talk) 20:02, 29 June 2015 (CEST)
