Difference between revisions of "Talk:Aufgaben:Problem 4"

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[[User:Djanine|Djanine]] ([[User talk:Djanine|talk]]) 11:10, 22 June 2015 (CEST)
 
[[User:Djanine|Djanine]] ([[User talk:Djanine|talk]]) 11:10, 22 June 2015 (CEST)
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Great, thanks.
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Not sure about the injectivity in the proposed shorter solution. What is your reasoning for \(e^{tA} =e^{tB} \Rightarrow A = B\)
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[[User:Carl|Carl]] ([[User talk:Carl|talk]]) 11:32, 22 June 2015 (CEST)

Revision as of 09:32, 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)