# Linear algebra, a question.

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If I have a linear operator: T:V->V The operator of the complexity of V is: ^T(u+iv) = T(u)+iT(v) Then ^(T*) = (^T)* However, I dont understand why it is true, because: ^(T*)(u+iv) = T*(u)+iT*(v) (^T)*(u+iv) = (T(u)+iT(v))* = T*(u)-iT*(v) Where I am wrong? Thanks in advance.

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What do you mean by "^" operator?

And also:
(a + b)* = a* + b*
(a * b)* = a* * b*
So:
(T(v))* = T*(v*) != T*(v)
So I doubt your last equation is true.

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Quote:
 Original post by defferWhat do you mean by "^" operator?And also:(a + b)* = a* + b*(a * b)* = a* * b*So:(T(v))* = T*(v*) != T*(v)So I doubt your last equation is true.

^T, is what I defined:
Lets say ^T = S(T).
Then
S(T)(u+iv) = T(u)+iT(v)
u and v are reals so it doesnt matter if they have * or not.
The claim is that: (S(T))* = S(T*)

Its suppose to be an expantion of Real linear operator on a Real vector space into a complex linear operator on a complex vector space.

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Please justify this thread. Remember, the forum focuses on game development and not math in general. The question (and the other related thread) looks like school work, which violates forum policy. Review the Forum FAQ for details.

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Its not school HW, I am just trying to understand the material. but if it needs to be related to game development, then I wont ask such question anymore.

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Quote:
 (^T)*(u+iv) = (T(u)+iT(v))* = T*(u)-iT*(v)

Shouldn't that be
(^T)*(u+iv) = (T(u)+iT(v))* = T(u)-iT(v)

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