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    Q: Is the following statement true or false? If A^3 = I_n then det(A) = 1.

    My first instinct would be to say it is false, as proofs with determinants are *******s, but I checked for a general 2x2 matrix and it was true there, so I guess it may be true. Can anybody give a hint as to how to approach the problem? Unless I have forgotten something, the usual way of defining a determinant is \displaystyle det(A) = \sum_{\sigma \in S_n} sgn(\sigma ) \prod^{n}_{i=1} A_{i, \sigma (i)} , but I can't see how that will equal one when A^3 = I_n ? Is there a simpler way, or do I need to try and work with that beast?
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    Well, det(A^3) = [det(A)]^3

    A^3 = I_n => [det(A)]^3 = det(I_n) = 1...

    Capiche?
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    (Original post by JohnnySPal)
    Well, det(A^3) = [det(A)]^3
    Oops. That identity slipped my mind, thanks.
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    I don't see why you can't have det A = e^{2\pi i / 3}, however.
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    (Original post by DFranklin)
    I don't see why you can't have det A = e^{2\pi i / 3}, however.
    Yeah, that's true. I didn't even spot that - whenever I see "Linear Algebra" I just assume that all the matrices will be nice and real. Ceratinly if A is real then it must have determinant 1.

    Good luck finding a counter-example over \mathbb{C} :p:
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    Not exactly hard. There's a 1x1 matrix that springs to mind...
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    (Original post by DFranklin)
    Not exactly hard. There's a 1x1 matrix that springs to mind...
    You are such a **** :p:

    Love you really xx
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    Hmm, Dave has thrown his spanner and now I'm confused. I have A as a matrix over a field \mathbb{K} containing 1/2, but I thought det(AB) = det(A)det(B) was true for all A,B over an arbitrary field. What's up?
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    Not sure what you're asking. If A is the 1x1 matrix \Big( e^{2\pi i /3} \Big), then det A = e^{2\pi i /3} and A^3 = I, but we never break the det(AB) = det A det B rule (which is always true, as you say).

    Edit: this may be clearer. It is completely true that if A^3 = 1, then det(A^3) = 1 and so det(A)^3 = 1. But over a complex field, you can't deduce from X^3 = 1 that X = 1, it might be one of the other complex cube roots of unity.
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    Oh right. My "proof" was: det(A^3) = (det(A))^3 \Rightarrow (det(A))^3 = 1 \Rightarrow det(A) = 1 , but of course z^3 = 1 has complex solutions other than 1. I wonder if that was actually the intention of the problem-setter or not, but it's a nice spot. :dontknow:
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    Next question I'm having trouble with: Let T:V->V be a linear map with dim(V) = n, T^n = 0, and let there exist a vector v \in V such that T^{n-1}(v) = 0_V. My object is to show that v, T(v), T^2(v), ..., T^{n-1}(v) are linearly independent, and that the nullity is 1.

    I can't see how to get at the problem. I can see that this sequence of vectors will be a basis of V, and that T^n(v) = 0 \Rightarrow T(T^{n-1}(v)) = 0, so for all v in V, T(T^{n-1}(v)) = 0, and there exists v such that T^{n-1}(v) = 0, but I'm not sure how to put this together. Any hints?
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    Do you mean T^(n-1)(v) is not zero?
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    (Original post by DFranklin)
    Do you mean T^(n-1)(v) is not zero?
    No. It's P7 here in case I've made a mistake in the question (but I don't think I have): http://www.warwick.ac.uk/~masdf/alg1/p1.pdf
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    Well, I think the question is wrong. Mainly because the existence of a v with T^{n-1}(v)\neq 0 is what "makes sense" if you know what the question is getting at.

    But as a clincher, if T^(n-1)(v) = 0, then the question is claiming that a set of vectors including the zero vector is linearly indepedent, which is clearly nonsense.
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    That question's just wrong, surely? Let V = R^7, T(v) = 0 for all v. Then T^7 = 0, and there exists v such that T^6(v) = 0, but T(v) and T^2(v) aren't linearly independent. I think it should be T^(n-1)(v) =/= 0 too.

    Edit: beaten.
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    Thanks for the correction, guys. However, I still can't find what kind of thing I should be doing to answer the corrected question. :blushing: Any hints?
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    Well, suppose v, T(v), T^2(v),...,T^n-1(v) isn't a basis. Then we can find a0,a1,...,a_n-1 not all zero with \sum_0^{n-1} a_k T^k(v) = 0. Now think about what happens if you apply a suitable power of T to that equation...
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    (Original post by Kolya)
    No. It's P7 here in case I've made a mistake in the question (but I don't think I have): http://www.warwick.ac.uk/~masdf/alg1/p1.pdf
    Woah even Dmitri's Algebra 1 assignments look scary. Be glad you didn't have him for Algebra 2. It was basically a disaster for our year.

    I might actually have answers to some of these problems. It depends on whether he's changed the sheets much since Derek Holt set them for us a couple of years ago.
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    (Original post by DFranklin)
    Well, suppose v, T(v), T^2(v),...,T^n-1(v) isn't a basis. Then we can find a0,a1,...,a_n-1 not all zero with \sum_0^{n-1} a_k T^k(v) = 0. Now think about what happens if you apply a suitable power of T to that equation...
    So can I say
    \sum_0 ^{n-1} a_k T^k(v) = 0 \Rightarrow T^{n-1} (\sum_0 ^{n-1} a_k T^k(v)) = T^{n-1}(0) = 0
    \Rightarrow \alpha _0 T^{n-1}(v) = 0 , which contradicts our assumption that there exists a v such that T^{n-1}(v) \neq 0 and a_0 \neq 0, therefore our sequence v, T(v),...,T^{n-1}(v) is a basis?
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    (Original post by JohnnySPal)
    I might actually have answers to some of these problems. It depends on whether he's changed the sheets much since Derek Holt set them for us a couple of years ago.
    I don't think they've been changed much, if at all. The solutions to a fair few of the problems are online, but unsurprisingly there was no solution to the dodgy question!
 
 
 
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