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# A Little Linear Algebra watch

1. Q: Is the following statement true or false? If 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 , but I can't see how that will equal one when ? Is there a simpler way, or do I need to try and work with that beast?
2. Well, det(A^3) = [det(A)]^3

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

Capiche?
3. (Original post by JohnnySPal)
Well, det(A^3) = [det(A)]^3
Oops. That identity slipped my mind, thanks.
4. I don't see why you can't have , however.
5. (Original post by DFranklin)
I don't see why you can't have , 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
6. Not exactly hard. There's a 1x1 matrix that springs to mind...
7. (Original post by DFranklin)
Not exactly hard. There's a 1x1 matrix that springs to mind...
You are such a ****

Love you really xx
8. Hmm, Dave has thrown his spanner and now I'm confused. I have A as a matrix over a field 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?
9. Not sure what you're asking. If A is the 1x1 matrix , then det A = 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.
10. Oh right. My "proof" was: , but of course 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.
11. Next question I'm having trouble with: Let T:V->V be a linear map with dim(V) = n, , and let there exist a vector such that . My object is to show that 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 , so for all v in V, , and there exists v such that , but I'm not sure how to put this together. Any hints?
12. Do you mean T^(n-1)(v) is not zero?
13. (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
14. Well, I think the question is wrong. Mainly because the existence of a v with 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.
15. 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.
16. Thanks for the correction, guys. However, I still can't find what kind of thing I should be doing to answer the corrected question. Any hints?
17. 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 . Now think about what happens if you apply a suitable power of T to that equation...
18. (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.
19. (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 . Now think about what happens if you apply a suitable power of T to that equation...
So can I say

, which contradicts our assumption that there exists a v such that and , therefore our sequence is a basis?
20. (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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