A Little Linear Algebra

Well, det(A^3) = [det(A)]^3

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

Capiche?

Reply 2

JohnnySPal

Well, det(A^3) = [det(A)]^3

Oops. That identity slipped my mind, thanks.

Reply 3

I don't see why you can't have

det A = e^{2\pi i / 3}

, however.

Reply 4

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}

Reply 5

Not exactly hard. There's a 1x1 matrix that springs to mind...

Reply 6

DFranklin

Not exactly hard. There's a 1x1 matrix that springs to mind...

You are such a cock :p:

Love you really xx

Reply 7

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?

Reply 8

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.

Reply 9

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:

Reply 10

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. :frown:

Any hints?

Reply 11

Do you mean T^(n-1)(v) is not zero?

Reply 12

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

Reply 13

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.

Reply 14

generalebriety

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.

Reply 15

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?

Reply 16

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...

Reply 17

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.

Reply 18

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?

Reply 19