Linear Transformation Question

HarryHumpo
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Homework Statement



Let V = F^n
for some n ≥ 1. Show that there do not exist linear maps
S, T : V → V such that ST − T S = I.

The Attempt at a Solution



I used induction to prove that ST^n-T^nS = nT^n-1 and that S^nT-TS^n=nS^n-1, and I know I'm supposed to use that to come up with a contradiction to the fact that the space is finite dimensional, but I'm not sure how to approach that part of the problem.
Any help is appreciated
 
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HarryHumpo said:

Homework Statement



Let V = F^n
for some n ≥ 1. Show that there do not exist linear maps
S, T : V → V such that ST − T S = I.

The Attempt at a Solution



I used induction to prove that ST^n-T^nS = nT^n-1 and that S^nT-TS^n=nS^n-1
I don't understand why you're doing this; i.e., looking at Tn.

V is an n-dimensional vector space over some field F - that's what Fn means.

It seems to me that the most obvious way to approach this problem is to assume that there are linear maps from V to V such that ST - TS = I. If you get a contradiction, and you should, you can conclude that no such maps exist.
HarryHumpo said:
, and I know I'm supposed to use that to come up with a contradiction to the fact that the space is finite dimensional, but I'm not sure how to approach that part of the problem.
Any help is appreciated
 
The hint that was given was to prove that ST^n-T^nS = nT^n-1 (which I did) and see how it bring about a contradiction, but I don't see how the second part works.
 
HarryHumpo said:
The hint that was given was to prove that ST^n-T^nS = nT^n-1 (which I did) and see how it bring about a contradiction, but I don't see how the second part works.

You know every matrix satisfies its characteristic polynomial. Try working with that. The easy alternate way to do it is to take the trace of both sides. I think both might get into trouble if the field has finite characteristic. Can it?
 
Looking at the trace it seems so obvious now, thank you!
 
HarryHumpo said:
Looking at the trace it seems so obvious now, thank you!

That has some qualifications depending on what F is. In some finite fields, trace(I) is zero. You can also argue using the hint and the characteristic polynomial, but I think that has the same qualification. If the field is such that 1+1+...+1 (n times) isn't equal to zero, then it should work.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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