Your basis has to be a set of 3x3 matrices.
That is fair, I'm just wondering the method you used.
You've (assuming your calculations are correct) found what matrices in the subspace U "look like". Can you write that matrix as a linear combination of other matrices? (I haven't verified, but it...
What does it mean to be a basis?
I don't agree with your explanation of the dimension and basis of U. Column vectors (in this case) are ##3x1## matrices. You're claiming that you can generate a whole space of 3x3 matrices with linear combinations of 3x1 matrices. The basis that you are claiming...
That is a fair critique I suppose. If we wanted to be super rigorous we could fix an algebraic closure of ##\mathbb{F}## call it ##\mathbb{F}^*## with ##\mathbb{F} \subseteq \mathbb{K} \subseteq \mathbb{F}^*##.
But either way, while the argument is very short, it seems like a useful lemma that...
You might start by trying to prove some statements about parity. Like an odd number times an odd number is always odd. Even number times anything is always even. Those are pretty straight forward. If you feel you already feel comfortable with those. Then really you should just find a good...
Since ##\alpha## is a root of f and ##\mathbb{L}## is the splitting field of f. ##\mathbb{L}## must contain all roots of f, basically by definition of splitting field.
Three things: First, got to have some context. We can't help you if we don't know what the question is. Second, learn latex it's not hard at all, your equations are impossible to interpret with certainty. Third, are these your equations? If not, you may quote this message and see how it was...
Can you describe your intent with the attached tree? That is, how have you chosen your labels on your branches? Why do you know there is something wrong?
Also, if you scroll through the Calculus and beyond section you might find some exercises to try to prove.
Or.. you might prove that there are no positive integers ##a,b,## and ##c## such that for ##n \geq 3##
$$a^n + b^n = c^n$$.
:-p
This is a question that came about while I attempting to prove that a simple extension was a splitting field via mutual containment. This isn't actually the problem, however, it seems like the argument I'm using shouldn't be exclusive to my problem. Here is my attempt at convincing myself that...
It's not obvious to me why simply mapping the generators to generators should define a homomorphism.
Just to be sure I'm on the same page as you. Let ##a,b ## generate G, ##a',b'## generate H. Let ##\phi : G \rightarrow H## be defined by ##\phi(a)=a'## and ##\phi(b)##.
From just this, It's not...
If you are trying to show that two groups, call them H and G, are isomorphic and you know a presentation for H, is it enough to show that G has the same number of generators and that those generators have the same relations?