Master Linear Algebra Proofs: Dimension Theorem, Rank-Nullity Theorem, and More!

killpoppop
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Hey people. I find myself getting through my course but currently with not as much understanding as I would like. We've got to some proofs and i either vaguely understand them or do not know how to prove them.

Homework Statement


The first would be to prove the Dimension theorem that.

dimU + dimV = dim(U + V) + dim( U intersection V )

The second is if U,V are finite dimensional vector spaces over a field, and T: V -> W is a bijection

Show dimU = dimW

The last is to do with the Rank-Nullity Theorem and a basic linear transformation T: Rn -> Rm

That if m < n prove that T is not injective
And if m = n prove that T is injective iff T is surjective.

Some general insight into these proofs would be very very helpful. All feedback would be very much appreciated.

Thanks.
 
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killpoppop said:
The first would be to prove the Dimension theorem that.

dimU + dimV = dim(U + V) + dim( U intersection V )

Let {u1, ... , uk} be a basis for U and {v1, ... , vp} a basis for V. Now consider the intersection U\capV. If U\capV = {0}, what can you say about the linear depence of the set {u1, ... , uk, v1, ... , vp}? How about the situation U\capV \neq {0}?

The second is if U,V are finite dimensional vector spaces over a field, and T: V -> W is a bijection

Show dimU = dimW
If T: V -> W is a bijective map, then what can you say about its range and kernel? How is this related to the first problem?

The last is to do with the Rank-Nullity Theorem and a basic linear transformation T: Rn -> Rm

That if m < n prove that T is not injective
And if m = n prove that T is injective iff T is surjective.
Again, consider range and kernel.
 
killpoppop said:
Hey people. I find myself getting through my course but currently with not as much understanding as I would like. We've got to some proofs and i either vaguely understand them or do not know how to prove them.


Homework Statement


The first would be to prove the Dimension theorem that.

dimU + dimV = dim(U + V) + dim( U intersection V )
I would do this: subtract dim(U intersect V) from both sides to get the equivalent
dim(U+ V)= dimU + dimV- dim( U intersection V )
Choose a basis for U intersect V, then extend it to a basis of U. Extend that same basis for U intersect V to a basis for V. Show that the union of those two bases is a basis for U+ V.


The second is if U,V are finite dimensional vector spaces over a field, and T: V -> W is a bijection

Show dimU = dimW
Choose a basis for U, {u1, u2, ..., un} and show that {Tu1, Tu2, ... Tun} are independent. Thus, dimU\le dimW. Choose a basis for W, {w1, w2, ..., wm} and show that {T-1w1, T-1w2, ..., T-1wn} are independent.

The last is to do with the Rank-Nullity Theorem and a basic linear transformation T: Rn -> Rm

That if m < n prove that T is not injective
And if m = n prove that T is injective iff T is surjective.
Again choose a basis for Rn, {v1, v2, ..., vn} and look at {Tv1, Tv2, ..., Tvn}.

Some general insight into these proofs would be very very helpful. All feedback would be very much appreciated.

Thanks.
 
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Thanks for the feedback it all helps. But still doesn't clear a lot up for me.

Choose a basis for U intersect V, then extend it to a basis of U. Extend that same basis for U intersect V to a basis for V. Show that the union of those two bases is a basis for U+ V.

For this questions I've proved U+V is a subspace. What your saying is the next step but it doesn't make much sense to me. Can you explain further?
 
Also can anyone suggest a decent book or website where i can read up on these?
 
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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