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LuisRodg

Real Member

Registered: 2007-10-23

Posts: 322

linear algebra proof

I been trying to prove this for 2 days and I havent been able to. I have to prove this:

Could anyone help me with a hint?

Thanks.

Last edited by LuisRodg (2008-06-05 08:55:39)

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: linear algebra proof

I think in this case you could just look at each side individually and prove they're both always true.
There may be an easier way that I'm not seeing though.

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Why did the vector cross the road?
It wanted to be normal.

LuisRodg

Real Member

Registered: 2007-10-23

Posts: 322

Re: linear algebra proof

The thing is that each side individually is not always true. You have to assume one side and proof it implies the other side. Vice versa.

Is that what you meant?

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: linear algebra proof

Oh, I see what I've done. I managed to prove that R(A^2) is always a subset of R(A), and so took that to mean R(A) = R(A^2), which isn't necessarily true.

OK then, ignore me.

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Why did the vector cross the road?
It wanted to be normal.

LuisRodg

Real Member

Registered: 2007-10-23

Posts: 322

Re: linear algebra proof

For purposes outside of my initial question, can you post your proof that R(A^2) is always a subset of R(A)?

Ricky

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Registered: 2005-12-04

Posts: 3,791

Re: linear algebra proof

I believe to answer this question, you should know what it means to represent a linear transformation by a matrix and how matrix multiplication relates to linear transformations.  Do you?

Assuming you do.. here is a hint:

If R(A) = R(A^2), what does this mean, in mathematical-english?  (i.e. not strict terms but more the general "feel" of it?)

Taking this fact, where does something map to if it's not in R(A)?  Perhaps an easier way to see it is (and a more explicit hint...) "What if something not in the range doesn't map to 0?"

Once you get that part, the other way follows the same basic idea.  It may help to interpret R(A) U N(A) = R^n as "If it's not in the range, then it's in the null set."

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

LuisRodg

Real Member

Registered: 2007-10-23

Posts: 322

Re: linear algebra proof

If R(A) = R(A^2) that means that the set that A maps into is the same set that A^2 maps into as well. But im not really sure I get what does R(A) = R(A^2) really imply? What does it really mean that the range of A is the same as the range of A^2?

I tried to play with it for a while but I couldnt derive anything meaningful out of it. What does it mean if y = A(R(A)) etc?

If something is not in R(A) then it is in N(A) which is why R(A) U N(A) = R^n, I understand that part. I feel im missing "something" that just makes all this "click" together.

As I said, I been thinking and pondering about this for a couple days now but with my limited mathematical knowledge, I havent been able to derive anything. But its always good to have a problem in your head your always thinking about, I guess.

Last edited by LuisRodg (2008-06-05 14:57:47)

Ricky

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Registered: 2005-12-04

Posts: 3,791

Re: linear algebra proof

I'm not certain if you see this, but R(A) can be reached solely by applying A to elements in R(A).

Perhaps a good first step would be to prove the following face:

Start by picking a basis for R(A) (remember it is a vector space).

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: linear algebra proof

I don’t think the statement is true. At least, I don’t think

is true.

Take n = 2, and

. Then R(A) = R(A[sup]2[/sup]) since A = A[sup]2[/sup], but

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: linear algebra proof

The other implication is true:

Proof:

So the arrow only works one way, not both.

Sometimes, if you can’t prove that something is true, it helps to suspect that it may not be true at all.

Last edited by JaneFairfax (2008-06-06 00:02:48)

Ricky

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Registered: 2005-12-04

Posts: 3,791

Re: linear algebra proof

Ah, I was thinking this was equivalent to a similar proof:

But of course, it isn't.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

LuisRodg

Real Member

Registered: 2007-10-23

Posts: 322

Re: linear algebra proof

Thanks a lot guys. Today (friday) I sat down and went over everything Ricky told me to understand as well as went over Jane's proofs just so I could understand them.

Also, seems like I made a mistake. Its not "union" its "+" as in:

I thought in this case, plus was the same as union but thats not case which Jane proved so. Supposedly, the one with + is supposed to hold true.

What does + means in sets? Like, is addition defined in Set Theory?

Last edited by LuisRodg (2008-06-06 12:40:12)

Ricky

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Registered: 2005-12-04

Posts: 3,791

Re: linear algebra proof

Means:

and

So it's just a stronger condition than +, and this stronger condition is used a lot in more advanced topics.

However, remember that R(A) + N(A) = R^n means the same as R(A) - N(A) = R^n, which is how it came out naturally for me.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

LuisRodg

Real Member

Registered: 2007-10-23

Posts: 322

Re: linear algebra proof

Oh, so I guess my professor meant:

But he wrote it like a normal plus sign and not the "exclusive" sign like you said. Are they interchangeable when talking about sets?

Last edited by LuisRodg (2008-06-06 12:43:03)

LuisRodg

Real Member

Registered: 2007-10-23

Posts: 322

Re: linear algebra proof

When you say:

What do you mean by "+" ?

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: linear algebra proof

There is a difference between

and

.  The latter implies that the two sets only share one common element, 0.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

LuisRodg

Real Member

Registered: 2007-10-23

Posts: 322

Re: linear algebra proof

Where A and B are sets?

I never knew that sets had addition defined.

I just cant picture the addition of two sets? My mind keeps thinking addition means union.

what is {1,2,5} + {2,3,6,7}? Just so I could see how you add sets.

Last edited by LuisRodg (2008-06-06 18:46:10)

JaneFairfax

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Registered: 2007-02-23

Posts: 6,868

Re: linear algebra proof