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Kurre

Member

Registered: 2006-07-18

Posts: 280

Linear algebra

Need help with some problems in linear algebra:
1. Prove that the nullity is the same for two similar matrices.
My proof:
let

and assume there is a vector x such that Ax=0

Multiply both sides with P:

thus

The linear transformation P is injective since P is invertible, and since P is a mapping from all x in ker(A) to all y=Px in ker(B), P must be bijective, and thus ker(B) is isomorphic to ker(A) and Null(A)=Null(B).
Is this proof completely correct?

2. Let C and D be mxn matrices, and let B=(v1,...,vn) be a basis for a vector space V. Show that if

for all x in V, then C= D.
If this holds for all x, then we can choose x such that xi≠0, and multiply both sides with y= (1/x1,...,1/xn) and get C= D. What I dont get is why a different basis is introduced, does that change the problem somehow, or is my proof still correct?
3. If A^2 and B^2 are similar, must A and B be similar?
Im not sure, But I dont think so.

so det(A) could equal -det(B) which would contradict similarity, but this doesnt really prove anything. I maybe need to find a counterexample, but I dont really see how...

Last edited by Kurre (2008-08-18 08:02:48)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Linear algebra

and since P is a mapping from all x in ker(A) to all y=Px in ker(B), P must be bijective

Why is it surjective?

It seems to me that a much easier argument is to just go the standard set theory way of proving two sets are equal: Show they are each a subset of each other.  In fact, much of your argument works this way.

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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..."

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Linear algebra

You're lucky I just unpacked my books, I was unfamiliar with the notation in problem 2 as well.

The notation:

Means the coordinate relative to base B.  You are finding what values must be multiplied by each basis vector in B to get x.  For example, we shall take V to be the polynomials of degree 2 over the real numbers, and B = {1, x, x^2}.  Then we have a vector f(x) = 2 + 8x - 3x^2.  So:

Hopefully that makes sense.

Edit: B must be an ordered basis for this to work.

Last edited by Ricky (2008-08-18 14:01:01)

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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..."

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Linear algebra

3. Contrapositive is the way to go on this one.  Remember that if a is not equal to b, then a*a is certainly not equal to b*b.

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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..."

Kurre

Member

Registered: 2006-07-18

Posts: 280

Re: Linear algebra

hm I see now that me arguments for P is not enoguh for surjectivity, to prove that P is a mapping to all y in ker(B) I must prove that all y in ker(B) has the form y=Px, so far I have only proved that all Px lies in ker(B). Altough this is easy fixed by letting

and consider all y such By=0, then

multiplication from left by P^-1:

thus there is an x in ker(A) such that

Thus for all y in ker(B), there is an x in ker(A) such that y=Px and hence P is surjective.
But I dont get what you mean that they should be subsets of eachother. Then every x such that Ax=0, should also satisfy Bx=0,
so

must imply that Bx=0. I dont really get it

Kurre

Member

Registered: 2006-07-18

Posts: 280

Re: Linear algebra

2. I see now that my idea on 2 fails, so here is my second attempt:
If x is written in the basis B, then I can choose

such that all xi are non zero, except for xj=1 for any j. Thus Cx=Bx can be written as:

Now multiplying this out will get a nx1 matrix, the jth column of C, on the left side and the nx1 matrix that is the jth column of D on the right showing that cij=dij for all i. And letting j be 1,2,...,n shows that cij=dij for all j as well, and thus C= D.

Is that what

means, just the nx1 matrix obtained by the coefficients when x is written in base B?

Last edited by Kurre (2008-08-18 23:38:12)

Kurre

Member

Registered: 2006-07-18

Posts: 280

Re: Linear algebra

3. Contrapositive is the way to go on this one.  Remember that if a is not equal to b, then a*a is certainly not equal to b*b.

I dont get it, how does equality between a and b have anything do with similarity?

Kurre

Member

Registered: 2006-07-18

Posts: 280

Re: Linear algebra

Riiiiiiiiickyyyyyyyyyyyyyyyyyyyy where are youu!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!:o
I got exam tomorrow!!!!!!!!!!

TheDude

Member

Registered: 2007-10-23

Posts: 361

Re: Linear algebra

I'm not much help since it's been some time since I've worked with linear algebra, but thanks to Ricky's hints I think I can help you with #3.

In logic, the contrapositive means this: If A implies B, then Not B implies Not A.  In terms of question 3 it can be reworded like this: If A is not similar to B, then A^2 is not similar to B^2.  This is equivalent to proving that if A^2 is similar to B^2 then A is similar to B.  Note that you don't have to guess the correct answer from the start; either you do prove the assumption (A is not similar to B implies A^2 is not similar to B^2) or you find that it is wrong and prove the reverse by contradiction (my favorite kind of proof, btw).

Now that the idea is laid out, the only work to do is to actually prove the assumption is either right or wrong.  Ricky states outright that the assumption is true, but I don't know of any basic theorems that support that conclusion.  After all, it's not true for the real numbers.  I don't actually know the answer, but I assume you can work from there.

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Wrap it in bacon

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Linear algebra

Riiiiiiiiickyyyyyyyyyyyyyyyyyyyy where are youu!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

At orientation for my graduate school.

1. You've already argued that Px and P-¹x are both nonzero so long as x is nonzero  This means that if P-¹BPx = 0, then Bx = 0.

2.

then I can choose...such that all xi are non zero, except for xj=1 for any j.

Yes, you can, but this is a rather poor argument.  Why not just say you picked x=vj?  Now the next too lines of LaTeX (that I'm too lazy to copy) are rather poor notation, but other than that, the proof is correct.  It is very important that you see the x's you picked are in fact basis vectors, and it's the basis vectors themselves which force the equality of the matrices.

3.  Dang, it was late, I was tired.  Of course TheDude is right (even though he might not know it).  Take any matrix with a single non-zero entry and then multiply it by negative one.  They certainly aren't equal but their squares are.

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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..."

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Linear algebra

3. A = {{1, 0}, {0, 1}}, B = {{0, 1}, {1, 0}}

Sheesh that took way too much effort for such a simple solution.

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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..."

Kurre

Member

Registered: 2006-07-18

Posts: 280

Re: Linear algebra

Yes that struck me too, that if B=-A, then A^2=B^2, but A≠B. But still, A^2 is similar to B^2, but how do I prove there is no matrix P such that A=P-¹BP=-P-¹AP ?

"1. You've already argued that Px and P-¹x are both nonzero so long as x is nonzero  This means that if P-¹BPx = 0, then Bx = 0. "

Am I being stupid, or am i missing how that really is implying the other?

But anyway, I had my test today, and it was the most boring math test I have every done in my life (and my first at the university actually ). It was all about diagonalizing matrices and finding eigenvalues and other time-requiring counting exercises with loads of numbers, no "real" mathematics.. But I think I did well.

Last edited by Kurre (2008-08-20 07:27:44)