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Linear Algebra Problem
Demerdar
Registered User regular
Registered User regular
Hey ya'll, I was having a little difficulty with this problem for my Linear Algebra class, maybe someone can shed some light on it.
Given W is a subspace of R^n and has an orthogonal basis {u(1) .... u(p)}
Also, W' is its orthogonal compliment (perp W)
show that dim(W) + dim(perpW) = n.
Now, I'm thinking this has to do with the rank theorem where rank(A)+ nul (A) = n
If you were to put the vectors of W into a matrix A, then rank(A) = dim(W).
However I'm having a hard time connecting nul(A) with dim(perpW). Can anybody point me in the right direction? If clarification is needed let me know, this post feels a little disjointed.
Given W is a subspace of R^n and has an orthogonal basis {u(1) .... u(p)}
Also, W' is its orthogonal compliment (perp W)
show that dim(W) + dim(perpW) = n.
Now, I'm thinking this has to do with the rank theorem where rank(A)+ nul (A) = n
If you were to put the vectors of W into a matrix A, then rank(A) = dim(W).
However I'm having a hard time connecting nul(A) with dim(perpW). Can anybody point me in the right direction? If clarification is needed let me know, this post feels a little disjointed.
Demerdar on
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As I understand it, the Rank-Nullity theorem is the big piece you need.
Without spoiling the whole problem for you: I would review what perp(W) in fact means, and then what you can say about the vectors in W wrt perp(W) as a result.
Does that help?
Also, if I were to put W in A, then perp(colA) = Nul(A^t).
Is it true that for orthogonal sets dim(NulA^t) = dim(NulA)?
I think I'm just missing a piece of the puzzle. Maybe you can tell me where I'm going wrong?
Pretty sure this is correct. Here's why: You're right in saying that Perp(W) is the set of all vectors orthogonal to W. So you could if you like calculate this via the following matrix equation:
Ax=0, where A is the matrix made by the basis of W.
so if you were to find all x, you'd find a basis for vectors orthogonal to the vectors in A. The set of all linear combinations makes a nice space called Perp(W)!
As well Ax=0 is just solving for the Kernel of A.
is that more useful?
Anyone else please chime in if ive messed up
So if you solve Ax=0 with W as your matrix A, you're saying that Nul A forms a basis for perpW? In a theorem i'm reading it says that the orthogonal complement to RowA is NulA. And orthogonal complement of colA is Nul A^t.
Am I not seeing something here?
anyway: I'd do this by contradiction. Assume that dim(W)+dim(perp(W)) =/=n
this produces 2 cases: when dim(W)+dim(perp(W)) > n or when dim(W)+dim(perp(W)) < n
based on the fact that you have n basis vectors in your original space W you should get a contradiction in either case pretty quick
do you have access to the dimension theorem? If you do there's a pretty slick proof there as well
Yup. I ran this by another student in my office and he agreed, for what its worth. So as I said solving Ax=0 is finding all the vector perpendicular to the vectors that make the matrix A. Since A is made of basis vectors, we are finding a basis for the set of vectors perpendicular to those in A.
So That's a basis for the orthogonal complement, as we are doing our ops with a basis.
As well, Ax=0 is solving for the kernel. So Dim(Ker(A){the Nullity} should =dim(Perp(A)).
So then by the rank nullity theorem Dim(Im(A)) plus Dim(Ker(A)), which is the same as dim(Perp(A)), have to add up to N.
remember the matrix A is the matrix formed by the basis vectors of W, the subspace you started with.
Sound good?