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Could someone please explain the answers forColumn spaceandNullspace of $\displaystyle A^{T}$?

Thanks.

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- Feb 12th 2011, 07:58 PMalexmahoneDimensions of the Four Subspaces
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Could someone please explain the answers for**Column space**and**Nullspace of $\displaystyle A^{T}$**?

Thanks. - Feb 12th 2011, 09:21 PMdwsmith
- Feb 12th 2011, 09:26 PMalexmahone
The

*dimension*of the column space is equal to that of the row space. But the column space is not equal to the row space in this problem.

I know that. But I don't know how the textbook has found**Nullspace of $\displaystyle A^{T}$**in this particular problem. - Feb 12th 2011, 09:27 PMdwsmith
- Feb 12th 2011, 09:31 PMdwsmith
Are you sure the column space of A^T isn't the basis of the row space of A? I would think it is that.

It is stated on http://www.ltcconline.net/greenl/cou...ctors/rank.htm that that the column space of A^T is the row space of A.

I don't see why it wouldn't be the same the other way around. - Feb 13th 2011, 04:23 AMHallsofIvy
I cannot see the "A" that apparently was posted so I don't know if we

**are**talking about matrices or general linear transformations or what dimensions. It is true that the "row space" of a linear transformation A is the same as (strictly speaking "is isomorphic to") the "column space" for**finite**dimensions. For infinite dimensions (such as function spaces) two spaces can have the same dimension but not be isomorphic. - Feb 13th 2011, 07:27 AMalexmahone
- Feb 13th 2011, 09:22 AMLoblawsLawBlog
I think he's using rank-nullity and then just making sure he gets the correct number of independent vectors. He knows the column space of $\displaystyle A$ is two, and the first and third columns of $\displaystyle A$ are the ones he lists and they're independent. I'm not sure if this is obvious from $\displaystyle E^{-1}$, but I suspect it might be because of the nice factorization. I forget all the details about the relationships between these bases and how they change when you do elementary operations.

As for the nullspace of $\displaystyle A^T$, it looks like he just interprets matrix multiplication as acting on the rows of $\displaystyle A$ and then finds a vector in the nullspace of $\displaystyle A^T$. It's in this nullspace because he considers left multiplication on $\displaystyle A$. Since this nullspace has dimension 1, there's a basis.

Is this Strang? I know he talks about this in one of his MIT lectures online. That should answer everything. - Feb 13th 2011, 09:22 AMdwsmith