Figuring Out When to Find dimH

[ElliottG]

Hi guys

So I'm studying for my test tomorrow and I understand how to find the dimensions of ColA and NulA, but what I don't understand is when they give you for example a matrix H, and then they just say Find dimH.

What am I suppose to be solving for? We did two examples in class and in one question my teacher obviously solved for the dim of the column space, and in the second question she solved for the dimension of the nul space.

So when do you solve for each and how do you know which one to solve for?
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Secondly, I've seen a question that says "when does (a value) in the set (v1, v2, v3) not form a basis for R^3?" and the idea here was that you had to solve so that the determinate was not equal to 0.

DOes this have something to do with the IMT stating that detA cannot = 0 for a matrix to be invertible and thus cannot be a basis? I'm kind of confused on the relationship here between the determinate and whether or not it is a basis.

Thanks,
Elliott

 

[homeomorphic]

There has to be more info.

I don't know what dim H is. That's meaningless unless you say what H is. What's H?

 

[ElliottG]

I forgot to say that the matrix they give is called H.

I edited my first post.

 

[homeomorphic]

They mean the rank of the matrix. Which is the dimension of the column space (and it's a theorem that that is equal to the dimension of the row space--so it's sort of combined into one word, which is the rank).

So, it's clear what your teacher did in the first example. 2nd one: if you know the dimension of the null space, you can easily find the dimension of the column space from that. Doesn't matter how you do it.

Determinant measures the volume of a parallelepiped spanned by the columns (any prof who fails to mention this should be shot!). Volume zero for 3 vectors means they must lie in the same plane (or a line or a point), so they don't span R^3, so it's not a basis. In other words, the columns are not linearly independent.

The relation to whether it's invertible is that, if you think of it as a mapping, if the columns are not linearly independent, the map will not be one to one.

 

[ElliottG]

Thanks for your help I pretty much get it now. But in the second question I only see her solving for the dim of the null space and not looking @ the rank or dim of the column space as you said (I understand why you would do that now).

Here's a pic of it in my notes:

[URL]http://74.52.147.194/~devilthe/uploads/1320673566.jpg[/URL]

EDIT: The bottom got cut off but it says dimH = 3.

Is it because the column space doesn't exist as there is only 1 row? If that were true, why would you then substitute the dimension of the null space in for the rank? I guess it's sort of a special question...

 

[Deveno]

H doesn't look like a matrix at all, but a subspace of R⁴.

so you're not looking for "the null space" but just a basis for H.

now, you can regard H as the null space of the matrix:

[1 5 -3 0], it's pretty clear this can only be of rank 1,

and by a well-known theorem (the rank-nullity theorem),

the null space of this has to be of dimension 3.

in general, you can think of an mxn matrix as a way of turning n-vectors

into m-vectors. n-vectors live in a space of dimension n.

you usually write them as nx1 matrices (column vectors).

if there are only k linearly independent columns of a matrix M,

then dim(M(Rⁿ)) (the column space) is k (this will be ≤ m).

the rest of the dimensions of Rⁿ, (n-k), M must shrink down to 0,

these n-k dimensions are the null space of M.