# Math Help - Question about column space

1. ## Question about column space

Here's the question I'm trying to answer:

Suppose a 6 x 8 matrix A has four pivot columns. What is dim Nul A? Is Col A = R4? Why or why not?

I know that dim Nul A = 4 because there are 4 columns that are not pivot columns. But I'm unsure about the column space. I know that the four pivot columns consisting of Col A are linearly independent, but the column vectors of A exist in R6, not R4, because A is 6 x 8. Is having 4 linearly independent column vectors enough to say that they span R4, or is this statement untrue if the vectors are not in R4 to begin with?

I'd appreciate any insight!

2. ## Re: Question about column space

Originally Posted by Chaobunny
Here's the question I'm trying to answer:

Suppose a 6 x 8 matrix A has four pivot columns. What is dim Nul A? Is Col A = R4? Why or why not?

I know that dim Nul A = 4 because there are 4 columns that are not pivot columns. But I'm unsure about the column space. I know that the four pivot columns consisting of Col A are linearly independent, but the column vectors of A exist in R6, not R4, because A is 6 x 8. Is having 4 linearly independent column vectors enough to say that they span R4, or is this statement untrue if the vectors are not in R4 to begin with?
I'd appreciate any insight!
Well, each column vector has 6 components so to say they span R4 is incorrect. Vectors in R4 have 4 components. What you have is a 4 dimensional span of 6 dimensional vectors in an 8 dimensional space. You can think of it as a 4 dimensional 'plane' in a 8 dimensional space if that helps.

All linear combinations of 4 linearly independant columns will span a 4 dimensional column space, so the dimension of column space of A must be 4.

I'm a lttle confused myself so maybe other posters can be more helpful

3. ## Re: Question about column space

no, col(A) is not R4. col(A) is a 4-dimensional subspace of R6, which is isomorphic to R4, but not equal to it.

col(A) is a space spanned by 4 columns of A. the pivot columns are a canonical choice, but any 4 linearly independent columns will do (bases are not unique).

but there is NO subset of R4 which contains ANY elements of col(A), no matter how linearly independent they are.

4. ## Re: Question about column space

As others have pointed out, strictly speaking, the column space is a four dimensional subspace of ] $R^6$, not $R^4$. The matrix has rank 4 and maps $R^8$ to a four dimensional subspace of $R^6$. By the "rank-nullity" the rank and nullity must add to the dimension of the domain space, 8, so the nullity is also 4.

5. ## Re: Question about column space

Thanks everyone this makes a lot more sense now!