# Thread: Vectors in Column Spaces

1. ## Vectors in Column Spaces

Hi, can anyone help me with this question?

Let A=

[2 2 6]
[-3 2 4]
[5 -2 8]

Determine whether each the following vectors is in the column space of A, the null space of A, or neither.

1.
[8]
[-9]
[7]

2.
[2]
[-8]
[12]

3.
[-2]
[-6]
[-5]

4.
[-4]
[-2]
[2]

How can you tell? Thanks!

2. ## Re: Vectors in Column Spaces

This matrix A, are the columns linearly independent? Do you know how to find out? I'll tell you if you don't.

Assuming they are linearly independent, what is the null space of A?

What is the span of the three columns, i.e., how can you describe the set of vectors that can be expressed as linear combinations of those three?

I could be wrong, but I get the impression you don't understand what the terms "column space" and "null space" mean. If you don't, find out.

3. ## Re: Vectors in Column Spaces

Thanks for your reply zhandele. Hahaha I was just looking at the wikipedia pages on "column space" and "null space" an hour ago, and I think I understand the basics of it now.
However, I'm kind of hesitant on how to find whether or not the vectors are linearly independent.

The null space just refers to the solution Ax = 0 right?

4. ## Re: Vectors in Column Spaces

Yes, the null space is the set of vectors that are solutions to Ax=0.

Now, let's suppose that the columns of A are linearly independent. In that case, what's the solution to Ax=0 ?

You probably learned a definition of linear independence that used the term "trivial solution." An alternative and equivalent definition would be "the columns of A are linearly independent if the null space of A contains only the zero vector." So then none of the vectors you list would be in the null space.

As for the column space, well, three linearly-independent three-element vectors are sufficient to span R^3 (or V^3, if you prefer that notation). So all the vectors you list are in that column space.

Provided, of course, that the columns are linearly independent. How can we tell?

There are at least two ways. One is to take the determinant. If the determinant is not equal to 0, the columns are linearly independent. Perhaps you're not quite ready to understand that yet? If not, you will soon.

The other way is to row-reduce. If A reduces to the identity matrix, then the columns are linearly independent. You probably know why already. If not, look it up.

I reduced your matrix, and I found that the three columns are linearly independent. But you should check for yourself. If it turns out that they are not linearly independent and determinant is 0, then your question is more interesting.

5. ## Re: Vectors in Column Spaces

You don't really need to know if the vectors are independent or not. Though it helps: if all three "column vectors" are independent, then the null space is trivial and none of those vectors are in it.

But, straight forwardly, a vector <x, y, z>, is in the column space of A if and only if it is a linear combination of the columns: if and only if there exist numbers, A, B, C such that <x, y, z>= A<2, -3, 5>+ B<2, 2, -2>+ C<6, 4, 8>= <2A+ 2B+ 6C, -3A+ 2B+ 4C, 5A-2B+ 4C>= <x, y, z> or, equivalently, 2A+ 2B+ 6C= x, -3A+2B+ 4C= y, and 5A- 2B+ 4C= z.

So, for example, <8, -9, 7> will be in the column space if and only if there exist numbers A, B, and C such that 2A+ 2B+ 6C= 8, -3A+ 2B+ 4C= -9, 5A- 2B+ 4C= 7. Can you solve those three equations?

Similarly, a vector, v= <x, y, z> is in the null space of A if and only if Av= 0. For this A, that is $\displaystyle \begin{bmatrix}2 & 2 & 6 \\ -3 & 2 & 4 \\ 5 & -2 & 8 \end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}2x+ 2y+ 6z \\ -3x+ 2y+ 4z \\ 5x- 2y+ 8z\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}$.

That is, if and only if 2x+ 2y+ 6z= 0, -3x+ 2y+ 4z= 0, and 5x- 2y+ 8z= 0. Is that true, say, for <x, y, z>= <8, -9, 7>?