# Homework Help: Linear Combinations and Span (Concept Question)

1. Feb 28, 2015

### _N3WTON_

1. The problem statement, all variables and given/known data
Let $A$ be an $m \hspace{1 mm} x \hspace{1 mm} n$ matrix, and let $\vec{b}$ be a vector in $\mathbb{R}^{m}$. Suppose that $\vec{b}$ is a linear combination of the columns of A. Then the columns of A span $\mathbb{R}^{m}$
2. Relevant equations

3. The attempt at a solution
I said that this statement was true using the following theorem from my textbook:
Let $A$ be an $m \hspace{1 mm}x \hspace{1 mm}n$ matrix. Then the following statements are logically equivalent.
a) For each $\vec{b}$ in $\mathbb{R}^{m}$, the equation $A \vec{x} = \vec{b}$ has a solution
b) Each $\vec{b}$ in $\mathbb{R}^{m}$ is a linear combination of the columns of A
c) The columns of A span $\mathbb{R}^{m}$
d) A has a pivot position in every row
However, my book says that this statement is false and I am not sure why. I think I am probably missing something obvious, but I'm not sure what.

Last edited: Feb 28, 2015
2. Feb 28, 2015

### Fightfish

In the statement, $\vec{b}$ is a particular vector, and not any arbitrary vector in $\mathbb{R}^{m}$; that is to say, it is not necessarily true that any vector in $\mathbb{R}^{m}$ can be expressed as a linear combination of the columns of A.

3. Feb 28, 2015

### _N3WTON_

ok, so I was wondering if this counter example would be a good way to verify that it is false? I picked an arbitrary matrix and an arbitrary vector:
$$A = \begin{bmatrix} 0 & 3\\ 1& 5\\ 2 &8 \end{bmatrix}$$
$$\vec{b} = \begin{bmatrix} 1\\ 2 \\5 \end{bmatrix}$$
I reduced A and found that there is not a pivot in every row, so I said that the columns of A do not span $\mathbb{R}^{m}$. Is this a sufficient counter example?

4. Mar 1, 2015

### Staff: Mentor

I'm not sure it meets the conditions of the original problem, which states that $\vec{b}$ is a linear combination of the columns of A. In any case, the condition for $\vec{b}$ seems to me to be something of a red herring. Your 3 x 2 matrix clearly (I hope) can't span R3, since there are only two columns.

5. Mar 1, 2015

### _N3WTON_

You're right that my example doesn't meet the given conditions. However, if I were to find an example that does meet the required conditions using a 3x2 matrix, could I then use that counter example to prove that the statement is false? Basically I'm still a little confused about where to go here...

6. Mar 1, 2015

### Staff: Mentor

Yes, I believe so.

7. Mar 1, 2015

### _N3WTON_

Awesome. I was thinking in this case it may be easier to use a matrix made up of stars and squares(like the kind used to determine echelon forms) rather than actually come up with a linear combination.

8. Mar 1, 2015

### Staff: Mentor

No, I would use a specific matrix.

9. Mar 1, 2015