Linear Combinations and Span (Concept Question)

_N3WTON_
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Homework Statement


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}

Homework Equations

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.
 
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_N3WTON_ said:
b) Each \vec{b} in \mathbb{R}^{m} is a linear combination of the columns of A
_N3WTON_ said:
let \vec{b} be a vector in \mathbb{R}^{m}. Suppose that \vec{b} is a linear combination of the columns of A.
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.
 
Fightfish said:
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.
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?
 
_N3WTON_ said:
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?
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.
 
Mark44 said:
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.
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...
 
_N3WTON_ said:
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?
Yes, I believe so.
 
Mark44 said:
Yes, I believe so.
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.
 
_N3WTON_ said:
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.
No, I would use a specific matrix.
 
Mark44 said:
No, I would use a specific matrix.
Ok, thanks for the advice
 
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