Linear Combinations and Span (Concept Question)

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The discussion centers on the relationship between a matrix A and a vector b in terms of linear combinations and span. It is clarified that while b can be a linear combination of the columns of A, this does not imply that all vectors in R^m can be expressed as such, which is a key point of confusion. A counterexample involving a specific 3x2 matrix is proposed to demonstrate that the columns of A do not span R^m, as there are not enough columns to cover the space. The participants agree that a valid counterexample must meet the original problem's conditions to effectively challenge the statement's validity. Ultimately, the conversation emphasizes the importance of understanding the implications of linear combinations and the span of a matrix's columns.
_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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