[Math] A vector $b$ is a linear combination of the columns of a matrix $A$ if and only if the equation $Ax=b$ has at least one solution.

linear algebramatricesproof-verificationproof-writing

I'd just like verification/critique of my proof of this statement.

Consider that the matrix $A$ has form:
$$A_{mn}=
\left[
\begin{array}{cccc}
A_{11}&A_{12}&\cdots&A_{1n}\\
A_{21}&A_{22}&\cdots&A{2n}\\
\vdots&\vdots&\ddots&\vdots\\
A_{m1}&A_{m2}&\cdots&A_{mn}
\end{array}
\right]$$
We'll first prove the forward direction.

$\Rightarrow$
Let
$\textbf{b}$
be a linear combination of the columns of $A$, and let us index the columns of $A$ by
$a_1,\cdots,a_n$.
Then,
$\textbf{b}=c_1a_1+c_2a_2+\cdots+c_na_n$,
for some
$c_1,\cdots,c_n\in\mathbb{R}^n$.
Then,
$Ax=\textbf{b}$
has the solution
$x=
\left[
\begin{array}{cccc}
c_1&c_2&\cdots&c_n\end{array}\right]^T$.
Now we prove the backward direction.

$\Leftarrow$
Assume
$Ax=\textbf{b}$
has a solution. Then,
$\textbf{b}=
\left[
\begin{array}{c}
A_{11}x_1+A_{12}x_2+\cdots+A_{1n}x{n}\\A_{21}x_1+A_{22}x_2+\cdots+A_{2n}x{n}\\
\vdots\\
A_{m1}x_1+A_{m2}x_2+\cdots+A_{mn}x{n}
\end{array}
\right]$.
Thus,
$\textbf{b}$
is a linear combination of the columns of $A$ with coefficients
$x_1,\cdots,x_n$.

Best Answer

The column space of $A$ is a subspace of $\Bbb R^n$ that is spanned by the column vectors. The equation $Ax=b$ is consistent if and only if $b$ is in the column space of $A$. This essentially means that $b$ can be written as a linear combination of the column vectors of $A$. So if I understand correctly, your statement is correct because $x_i \forall i$ are constants so if those constants exist, then $b$ is in the column space and the system of equations is consistent.

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