Prove that a system of linear equations $Ax=b$, where $A_{m\times n}$ has a solution iff $b \in Span(\text{col} (A))$.

If $b \in Span(\text{col} (A))$ then $b$ can be expressed uniquely with the vectors that form the columns of $A$, i.e., $b$ is a a linear combination of the columns of $A$ hence it is necessarily a solution?

If $Ax=b$ has a solution then necessarily $b$ belongs to the span of $\text{col} (A)$?

What is the geometric/intuitive meaning of these statements and how to show their correctness?

Thank you!

  • 1
    $\begingroup$ That is quite simply because the column vectors of $A$ are the images of the basis of $\mathrm R^n$ by the linear map associated with the matrix, and any vector of $\mathrm R^n$ is in the span of the basis (by definition), hence its image is in the span of the images of the basis. $\endgroup$ – Bernard Dec 14 '15 at 1:11

This is a pretty straightforward application of definitions.

Define vectors $\def\*#1{\boldsymbol{#1}} \*{v_1}, \*{v_2}, \ldots, \*{v_n}.$ to be the columns of $A$. That is, $A = \left[ \*{v_1}, \*{v_2}, \ldots, \*{v_n} \right] $

Let $\*x$ be a vector $\left[\begin{array}{c} x_1 \\x_2\\ \ldots \\x_n \end{array} \right]$. Then $ A \*x = x_1 \*{v_1} + x_2 \*{v_2} + \ldots + x_n \*{v_n}$.

What is the span of $ \*{v_1}, \*{v_2}, \ldots, \*{v_n}$?

$$\mathrm{Span}\left( \*{v_1}, \*{v_2}, \ldots, \*{v_n} \right) = \left\{ \left. \lambda_1 \*{v_1} + \lambda_2 \*{v_2} + \ldots + \lambda_n\*{v_n} \; \right| \; \lambda_1,\lambda_2, \ldots, \lambda_n \in \mathbb{R} \right\} $$

From this, it should be straightforward to show that $A\*x = \*b$ has a solution iff $\*b$ belongs to span of the columns of $A$.


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