column space of A is the answer to Ax=v

Question

What is the proof for the theorem mentioned here? What is the intuition behind the proof? https://www.math.upenn.edu/~moose/240S2013/slides7-22.pdf

Answer 1

This is to spell out rpa's comment/explanation above:

Write the matrix as a 'row of columns': $$ A = \left(c_1, \cdots , c_n \right).$$ By definition, the column space is the set (in fact, the vector space) of vectors $v$ which can be written as a linear combination of the columns: it is the set of $v$, for which there are some $x_1,\cdots,x_n \in {\mathbb R}$, such that $$ v = x_1c_1+ \cdots + x_nc_n.$$ (The $x_i$ depend of course on which $v$!)

Now, write the $x_1,\cdots, x_n$ as a (column) vector $x$: $$ x = \pmatrix{x_1 \\ \vdots \\ x_n}. $$ Then, by definition of matrix multiplication, $$ A x = (c_1, \ldots , c_n)\, x = (c_1, \ldots , c_n)\pmatrix{x_1 \\ \vdots \\ x_n}= x_1c_1+ \cdots + x_nc_n.$$

So $v = Ax $ for some $x$ — i.e., "the linear system $Ax =v$ is consistent" — precisely if there are some $x_1,\cdots,x_n \in {\mathbb R}$ such that $v = x_1c_1+ \cdots + x_nc_n$ — i.e., precisely "if $v$ is in the column space of $A$."

Hope this helps!