This set of n columns vectors, namely :

#A = \{ mathbf v_1, mathbf v_2, ...., mathbf v_n \}#

..where:

#\mathbf v_k = ((a_(1k)),(...),(a_(nk)))#

....forms this nxn matrix:

#A = ((a_(11)......a_(1n)),(..........),(a_(n1).....a_(n\n)))#

.

The set's span is all possible linear combinations of the column vectors, ie

#sum_(i=1)^n c_i v _ i# where #c_i in mathcalR, k in mathcalZ#

Taking 2 simple examples **to illustrate** how the import of that turns completely on the **singularity** or otherwise of matrix A, and also how practically to go about working a span:

#A = ((1,2),(3,5)) #

#Span(A)# follows from the set of all solutions to:

#c_1 ((1),(3)) + c_2 ((2),(5)) = ((x),(y))#

Which is the same as solving:

# ((1,2),(3,5)) ((c_1),(c_2)) = ((x),(y))#

And which solves as: #c_1 = -5 x+2y, c_2 = 3x-y#

In other words, for any #langle x, y rangle# you specify, there exists a #(c_1, c_2)# pairing that includes that point within the span of A. #Span (A) = mathcal R^2#.

And:

# ((1,2),(3,5)) ((c_1),(c_2)) = ((0),(0)) implies (c_1, c_2) = (0,0)#

**It is no accident that A is non-singular.** #A \mathbf c = mathbf x# has only one solution which is #\mathbf c = A^(-1) mathbf x#. #A \mathbf c = mathbf 0# has one solution which is #\mathbf c = mathbf 0#

**Example 2**

#A = ((1,2),(3,6)) #

This is clearly singular as column 2 is a multiple of column 1, ie: #A = ((1,(1 times 2)),(3,(3 times 2))) #

Without even trying to solve it, we know that #langle x, y rangle#'s will lie only along direction: #((1),(3))#

Because A is singular: #A \mathbf c = mathbf x# will have no solution or infinitely many: #det (A) = 0# so #A^(-1)# does not exist.

In addition, #A \mathbf c = mathbf 0# will have infinitely many solutions:

# ((1,2),(3,5)) ((c_1),(c_2)) = ((0),(0)) implies c_1 = -2 c_2#

#Span(A) = mathcal R#.