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.
