If we want to see if a set of vectors spans a vector space $V$, then lets say the set $A$ spans a vector space $V$ only If every linear combination of $A$ produces $V$, then $\text{Span}(A) = V$
Edit: if we forme the coeficient matrix of the system formed by $c_1s_1+..+c_ns_n =u$ where $s_i$ are the vectors in set $A$ and u is any vector in $V$ , if the determinant is zero, then there is at least one choice of u for which this system will not have a solution and hence can not be written as a linear combination of these vectors?why is this?
Note: i know that if determinant is non-zero then it will have exactly on solution for each u.i know that if the determinant is zero then it can have infinitily many solutions or no solution.
Supose that the coefficient matrix is an n*n matrix
In other words, if the determinant is a non-zero each u will have exactly one solution if it is zero what happens?and why?
Best Answer
And if $A$ (matrix) isn't a square matrix? The true condition is $A$ (set) spans $V$ iff $\text{rank }A =\dim V$ (iff some $\dim V\times\dim V$ submatrix of $A$ has $\det\ne 0$).
EDIT: let be the columns of $A$ (matrix) the elements of $A$ (set). The span of $A$ (set) can be written as the set of all the $Ax$ with $x$ column vector, i.e., the image of $A$. You want $A$ surjective, i.e., $\dim\text{Im}(A)=\dim V$.