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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?

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  • $\begingroup$ If you have n vectors of dimension n and det(A) = 0, where A is the matrix whose rows are those vectors, then at least one of those vectors is a linear combination of the others. Thus there are at most n-1 independent vectors and if your space is n dimensional they cannot span it. $\endgroup$ – Paul Dec 11 '14 at 10:52
  • $\begingroup$ @Paul i understand your point but didnt quite anwser my question.i have edit the question also $\endgroup$ – user173788 Dec 11 '14 at 17:22
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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., $\dim\text{Im}(A)=\dim V$.

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  • $\begingroup$ I dont quite follow you , because i don`t get to that part yet. I have edit the question can you check it? $\endgroup$ – user173788 Dec 11 '14 at 17:20
  • $\begingroup$ @KellyBlunie, you keep supposing that $A$ is a square matrix. In this case injective $\iff$ surjective $\iff$ $\det A\ne 0$. $\endgroup$ – Martín-Blas Pérez Pinilla Dec 11 '14 at 18:09
  • $\begingroup$ I supose my question is confuse,also i do not quite follow you, my question is if we have det $A$ non-zero then every vector u has exactly one solution what happens when determaniant is zero and why? $\endgroup$ – user173788 Dec 11 '14 at 18:23
  • $\begingroup$ @KellyBlunie, you know the rank-nullity theorem (en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem)? $\endgroup$ – Martín-Blas Pérez Pinilla Dec 11 '14 at 18:27
  • $\begingroup$ Not really but the book i am reading i ,did not get to that part,however make this claim: there is at least one choice of u for which this system will not have a solution and so u can not be written as a linear combination of these vectors. And that there are in fact infinitely many choices of u that will not yield solutions. $\endgroup$ – user173788 Dec 11 '14 at 18:37

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