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I know there's a theorem in linear algebra that has a list of statements that are equivalent, as follows:

  1. $\det(A) \ne 0$
  2. $Ax = 0$ has only the trivial solution
  3. $Ax = b$ is consistent for every matrix $b$
  4. $Ax = b$ has exactly one solution for every $n \times 1$ matrix $b$

(and several other which I do not need for the question that I'd like to ask).

So we've been learning about linear independence, span and bases, and I was wondering the following:

Is statement 2 the same as saying a set of vectors are linearly independent?

Is statement 3 the same as saying the set of vectors span a vector space? (Meaning that every single vector $b$ in the vector space can be made from some combination of $Ax$, where $A$ would represent that vectors and $x$ would be all of the scalar multiples of the vectors?

Is statement 4 the same as saying the set of vectors is a basis? (since that one solution must be the trivial solution, it must be linearly independent, and since it is consistent for every $b$, it spans the vector space?)

It seems pretty obvious that it's all related to each other, but I'm just wondering if they're all exactly the same.

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  • $\begingroup$ You write about "a set of vectors" and "the set of vectors", but you never tell us which set of vectors you mean. $Ax=0$ and $Ax=b$ are equations, they aren't sets of vectors. If you're going to get anywhere in Linear Algebra, you have to be a lot more precise in your use of the language. $\endgroup$ – Gerry Myerson Mar 29 '16 at 8:36
  • $\begingroup$ To be honest, while writing this, I thought I WASN'T being that precise, but I just didn't know how to be precise. Usually in the problems I'm given a set of vectors...which I write them as a system of linear equations. The coefficient matrix A seems to represent the vectors in the given set, and the column vector x represents the scalars used to create a linear combination...so would this be more precise: statement 2 is equal to saying that the set of vectors represented by the coefficient matrix A in the equation Ax = b is linearly independent because the solution matrix x is 0 ? $\endgroup$ – FrostyStraw Mar 29 '16 at 8:56
  • $\begingroup$ It's not more precise, until you explain what you mean by a matrix representing a set of vectors. $\endgroup$ – Gerry Myerson Mar 29 '16 at 10:16
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    $\begingroup$ I'm voting to close this question as off-topic because OP has abandoned it. $\endgroup$ – Gerry Myerson Apr 3 '16 at 7:13
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    $\begingroup$ OK, so, the vectors you are talking about are the columns of the matrix you are talking about. That's easy to say, and quite precise. $\endgroup$ – Gerry Myerson Apr 3 '16 at 11:38
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(2) is the same as saying the columns of $A$ are linearly independent.

(3) is the same as saying the columns of $A$ span ${\bf R}^n$, where $n$ is the number of rows of $A$.

(4) is the same thing as saying the columns of $A$ are a basis for ${\bf R}^n$. [I assume $Ax-b$ is a typo for $Ax=b$.]

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Your descriptions for (2) and (3) sound good. I wouldn't say (4) is quite a direct translation of the statement that the columns of $A$ form a basis, although it's of course equivalent and it's easy to see that (4) implies (2) and (3).

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