Let $v_1, v_2, v_3$ be three vectors in $\mathbb{R}^3$ that span $\mathbb{R}^3$. Does this imply that $v_1, v_2, v_3$
are also linearly independent?
Correct answer:
Yes. Suppose that the vectors $v_1, v_2, v_3$ span $\mathbb{R}^3$ and let $A$ be a 3 × 3 matrix with
columns $[v_1 | v_2 | v_3]$. The system $Ax = b$ must be consistent for all b in $\mathbb{R}^3$, so rref(A)
must have a leading 1 in each row. Since $A$ is a square matrix, it follows that $A$ has a
leading 1 in each column as well, so there are no free variables. Therefore the system
$Ax = 0$ has a unique solution $x = 0$, which means precisely that the columns of $A$ are
linearly independent.
Me: I'm confused on how this yes can be said definitely because this example uses the case where it was a square matrix. But if it wasn't then none of these aforementioned arguments apply because then $A$ wouldn't have a pivot position in every row, so for each $b$ in $\mathbb{R}^{row}$, the equation $Ax=b$ might not have a solution. So the matrix could be inconsistent in some cases.
Best Answer
Any set of linearly independent vectors can be said to span a space.
If you have linearly dependent vectors, then there is at least one redundant vector in the mix. You can throw one out, and what is left still spans the space. So if we say $v_1,v_2, v_3$ span some space $V$ then it is implied that they are linearly independent.
Suppose $A$ is not a square matrix.
Suppose A is a $4\times 3$ matrix. That is A takes a vector in $\mathbb R^3$ to a subset of $\mathbb R^4$
The range of $A$ then is the space spanned by the linearly independent column vectors of $A$.