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Assume that I have a $3 \times 3$ matrix $A$ with columns $A_1$, $A_2$, and $A_3$ that are linearly independent. Say that I want to find the column space of A. Isn't it possible for me to find some combination of $A_1$, $A_2$, and $A_3$ such that I can come up with $ \left[ {\begin{array}{c} 1 \\ 0 \\ 0 \end{array} } \right]$, $\left[ {\begin{array}{c} 0 \\ 1 \\ 0 \end{array} } \right]$ and $\left[ {\begin{array}{c} 0 \\ 0 \\ 1 \end{array} } \right]$ and just say that the columns of $A$ span all of $\mathbb{R}^3$?

Is anything stopping me from doing this? What is it that limits column spaces?

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    Another idea is that a matrix $A$ with linearly independent columns are invertible, so $\forall\ v \in \mathbb{R}^n, v = A (A^{-1} v)$, so $v \in Ran(A)$.2018-10-03

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Column space is, at least according to my understanding, the span of the columns of the associated matrix. If the three rows are linearly independent, then Gauss-Jordan elimination will leave you with an identity matrix if the algorithm is taken through to completion. From here, you can conclude that the span is all of $\mathbb R^3$.

Note that this is only true for linearly independent columns. If a column can be expressed as the sum of two other columns, then you have only two columns to work with in the span (for a 3x3 matrix). Since two columns form a plane instead of a 3D space, you would have to pick two columns, it doesn't matter which two (think about why?), and use that as your span.