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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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    I don't know what "full pivots" means. If the columns are linearly dependent then you won't be able to find the linear combinations you want, and the columns won't span ${\bf R}^3$. If "full pivots" prevents linear dependence, then, yes, the columns span ${\bf R}^3$.2012-05-01
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    Hint: If you have three linearly independent vectors, the dimension of the space that they span is ... $x$ ? How many subspaces of $\mathbb{R}^3$ are $x$-dimensional?2012-05-01
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    Alright. It just seemed odd to me that absolutely any square matrice with independent columns will be able to span $\mathbb{R}^n$ but I understand.2012-05-01
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    The singular of "matrices" is "matrix".2012-05-01
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    @LethalDiversion If you consider the case when $n=2$, it might be a little clearer why this works. A column vector describes a line through the origin, and linear dependence in $\mathbb{R}^2$ means that the two vectors generate the same line. As long as you're two column vectors don't generate the same line, you can mark every spot in the plane as a linear combination of the two vectors. We use $x$- and $y$-axes for convenience, not because they're the only ones that work.2012-05-01
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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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