# Column Space of Square Matrix

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 ${\bf R}^3$. – Gerry Myerson May 1 '12 at 7:41
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? – Jyrki Lahtonen May 1 '12 at 7:45
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. – LethalDiversion May 1 '12 at 7:47
The singular of "matrices" is "matrix". – joriki May 1 '12 at 7:53
@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. – chris May 1 '12 at 13:33

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 $\mathbb R^3$.