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Definiton: The column space of an $m \times n$ matrix $A$, written as $\operatorname{Col} A$, is the set of all linear combinations of the columns of $A$. If $A = [ a_1 \ldots a_n ]$, then $\operatorname{Col} A = \operatorname{Span}\{a_1,\ldots,a_n\}$.

$$A= \begin{bmatrix} 2 & 4 & -2 & 1 \\ -2 & -5 & 7 & 3 \\ 3 & 7 & -8 & 6 \end{bmatrix}$$

Find a nonzero vector in $\operatorname{Col} A$.

Solution: It is easy to find a vector in $\operatorname{Col} A$. Any column of $A$ will do.

I'm confused on why any column of $A$ works because $A$ in rref form is:

$$\begin{bmatrix} 1 & 0 & 9 & 0 \\ 0 & 1 & -5 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

So wouldn't only columns 1, 2, and 4 work because column 3 is linearly dependent, and thus not a part of the span?

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  • $\begingroup$ You could also say column 2 is "linearly dependent" since it's a linear combination of columns 1 and 3. Or column 1 is "linearly dependent" since it's a linear combination of columns 2 and 3. "Linear dependence", as usually defined, is a property of a set of vectors, not of a vector. $\endgroup$ Apr 14 '17 at 2:07
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The third column is also in the span because it can be written as $$Col 3=0.Col1+0.Col2+1.Col3+0.Col4$$ What the rref tells you is that the columns $1, 2$ and $4$ not only span $Col A$ but they are also linearly independent. Hence $Col 3$ could alternately be written as $$Col3=a.Col1+b.Col2+c.Col4$$ for some $a, b, c$.

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As you stated the column space is the set of all linear combinations of the columns of $A$. So if the matrix is not the zero matrix, you will be able to find some non zero vector in $Col A$. Colum 3 is non zero and in fact your column space is the set of all vectors in 3D because $x\times col1 + y \times col2 + z\times col3$ gives any vector in $\mathbb{R}^3$.

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