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Here in my notes I have an example of getting the columnspace of a matrix A. The answer is all linear combinations of (1,0,0,1) and (-2,1,0,0).

But I know that you can get the columnspace by reducing a matrix to echelon form and then looking at the pivot columns - the columns in the original matrix that correspond to the pivot columns will be a basis for the columnspace. If I do that in the case of the example below I get all linear combinations of (1,0,0,1) and (0,1,0,2). These vectors seem to span the same space as the vectors above - (1,0,0,1) and (-2,1,0,0). So are there two ways of getting the columnspace for a matrix?

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Yes, it is easy to see that $$\operatorname{span}\{ \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \\ 2 \end{bmatrix} \} = \operatorname{span}\{ \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} -2 \\ \phantom{-}1 \\ \phantom{-}0 \\ \phantom{-}0 \end{bmatrix} \} $$ To see this, note that one vector is in both bases, and we have $$ 2 \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix} + \begin{bmatrix} -2 \\ \phantom{-}1 \\ \phantom{-}0 \\ \phantom{-}0 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 2 \end{bmatrix}. $$ So yes, you could say that there are (at least) two ways of getting the column space. One of them is, as you've mentioned, reducing the matrix to echelon form and picking out column vectors in the original matrix corresponding to pivot columns. This one is the easiest to apply generally.

The other is to use the definition directly, and try to find the answer by inspection - one can easily see that in the matrix $A$, the third column is simply the sum of the first two, and furthermore, the first two columns are linearly independent. Hence the first two columns form a basis for the column space of $A$.

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Yes. ${}{}{}{}{}{}{}{}{}{}{}{}$

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