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From Gilbert Strang's textbook, the augmented matrix $$ \begin{bmatrix} 1 & 2 & 3 & 5 & \mathbf{b_1}\\ 2 & 4 & 8 & 12 & \mathbf{b_2}\\ 3 & 6 & 7 & 13 & \mathbf{b_3} \end{bmatrix} $$ reduces to upper-triangular $$ \begin{bmatrix} 1 & 2 & 3 & 5 & \mathbf{b_1}\\ 0 & 0 & 2 & 2 & \mathbf{b_2-2b_1}\\ 0 & 0 & 0 & 0 & \mathbf{b_3+b_2-5b_1} \end{bmatrix} $$

As a first description of the column space, he writes that the column space is all linear combinations of the pivot columns $(1 ,2 ,3)$ and $(3 ,8 ,7)$. As a second description. he writes that the column space is all vectors that satisfy the bottom equation $0=b_3+b_2-5b_1$.

I believe that those pivot columns are a basis for the column space, and I also checked that they satisfy that bottom equality. I just don't understand why the two pivot columns satisfy that equation.

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Suppose we plug in the first pivot column $1,2,3$ for $b_1, b_2, b_3$. We get the matrix $$\begin{bmatrix}1 & 2 & 3 & 5 & 1 \\ 2 & 4 & 8 & 12 & 2 \\ 3 & 6 & 7 & 13 & 3\end{bmatrix}$$ and as we row-reduce it, it will always be true that the first column is the same as the last. (Row operations cannot change this property!) Therefore we must end with $$\begin{bmatrix}1 & 2 & 3 & 5 & 1 \\ 0 & 0 & 2 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0\end{bmatrix}.$$ In other words: when we start with $(b_1, b_2, b_3) = (1,2,3)$, we get $(b_1, b_2-2b_1, b_3+b_2-5b_1) = (1,0,0)$. In particular, the condition that the last row is $0$ must be true of the column space.

The same will be true if we set $(b_1, b_2, b_3)$ equal to the other pivot column $(3,8,7)$. Then the last column will always equal the third column as we row-reduce. So when we're done, the last entry of the third column ($0$) will equal the last entry of the last column ($b_3+b_2-5b_1$).

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Well the two vectors above define a basis for the column space and the second condition is basically giving what type of vectors are in the column space. Hence eventually the basis columns will also satisfy that condition.

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