# How do the 2 descriptions of column space describe the same subspace of a matrix?

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.

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 ($$b_3+b_2-5b_1$$).