Find the basis for column space $$A=\begin{bmatrix}1&-1&3\cr 5&-4&-4\cr 7&-6&2\end{bmatrix}.$$
I'm quite confused because I thought there were two methods: using the transpose or not.
By not using the transpose of $A$ and computing the RREF, I would get $$A=\begin{bmatrix}1&0&-16\cr 0&1&-19\cr 0&0&0\end{bmatrix},$$ for which I find that columns 1 and 2 have leading 1s. Therefore, I would go to the original matrix's columns 1 and 2 and say the basis for the column space is $\begin{bmatrix}1 & 5 & 7\end{bmatrix}$ and $\begin{bmatrix}-1 & -4 & -6 \end{bmatrix}.$
However, my textbook solution did the transpose method to get $$A^T=\begin{bmatrix}1&5&7\cr -1&-4&-6\cr 3&-4&2\end{bmatrix},$$ and the reduced form is $$A=\begin{bmatrix}1&5&7\cr 0&1&1\cr 0&0&0\end{bmatrix}.$$ The book got $\langle 1,5,7 \rangle$ and $\langle 0,1,1 \rangle;$ however, if we reduce it further, we get $\langle 1,0,2 \rangle$ and $\langle 0,1,1 \rangle.$
- I'm not sure whether the textbook sol of $\langle 1,5,7 \rangle$ and $\langle 0,1,1 \rangle$ vs. $\langle 1,0,2 \rangle$ and $\langle 0,1,1 \rangle$ is right.
- And why are all these numbers different from finding the corresponding columns from the original matrix? Is this method wrong to begin?
Both of these solutions give rise to the same basis. Observe that for $v_1 = (1, 5, 7)$ and $v_2 = (-1, -4, -6),$ we have that $v_3 = v_1 + v_2 = (0, 1, 1),$ from which it follows that $\operatorname{span}_k \{v_1, v_2 \} = \operatorname{span}_k \{v_1, v_3 \}.$ Of course, we could go one step further to see that $v_4 = v_1 - 5 v_3 = (1, 0, 2)$ so that $\operatorname{span}_k \{v_1, v_3 \} = \operatorname{span}_k \{v_3, v_4 \}.$