In Hofmann and Kunze Linear algebra book exercise 2.4.4
- Let $W$ be the subspace of $\mathbb {C}^3$ spanned by $a_1= (1, 0, i)$ and $a_2 =(1 + i, 1, - 1)$.
Show that the vectors $b_1= (1, 1, 0)$ and $b_2 = (1, i, 1 + i)$ are in $W$ and form another basis for W.
When I saw the solution in the answers here the answer used augmented matrix to solve this question
$$ \left(\begin{array}{@{} cc|cc @{}} 1 & 1+i & 1 & 1 \\ 0 & 1 & 1& i \\ i & -1 & 0 & 1+i \\ \end{array}\right) $$
and the row reduced it to
$$ \left(\begin{array}{@{} cc|cc @{}} 1 & 0 & -i & 2-i \\ 0 & 1 & 1& 1 \\ 0 & 0 & 0 & 0 \\ \end{array}\right) $$
Then he concluded that $b_1= -i a_1 +a_2$ and $b_2 =(2-i)a_1 + a+2$
I want to ask how this work ? and how to use this in general cases i.e with $m \times n $ ?
I realised that if $AX=Y$ and $P$ is some inheritable matrix that make $A$ row reduced encholm matrix then $RX=PY$ has the same solutions as the original system
I only need to prove that $R$ has only ones of the non zero rows (which in the first of the row) and zeros
if not then I have no idea how we could use $R$ to solve the the equation in this example it is very simple since all it contian is the ones of the non zero rows
You have the system of equations $$x+(1+i)y=1, y=1,ix-y=0$$ which you solve by doing row reduction on the augmented matrix. You also have the system of equations $$x+(1+i)y=1, y=i,ix-y=1+i$$ which you can also solve by doing row reduction on the augmented matrix.
However the only change to the system is in the augmented section so it's the same row operations to solve both systems. That means you can place them next to each other and the row operations will solve them both simultaneously. This is because all the decisions we make to solve the system are made by the information on the left of the partition without consideration to the augmented part. So while it seems like magic that you're solving for column vectors using row operations it's just a system of linear equations in disguise.