Ok, so I have a question which is:
$$ A = \left ( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{array} \right ) $$
Find non-singular matrices P,Q such that $PAQ$ is a diagonal matrix of 1's and 0's, with the 1's appearing before the 0's, via a change of basis.
Now I can do this simply with row/column operations to get that:
$$P=I$$ and $$Q= \left ( \begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{array} \right ) $$
But I am unsure how to do this via a change of basis?
I can see that $null(A)=span(\{(1,1,-1)\})$ and $range(A)=span(\{(1,0,0),(0,1,0)\})$ and that a change of basis matrix to this basis is given by:
$$ \left ( \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & -1 \end{array} \right ) $$
and I can see that this does the job for Q, but I'm not really sure what I have done here (I've just blindly changed basis and this matrix has popped out!)
any help as to how I should be going about this and what it is I am actually doing would be great.
Thanks very much for any help
Best Answer
The procedure in general is as follows:
The two bases you should consider are $\mathcal B_1 \cup \mathcal B_2$ and $\mathcal B_3 \cup \mathcal B_4$ (try to understand why).
In your case $\mathcal B_1=\{(1,1,-1)\}$, you extended this randomly to a basis adding $\{(1,0,0),(0,1,0)\}$. The magic is that $A(1,0,0)=(1,0,0)$ and $A(0,1,0)=(0,1,0)$, so that $\mathcal B_3=\mathcal B_2$. The last step is to complete $\mathcal B_3$ to a basis, which can be done adding $(0,0,1)$.
The final result is: the first basis is the one you wrote down and the second basis is the canonical basis (hence $P=I$).