Find the change of basis matrix $P$ (say) for the change from the basis
$B=\{(1,0,0),(0,1,0),(0,0,1)\}$ to
$B_1=\{ (1,1,0),(0,0,1),(0,1,1)\}.$
I am familiar with change of coordinate matrix.
My definition of change of coordinate matrix is:
Let $V$ be a finite dimensional vector space such that, $\beta$ and $\beta'$ are ordered bases for $V$ and $I_v$ is the identity linear transformation, i.e $I_V:V\to V$ such that $I_v(x)=x,\forall x\in V.$ The change of coordinate matrix is defined to be the invertible matrix $Q=[I_v]_{\beta'}^{\beta}.$
But till now, never have I heard anything like, "change of basis matrix".
Assuming it to be synonymous with the notion of "change of coordinate matrix" of a vector space, my solution goes like this:
Let $e_1=(1,1,0),e_2=(0,0,1)$ and $e_3=(0,1,1).$
We note that, $$I_v(1,0,0)=(1,0,0)=1e_1+1e_2-e_3,$$$$I_v(0,1,0)=0.e_1-e_2+e_3,$$ and $$I_v(0,0,1)=0e_1+1e_2+0e_3.$$ Thus, $P=[I_v]_{B}^{B_1}=\begin{pmatrix} 1 && 0 && 0\\ 1 && -1 && 1\\ -1 && 1 && 0\end{pmatrix}.$
Now, I want to know whether my understanding along with my solution is valid? I need some clarification regarding this issue.
Best Answer
Your solution is valid, to check whether the result is correct you can consider the inverse map and its associated matrix: $$Q:=[I_v^{-1}]_{B_1}^B=\begin{pmatrix} 1 && 0 && 0\\ 1 && 0 && 1\\ 0 && 1 && 1\end{pmatrix} $$ Since $I_v\circ I_v^{-1}=Id=I_v^{-1}\circ I_v$, you should have $PQ=I=QP$, which is true according to your computation.