Let $f$ be an endomorphism of $\mathbb{R^3}$ which the matrix in the standard basis is $A=\begin{bmatrix}3&2&0\\-1&0&0\\0&0&1\end{bmatrix}$
Let $F_1=Ker(f-Id_\mathbb{R^3})$ , $F_2=Ker(f-2Id_\mathbb{R^3})$
Let $p$ be the projection on $F_1$ parallel to $F_2$
Determine the matrix of the projection $p$.
Method 1:
$\mathbb{R^3}=F_1 \oplus F_2$
$\forall v \in \mathbb{R^3} , v=v_1+v_2 $ with $v_1 \in F_1 , v_2 \in F_2$
$f(v)=f(v_1)+f(v_2)$
($f-Id)(v_1)=0$
Thus:
$f(v_1)=v_1$
$(f-2Id)(v_2)=0$
Thus
$f(v_2) = 2v_2$
$f(v)=v_1+2v_2=p(v)+2(v-v_1)=p(v)+2(Id(v)-p(v))=2Id(v)-p(v)$
This leads to : $p(v)=2Id(v)-f(v)$
And the matrix of $p$ is $P=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}-\begin{bmatrix}3&2&0\\-1&0&0\\0&0&1\end{bmatrix}=\begin{bmatrix}-1&-2&0\\1&2&0\\0&0&1\end{bmatrix}$
Method 2:
$p$ is the projection on $F_1$ parallel to $F_2$ hence $F_2$ is in the kernel of $p$
$F_1=span${$v_1=(1-1,0),v_2=(0,0,1)$}
$F_2=span${$v_3=(-2,1,0)$}
Let's determine $p$ in the basis $B=${$v_1,v_2,v_3$}
We have $p(v_3)=0$ , $p(v_1)=v_1$ ($v_1 \in F_1$) , $p(v_2)=v_2$ ($v_2 \in F_1$)
Hence the matrix of $p$ in the basis B is $P_B=\begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix}$
The transition matrix from $B$ to the standard basis is $T=\begin{bmatrix}-1&-2&0\\0&0&1\\-1&-1&0\end{bmatrix}$,$T^{-1}=\begin{bmatrix}1&0&-2\\-1&0&1\\0&1&0\end{bmatrix}$
And the matrix of $p$ in the standard basis would be equal to :
$P$=$T . P_B . T^{-1}=\begin{bmatrix}-1&-2&0\\0&0&1\\-1&-1&0\end{bmatrix} . \begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix} . \begin{bmatrix}1&0&-2\\-1&0&1\\0&1&0\end{bmatrix}=\begin{bmatrix}1&0&0\\0&0&0\\0&0&1\end{bmatrix}$
How come both of these methods give out different solutions ?
Where did i go wrong ?
Thanks in advance!
Your issue is that you've mixed up your transition matrices. Remember, you're only supposed to multiply $P_B$ by coordinate column vectors with respect to $B$. What you've written as $T^{-1}$ takes column vectors with respect to $B$ and produces the same vector as a column vector with respect to the standard basis. That is, $T^{-1} v$ is a standard basis coordinate column vector, not with respect to $B$. If you multiply the other way, $$\begin{bmatrix}1&0&-2\\-1&0&1\\0&1&0\end{bmatrix} \begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix} \begin{bmatrix}-1&-2&0\\0&0&1\\-1&-1&0\end{bmatrix}=\begin{bmatrix}-1&-2&0\\1&2&0\\0&0&1\end{bmatrix},$$ as required.