Let $T : R^3 → R^3$ be the linear map $T(x, y, z) = (y, −x, z)$ for all $(x, y, z) ∈ R^3$. Let $E$ be the standard basis of $R^3$, and let $F$ be the basis $\{f_1, f_2, f_3\}$ where $f_1 = (1, 1, 1), f_2 = (1, 1, 0)$ and $f_3 = (1, 0, 0).$
(a) Calculate the matrix $A$ of $T$ with respect to the standard basis $E$ of $R ^3$.
(b) Calculate (directly) the matrix $B$ of $T$ with respect to the basis $F$.
(c) Let $I$ be the identity map of $R^3$. Calculate the matrix $P$ of $I$ with respect to the bases $E, F$ and the matrix $Q$ of $I$ with respect to the bases $F, E$, and check that $PQ = I_3$
This is what I did:
My result for (a) is
$$
\begin{bmatrix}
0 & 1 & 0 \\
-1 & 0 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}$$
the standard basis is
$$
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}$$
A=$U^{-1}$BU and $B=UAU^{-1}$
I get
$$B=
\begin{bmatrix}
0 & -1 & 0 \\
-2 & -1 & -1 \\
0 & 2 & 0 \\
\end{bmatrix}$$
I am confusing about the (c). Considering identity matrix map to itself, then I believe I the identity matrix times the basis gives the standard basis but it is wrong. Where am I wrong?
$$\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{bmatrix}$$ $$\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0 \\ \end{bmatrix}$$
$I(1,0,0)=(1,0,0)=f_3$
$I(0,1,0)=(0,1,0)=f_2-f_3$
$I(0,0,1)=(0,0,1)=f_1-f_2$
$P=[I]^E_F = \begin{pmatrix} 0 & 0 &1 \\ 0 &1 &-1 \\ 1& -1 &0 \end{pmatrix} $
$I(1,1,1)=(1,1,1)=e_1+e_2+e_3$
$I(1,1,0)=(1,1,0)=e_1+e_2$
$I(1,0,0)=(1,0,0)=e_1$
$Q=[I]^F_E = \begin{pmatrix} 1 & 1 &1 \\ 1 &1 &0 \\ 1&0 &0 \end{pmatrix} $
$PQ=I_3$