Notation:
Let $V$ be a vector space. Given $T$ in $Hom(V)$ and a direct sum decomposition $V=\bigoplus_1^n V_i$, with corresponding projections $\{P_i\}_1^n$
we can consider the maps $T_{ij}=P_i\circ T\circ P_j$. Although $T_{ij}$ is from $V$ to $V$, we may also want to consider it as being from $V_j$ to $V_i$.
We picture the $T_{ij}$'s arranged schematically in a rectangular array similar to a matrix, as indicated below for $n=2$
\begin{array} {r|r} T_{11} & T_{12} \\ \hline T_{21} & T_{22} \\ \end{array}
Furthermore, since $T=\sum_{i,j}T_{ij}$, we call the doubly indexed family the block decomposition of $T$ associated with the given direct sum decomposition of $V$.
Question:
Show that if $V=V_1\oplus V_2 \oplus V_3$ and $T\in Hom(V)$, then the subspaces $V_i$ are all invariant under $T$ if and only if the block diagram for $T$ is
$$\begin{bmatrix} T_{11} & 0 & 0 \\ 0 & T_{22} & 0 \\ 0 & 0 & T_{33} \end{bmatrix}$$ Show that $T$ is invertible if and only if $T_{ii}$ is invertible (as an element of $Hom (V_i)$) for each $i$.
I was able to solve the first part of the question but the second part doesn't seem right.
Here is a counter example for the second part:
Let $$V=\mathbb{R}^3$$
$$\begin{align}
V_1=\{\langle x,0,0 \rangle \space | \space x\in\mathbb{R}\} \\
V_2=\{\langle 0,y,0 \rangle \space | \space y\in\mathbb{R}\} \\
V_3=\{\langle 0,0,z \rangle \space | \space z\in\mathbb{R}\}
\end{align}$$
notice that $$V=V_1\oplus V_2 \oplus V_3$$
Let $T\in Hom(V)$, such that $$T(\langle x,y,z \rangle) = \langle y,x,z \rangle$$
$T$ is invertible ($T^{-1}=T$).
Let $P_i$ be the projection of $V$ onto $V_i$, $i\in\{1,2,3\}$.
For example $P_1(\langle x,y,z \rangle)=\langle x,0,0 \rangle$
We have defined $T_{11}\in Hom(V_1)$ as $T_{11}=P_1\circ T\circ P_1$.
Let $\langle x,0,0 \rangle \in V_1$.
We have, $$T_{11}(\langle x,0,0 \rangle) = P_1\circ T \circ P_1(\langle x,0,0 \rangle)= P_1 \circ T(\langle x,0,0 \rangle) = P_1(\langle 0,x,0 \rangle)=\langle 0,0,0 \rangle$$
Hence $T_{11}=0$ and not invertible. but $T$ is invertible therefore the claim doesn't hold.
Is there something wrong with my counter example or the question is wrong?