Let a field $\mathbb{F}$ and $n_1,\ldots,n_l$, natural numbers. For all $1\le i\le l$ Let $A_{ii} \in M_{n_i}(\mathbb{F})$.
Let $$A = \left( {\matrix{ {{A_{11}}} & {{A_{12}}} & \cdots & {} & {} & {{A_{1n}}} \cr 0 & {{A_{22}}} & {{A_{23}}} & \cdots & {} & {} \cr \vdots & {} & \ddots & {} & {} & {} \cr {} & {} & {} & {} & {} & {} \cr {} & {} & {} & {} & {{A_{(l - 1)(l - 1)}}} & {{A_{(l - 1)l}}} \cr 0 & \ldots & {} & {} & 0 & {{A_{ll}}} \cr } } \right)$$
Prove that $A_{ii}$ is similar to a triangular matrix iff $A$ is similar to a triangular matrix
My Try:
Lets assume $A$ is similar to a triangular matrix.
We know that the characteristic polynomial of $A$ is $f_A(x) = f_{A_{11}}(x)\cdot\ldots \cdot f_{A_{ll}}(x)$
Since $A$ is similar to $B$, an upper triangular matrix they have the same characteristic polynomial. Hence, $$ f_B(x) = f_{A_{11}}(x)\cdot\ldots f_{A_{ll}}(x)$$
But, we know that $B$ is diagonalizable (since it's similar to $A$). Hence, the characteristic polynomial can also be written as:
$$ f_B(x) = (x-\lambda_1)\cdots \ldots\cdot (x-\lambda_l) $$
So for every $f_{A_{ii}}$ there is a $j$ such that $$f_{A_{ii}} = (x-\lambda_j)$$
Can I conclude at this point that $f_{A_{ii}}$ is similar to an upper triangular matrix?
Notice that if exists $B$ such that $A=BTB^{-1}$, where $T\in M_n(\mathbb{F})$ is an upper triangular matrix then the eigenvalues of $A$ belong to $\mathbb{F}$.
Now, if all eigenvalues of $A$ belong to $\mathbb{F}$ then exists $B\in M_n(\mathbb{F})$ such that $A=BJB^{-1}$, where $J$ is the jordan canonical form of $A$, which is upper triangular. See this link for more details.
Thus, $A\in M_n(\mathbb{F})$ is similar to a triangular matrix $T\in M_n(\mathbb{F})$ if and only if the eigenvalues of $A$ belong to $\mathbb{F}$.
Since the eigenvalues of $A$ are the eigenvalues of $A_{ii}$, $1\leq i\leq n$, then $A$ is similar to a triangular matrix if and only if the eigenvalues of $A_{ii}$, $1\leq i\leq n$, belong to $\mathbb{F}$ if and only if each $A_{ii}$ is similar to a triangular matrix.