Let $n>1$ and $A_1,A_2,A_3 \in M_n(F)$. Let $A$=\begin{bmatrix}A_{1}&A_{2}\\0_{n}&A_{3}\\\end{bmatrix} Prove the following.
1.) If $A_1$ and $A_3$ are diagonalizable, then $A$ is diagonalizable.(Does the converse of this is true?)
2.) If $\lambda$ is an eigenvalue of $A_1$ and of $A_3$, then $\lambda$ is an eigenvalue of $A_1 + A_3$.
Any and all help would be appreciated, please help. Thank you.
Part (1) is wrong $$ \left( \begin{array}{rrrr} 0&1&1&0 \\ -1&0&0&1 \\ 0&0&0&1 \\ 0&0&-1&0 \end{array} \right) $$ The characteristic polynomial and the minimal polynomial are both $(x^2+1)^2$
Or, all real characteristic values: $$ \left( \begin{array}{rrrr} 1&0&1&0 \\ 0&3&0&1 \\ 0&0&1&0 \\ 0&0&0&3 \end{array} \right) $$ The characteristic polynomial and the minimal polynomial are both $(x-1)^2 (x-3)^2$
Both your statements are false!
Take $A_1=A_3=0_n$ and $A_2$ to be any non-zero matrix.
Take $A_1=\begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}$ and $A_3=-A_2$. Then 1 is an eigenvalue of $A_1$ and $A_3$ but $A_1+A_3=0_2$.
