$C= A\oplus B$. Then prove that $C$ is diagonalisable if and only if both $ A$ and $B$ are diagonalisable. 
Let $A\in M_n(R)$ and $B\in M_m(R)$. Suppose $C= A\oplus B$. Then
  prove that $C$ is diagonalisable if and only if both $ A$ and $B$ are
  diagonalisable.

Let $C$ be diagonalisable, so $\exists P=P_1 \oplus P_2$ an invertible matrix such that,
$P^{-1}CP=D$, where $D= D_n \oplus D_m$ is diagonal matrix 
$\rightarrow (P_1^{-1} \oplus P_2^{-1})( A\oplus B) (P_1 \oplus P_2) = D_n\oplus D_m $ 
$\rightarrow P_1^{-1}AP_1 = D_n$ and $P_2^{-2}BP_2=D_m$ so they both become diagonalisable. Is this correct?
Now if I suppose both $A$ and $B$ are diagonalisable, then there will exists invertible matrices $P_1$ and $P_2$ such that 
$P_1^{-1}AP_1 = D_n$ and $P_2^{-1}BP_2=D_m$
Then I will define $P= P_1 \oplus P$ which will give me $P^{-1}CP=D$
Is this correct?
 A: Okay. So we know that $\mathbb{R}^n$ has a basis $\{v_j\}_{1\leq j\leq n}$ of eigenvectors of $V$ with corresponding eigenvalues $\{\lambda_j\}_{1\leq j\leq n}.$
Indeed, each $v_j=v_{j,A}+v_{j,B}$, where $v_{j,A}\in Dom(A)$ and $v_{j,B}\in Dom(B)$. Note that this decomposition is unique.
Now, we see that
$$
\lambda_j v_{j,A}+\lambda_j{v_{j,B}}=\lambda_jv_j=Cv_j=Av_{j,A}+Bv_{j,B}
$$
By the uniqueness of the direct sum composition, we see that $v_{j,A}$ and $v_{j,B}$ are eigenvectors of $A$ and $B$ respectively corresponding to the same eigenvalue, provided that they aren't $0$. Since $v_j$ isn't $0$, at least one of the two must be an actual eigenvector.
Furthermore, since $\{v_j\}_{1\leq j\leq n}$ is spanning, we can thin $v_{j,A}$ to a basis $\{e_k\}_{1\leq k\leq dim(A)}$ of $Dom(A)$ and $v_{j,B}$ to a basis $\{f_l\}_{1\leq l\leq dim(B)}$ of $Dom(B)$. Now, by the above, these are bases of eigenvectors of $A$ and $B$ respectively. This implies that $A$ and $B$ are diagonalisable (you could of course build a new basis of $\mathbb{R}^n$ using the $e_k$ and $f_l$ to show that $P$ has the proper decomposition, but that seems fairly unnecessary at this point).
