diagonalization addition

Question

Suppose that A and B are diagonalizable and have the same eigenvectors. Prove that A+B is diagonalizable. I get $A=PD_AP^{-1}$ and $B=PD_BP^{-1}$, and A+B = $PD_AP^{-1}+PD_BP^{-1}$, but how do I combine these two terms?

Answer 1

$$A+B = PD_AP^{-1} + PD_BP^{-1} = P(D_A + D_B)P^{-1}$$

The last step is valid because: $$P(D_A + D_B)P^{-1}=(P(D_A + D_B))P^{-1} = (PD_A + P D_B)P^{-1} = PD_AP^{-1} + PD_BP^{-1}$$

Answer 2

then $(A+B)P,=P D_A + P D_B$ , then left multiply by $P^{-1}$, then revert the multiplications

Answer 3

If $A$ is diagonalisable then it means there must be a basis of eigenvectors which are shared by $B$. Call them $v_1,...,v_n$.

Since $A,B,A+B$ are linear it is sufficient to check the behaviour on a basis, in the case the eigenvectors.

Suppose the eigenvalues corresponding to the $v_k$ are $\lambda_k^A,\lambda_k^B$, then we have

$(A+B)v_k = (\lambda_k^A+\lambda_k^B) v_k$ and so we see that the $v_k$ are eigenvectors of $A+B$ and since they form a basis, $A+B$ is diagonalisable (using the matrix $V=\begin{bmatrix} v_1 & \cdots & v_n \end{bmatrix}$).