I have a $4\times 4$ matrix which I can write in $2\times 2 $ block form as
$$\begin{pmatrix}A&O\\O&B\end{pmatrix}$$
I was asked how many linearly independent eigenvectors it has? I checked that $A$ and $B$ have distinct eigenvalues, i.e the whole matrix has $4$ distinct eigenvalues, eigenvectors corresponding to distinct eigenvalues are linearly independent.
So $4$ is the answer. Is this correct?
Best Answer
I see what you mean, but I think a strict proof should be like this:
Since $A,B$ have distinct eigenvalues, we have decompositions: $$A=P\Lambda_1 P^{-1},\qquad B=Q\Lambda_2 Q^{-1},$$ where $\Lambda_1$, $\Lambda_2$ are diagonal matrices with elements being $A$'s eigenvalues and $B$'s eigenvalues respectively, and $P,Q$ are invertible matrices with columns being $A$'s eigenvectors and $B$'s eigenvectors respectively. Then we have $$\left(\begin{array}{ccc}A&0\\0&B\end{array}\right)=\left(\begin{array}{ccc}P&0\\0&Q\end{array}\right)\left(\begin{array}{ccc}\Lambda_1&0\\0&\Lambda_2\end{array}\right)\left(\begin{array}{ccc}P&0\\0&Q\end{array}\right)^{-1},$$ and $\left(\begin{array}{ccc}\Lambda_1&0\\0&\Lambda_2\end{array}\right)$'s diagonal elements are the eigenvalues of $\left(\begin{array}{ccc}A&0\\0&B\end{array}\right)$, and $\left(\begin{array}{ccc}P&0\\0&Q\end{array}\right)$'s columns are the corresponding eigenvectors.