# How many linearly independent eigenvectors does this matrix have?

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?

• Yup, you're correct. – Jim Aug 17 '13 at 5:56

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.
The algebraic multiplicity of the eigenvalues of $A$ and $B$ is $1$ since they are distinct, therefore we can write
$$\begin{bmatrix}A & 0 \\ 0 & B \end{bmatrix} = \begin{bmatrix} P_A & 0 \\ 0 & P_B \end{bmatrix}^{-1}\begin{bmatrix}\Lambda_A & 0 \\ 0 & \Lambda_B \end{bmatrix} \begin{bmatrix} P_A & 0 \\ 0 & P_B \end{bmatrix}$$, where $\Lambda_X$ and $P_X$ comes from the decomposition $X = P_X^-{1}\Lambda_XP_X$, where $X$ is an index in $\{A, B\}$.
Recall that the eigenvalues of $\Lambda = \operatorname{diag}[\Lambda_A, \Lambda_B]$ also have algebraic multiplicity $1$, therefore the direct sum of the columns of $P_A$ and $P_B$ is the image of the block matrix $\operatorname{diag}[A, B]$ and the null space of $\operatorname{diag}[A, B] = \{0\}$. This is equivalent to say that the eigenvectors of $\operatorname{diag}[A, B]$ are linearly independent.