Blocks and Diagonalization
Suppose $A$ is diagonalizable as $S \Lambda S^{-1}$. Diagonalize the block matrix \begin{align} B = \begin{bmatrix} A &0\newline 0 &2A \end{bmatrix} \end{align} and find its eigenvalues and eigenvectors.
Solution
We break $B$ into three component block matrices made up of $S$, $\Lambda$, and $S^{-1}$. \begin{align} B = \begin{bmatrix} A &0\newline 0 &2A \end{bmatrix} = \begin{bmatrix} S\Lambda S^{-1} &0\newline 0 &2S\Lambda S^{-1} \end{bmatrix} = \begin{bmatrix} S &0\newline 0 &S \end{bmatrix} \begin{bmatrix} \Lambda &0\newline 0 &2\Lambda \end{bmatrix} \begin{bmatrix} S^{-1} &0\newline 0 &S^{-1} \end{bmatrix} \end{align} We can verify that the above multiplications give back the original block matrix $B$. Further, \begin{align} \begin{bmatrix} S &0\newline 0 &S \end{bmatrix} \begin{bmatrix} S^{-1} &0\newline 0 &S^{-1} \end{bmatrix} = \begin{bmatrix} I &0\newline 0 &I \end{bmatrix} = I_{2n} \end{align}
Hence, the above form is the correct diagonalized form of $B$ with the eigenvectors \begin{align} \begin{bmatrix} S &0\newline 0 &S \end{bmatrix} \end{align} and eigenvalues $(\lambda, 2\lambda)$ where $\lambda$ are all the diagonal entries of $\Lambda$ (eigenvalues of $A$).