Exercises

1. Blocks and Diagonalization

Suppose \(A\) is diagonalizable as \(S \Lambda S^{-1}\). Diagonalize the block matrix

\[ \begin{aligned} B = \begin{bmatrix} A &0\newline 0 &2A \end{bmatrix} \end{aligned} \]

and find its eigenvalues and eigenvectors.

Solution

We break \(B\) into three component block matrices made up of \(S\), \(\Lambda\), and \(S^{-1}\).

\[ \begin{aligned} B = \begin{bmatrix} A &0\newline 0 &2A \end{bmatrix} = \begin{bmatrix} S\Lambda S^{-1} &0\newline 0 &2S\Lambda S^{-1} \end{bmatrix} = \begin{bmatrix} S &0\newline 0 &S \end{bmatrix} \begin{bmatrix} \Lambda &0\newline 0 &2\Lambda \end{bmatrix} \begin{bmatrix} S^{-1} &0\newline 0 &S^{-1} \end{bmatrix} \end{aligned} \]

We can verify that the above multiplications give back the original block matrix \(B\). Further,

\[ \begin{aligned} \begin{bmatrix} S &0\newline 0 &S \end{bmatrix} \begin{bmatrix} S^{-1} &0\newline 0 &S^{-1} \end{bmatrix} = \begin{bmatrix} I &0\newline 0 &I \end{bmatrix} = I_{2n} \end{aligned} \]

Hence, the above form is the correct diagonalized form of \(B\) with the eigenvectors

\[ \begin{aligned} \begin{bmatrix} S &0\newline 0 &S \end{bmatrix} \end{aligned} \]

and eigenvalues \((\lambda, 2\lambda)\) where \(\lambda\) are all the diagonal entries of \(\Lambda\) (eigenvalues of \(A\)).