Diagonalization and Commuting Matrices

Question

Attempt:

I have shown part (ii) and I have found the eigenvalues and eigenvectors for A, B respectively and shown they can be diagonalised. I need help with (iii) and (iv), for (iii) I can't show the last part that there exists a common basis transformation. Also, for (iv) I don't know how to find the matrix S (if someone could tell me the method that would be great). Thanks.

Answer 1

For $(iii)$

$Dx=\lambda x$ so we have $CDx= DCx = \lambda Cx$

• if $Cx = 0$ , the statement is true
• if $Cx \neq 0$, $Cx$ will be an eigenvector of $D$ associated to $\lambda$. this implies that $Cx$ and $x$ are collinear because $D$ has $n$ different eigenvalues therefor the eigenspace associated to an eigenvalue $\lambda$ is one dimensional space.

we have $ Cx= \beta x $, therefor $x$ is an eigenvector of $C$, and there are $n$ of them. this shows that $D$ and $C$ are diagonalizable by the same system of ($n$ different) eigenvectors.

Hint for $(iv)$

After verifying that the $A$ and $B$ are commutative. You have to find the three distinct eigenvectors of $A$ ( or $B$ ) name them $x_1$, $x_2$ and $x_3$

Take $S$ to be the matrix

$$ S :=[x_1 \ x_2 \ x_3 ] $$

then we have $$ A= S D_1 S^{-1} \ \ \ \ \ B = S D_2 S^{-1} $$

Where $D_1$ and $D_2$ are diagonal. And therefor

$$ D_1= S^{-1} A S \ \ \ \ \ D_2 = S^{-1} B S $$