Find a and b such that the matrix
$$
\left( \begin{array}{ccc}
1 & a \\
0 & b \\
\end{array} \right)
$$
is diagonalizable.
I know that
$$
D = S^{-1} A S
$$
where S is a matrix made of the eigenvectors and A is the original matrix.
I can't seem to find the eigenvectors properly, but I think I have the eigenvalues correct.
$$
det(
\left( \begin{array}{ccc}
1-\lambda & a \\
0 & b-\lambda \\
\end{array} \right)) = 0 \\
(1-\lambda)(b-\lambda)=0 \\
\lambda = 1 \\
\lambda = b
$$
I'm not really sure how to find the eigenvectors from here or how to invert S afterward.
Best Answer
If $b$ is anything other than 1 then the matrix is diagonalizable. If $b=1$ then the matrix is diagonalizable only if $a=0$ (its already in diagonal form when this is true, in fact), since if $a \neq 0$ then \begin{equation} A-I=\begin{bmatrix} 0 & a \\ 0 & 0 \end{bmatrix},\end{equation} the space of solutions of the homogenous system associated with the matrix above is then spanned by a single vector, namely \begin{equation} \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \end{equation} so that the eigenspace associated with eigenvalue 1 only has dimension 1 and is therefore not the full vector space (which is dimension 2).