What value of b make the following matrix diagonalisable

Question

$$ \begin{bmatrix} 1 & -1 \\ 1 & b \\ \end{bmatrix} $$

It seems like there's no two distinct eigenvalues to make it diagonalisable. Can anyone confirm that is true?

Answer 1

The characteristic polynomial of the matrix is

$$p(\lambda)=(b-\lambda)(1-\lambda) + 1 = \lambda^2 -(b+1)\lambda + (b+1)$$

with determinant

$$\Delta=(b+1)^2 - 4(b+1)= (b+1)(b-3)$$

When $\Delta>0$ we have two distinct real eigenvalues, so the matrix is diagonalisable, which happens for $b>3$ and $b < -1$. For $-1 < b < 3$ we have no real eigenvalues (so not diagonalisable). Leaves to check $b=-1$ and $b=3$. $b=3$ leaves $\lambda =2$ (double multiplicity, while we have one eigenvector $\begin{bmatrix}1\\1\end{bmatrix}$). $b=-1$ is a similar failure.

So my conclusion is that we can diagonalise the matrix, e.g. for $b>3$ and $b < -1$.