For which $\beta \in \mathbb{C}$ is the matrix $A=\bigl(\begin{smallmatrix} 0&1\\1&\beta \end{smallmatrix}\bigr)$ diagonalisable?

Question

I have got a question refering to the following problem.

Let $K=\mathbb{C}$. For which $\beta \in \mathbb{C}$ is this matrix diagonalisable? $$A=\pmatrix{0&1\\1&\beta}$$

I think that it is not diagonalisable in any case, because the determinant of $A-xI_n$ is $-1$. Am I right?

No, the formula giving the determinant of $A-xI_n$ includes $\beta$. — Augustin, May 26, 2015 at 11:37
Hint : $det(A-xI_n)=-x(\beta-x)-1=x^2-\beta x-1$. By the way, a good thing to avoid such mistakes is to remember that the characteristic polynomial of a matrix of size $n$ is always a polynomial of degree $n$... — Clément Guérin, May 26, 2015 at 11:37

marwalix · Accepted Answer · 2015-05-26 11:55:38Z

2

The characteristic polynomial is $P(x)=x^2-\beta x-1$ whose discriminant is $\Delta=\beta^2+4$

If $\beta\neq\pm 2i$ The eigenvalues are distinct and the matrix is diagonalisable.

If $\beta=\pm 2i$ the root $2$ is double and the matrix is not diagonalisable.

answered May 26, 2015 at 11:48

marwalix

16.7k2 gold badges34 silver badges50 bronze badges

$\begingroup$ how about the case when two eigenvectors come from one eigenvalue? $\endgroup$
– marco11
May 26, 2015 at 12:19
$\begingroup$ @marco11 A $2\times2$ matrix with a double eigenvalue is diagonalizable if and only if it is diagonal (and so scalar). This is specific to $2\times2$ matrices, however. $\endgroup$
– egreg
May 26, 2015 at 13:06

Add a comment |

1 Answer 1