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?

  • 1
    $\begingroup$ No, the formula giving the determinant of $A-xI_n$ includes $\beta$. $\endgroup$
    – Augustin
    May 26, 2015 at 11:37
  • $\begingroup$ 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 $n$... $\endgroup$ May 26, 2015 at 11:37

1 Answer 1


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.

  • $\begingroup$ 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

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