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I need to show that $A$ is a diagonizable matrix: $$A \in M_{n\times n}(\mathbb{R}) = \begin{bmatrix}0 & a & b\\ a & 0 & b\\ b & a & 0\end{bmatrix}; \space a\neq b; \space a,b \neq 0$$

Since I can't find the eigenvectors I think the best way to approach this is to see if, with the help of the characteristic polynomial, I can prove that $A$ has three different eigenvalues.

$$|A - \lambda I_{n}| = 0 \Leftrightarrow$$ $$\Leftrightarrow -\lambda (\lambda^{2} - ab) -a (-a \lambda -b^{2}) + b(a^{2} + b\lambda) = 0 \Leftrightarrow$$ $$ \Leftrightarrow -\lambda^{3} + (ab + a^{2} + b^{2})\lambda + ab^{2} + a^{2}b = 0$$

But I didn't come to any conclusion... Maybe I'm missing on how to factorize this polynomial, or maybe there is a better approach to the problem... Can some one guide me?

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    $\begingroup$ It is possible to factor(ize) this polynomial. Note that the constant term is $ab(a+b)$ and the roots sum to zero, and check the most obvious possibility... $\endgroup$
    – Micah
    Oct 27, 2020 at 22:07
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    $\begingroup$ $-a, -b, a + b$ are eigenvalues. That being said, your statement is wrong for $a = 1, b = -2$ -- the resultant matrix is not diagonalizable. $\endgroup$
    – Zhanxiong
    Oct 27, 2020 at 22:09
  • $\begingroup$ Thanks guys... I will try to do what @Micah said $\endgroup$
    – mozzarino
    Oct 27, 2020 at 22:09
  • $\begingroup$ Perhaps the $2,3$ entry should be an $a$ instead of a $b$. $\endgroup$
    – Ben Grossmann
    Oct 28, 2020 at 1:26

1 Answer 1

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The statement is wrong with $a = 1, b = -2$, whose Jordan form is \begin{align*} \begin{pmatrix} -1 & 1 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -2 \end{pmatrix}, \end{align*} hence it's not diagonalizable.

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    $\begingroup$ Hmm, then maybe the book is wrong... $\endgroup$
    – mozzarino
    Oct 27, 2020 at 22:18
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    $\begingroup$ @mozzarino It's not uncommon that a book has wrong times. $\endgroup$
    – Zhanxiong
    Oct 27, 2020 at 22:19

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