For which $a$ and $b$ is this matrix diagonalizable?

$$A=\begin{pmatrix} a & 0 & b \\ 0 & b & 0 \\ b & 0 & a \end{pmatrix}$$

How to get those $a$ and $b$? I calculated eigenvalues and eigenvectors, but don't know what to do next?


That is a symmetric matrix: it is always diagonalizable over a field with characteristic$\;\neq2\;$. Not only that: it is orthogonally diagonalizable.

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    $\begingroup$ What if $a,b \in \mathbb C$? $\endgroup$ – Rodrigo de Azevedo Jun 22 '16 at 23:36
  • $\begingroup$ @RodrigodeAzevedo then the result is still a normal matrix $\endgroup$ – Ben Grossmann Jun 23 '16 at 2:25
  • $\begingroup$ No, it's not always diagonalisable. It's diagonalisable if and only if $b=0$ or the field has characteristic $\ne2$. For a counterexample, consider $a=0$ and $b=1$ over $GF(2)$. $\endgroup$ – user1551 Jun 23 '16 at 9:04
  • $\begingroup$ @user1551 Thank you, though I think this is a very particular case, specially because this kind of questions are many times related to inner product spaces, which are usually the reals or the complex ones. Yet I will edit to cover even this case. $\endgroup$ – DonAntonio Jun 23 '16 at 9:56

Calculating the eigenvalues, $b, b+a, a-b,$ one can then easily calculate their respective eigenvectors. For eigenvalue $b,$ \begin{equation} \begin{pmatrix} a-b&0&b\\0&0&0\\b&0&a-b \end{pmatrix}\begin{pmatrix} 0\\1\\0\end{pmatrix} = \begin{pmatrix} 0\\0\\0\end{pmatrix}, \end{equation} There may be special cases, such as if $a = 0,$ then the vector $\begin{pmatrix} 1\\0\\1\end{pmatrix}$ works too.

Likewise for $a+b,$ \begin{equation} \begin{pmatrix} -b&0&b\\0&-a&0\\b&0&-b \end{pmatrix}\begin{pmatrix} 1\\0\\1\end{pmatrix} = \begin{pmatrix} 0\\0\\0\end{pmatrix}, \end{equation}

And last for $a-b,$ \begin{equation} \begin{pmatrix} b&0&b\\0&2b-a&0\\b&0&b \end{pmatrix}\begin{pmatrix} 1\\0\\-1\end{pmatrix} = \begin{pmatrix} 0\\0\\0\end{pmatrix}. \end{equation}

Because no assumptions about the values of $a$ and $b$ were made along the way, this operator must be diagonalizable (which we already knew from symmetry), with an eigenbasis \begin{equation} \begin{pmatrix} 0\\1\\0\end{pmatrix}, \begin{pmatrix} 1\\0\\-1\end{pmatrix}, \text{ and } \begin{pmatrix} 1\\0\\1\end{pmatrix}\end{equation}

Note that if one remembers that matrices of the form $\begin{pmatrix} a & b\\ b& a \end{pmatrix} $ have the eigenbasis $\begin{pmatrix} 1 & 1\\ 1& -1 \end{pmatrix},$ then from a quick glance at the original matrix, you can already "know" the eigenvectors since this $2 \times 2$ is hidden within it.


Matrix A has 3 eigenvalues a-b, a+b and b. In order for them to be distinguishable, following conditions should be met. $$b \ne 0$$ $$a \ne 0$$ $$a \ne 2b$$

  • $\begingroup$ Why the third one? How to get that? $\endgroup$ – Ana Matijanovic Jun 22 '16 at 20:58
  • $\begingroup$ $a-b \ne b$ leads to $a \ne 2b$ $\endgroup$ – user115350 Jun 22 '16 at 20:59
  • $\begingroup$ Okay. But from what you get that? I dont understand. $\endgroup$ – Ana Matijanovic Jun 22 '16 at 21:01
  • $\begingroup$ you first calculate eigenvalues $det( \lambda I-A)=0$. $\endgroup$ – user115350 Jun 22 '16 at 21:05
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    $\begingroup$ Of course this is not true. For example, the identity matrix has all of its eigenvalues equal, yet it is diagonalizable (it is already diagonal). What matters is that for each eigenvalue, the algebraic and geometric multiplicities are equal. $\endgroup$ – user258700 Jun 22 '16 at 21:40

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