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I'm trying to determine whether this matrix is diagonalizable:

$$A = \pmatrix{\sqrt3&0&-1\\0&-1&0\\1&0&0}$$

The characteristic polynomial is:

$$-\lambda^3 + (\sqrt3-1)\lambda^2 + (\sqrt3-1)\lambda -1 = 0$$

And so we get eigenvalues: $$\lambda_1=-1,$$ $$\lambda_2=\frac{\sqrt3}{2}+\frac{i}{2},$$ $$\lambda_3=\frac{\sqrt3}{2}-\frac{i}{2}$$

I know that if A has 3 distinct eigenvalues (i.e. the characteristic polynomial has 3 distinct roots), then A will be diagonalizable, since we'll have 3 linearly independent eigenvectors. So how do we know if the 3 eigenvalues are "distinct" or not?

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  • $\begingroup$ Is $\lambda_1=\lambda_2$? $\endgroup$ – Lord Shark the Unknown Mar 30 '18 at 16:14
  • $\begingroup$ "Distinct" just means "not equal". All of these three numbers are different numbers. $\endgroup$ – Billy Mar 30 '18 at 16:14
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Just check whether any $\lambda_i = \lambda_j$, where $i \neq j$.

It that happens, you do not have distinct eigenvalues.

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"Distinct" means "pairwise not equal".

For example, if the characteristic polynomial were $(\lambda-1)^3$, then the eigenvalues would not be distinct, as $\lambda=1$ would be a triple root.

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No two eigenvalues are equal. To put it differently, all eigenvalues have an algebraic multiplicity of 1.

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