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I want to check if the following matrices are diagonalizable in the field $K$.

(a) $A=\begin{pmatrix}2 &1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{pmatrix}, \ K=\mathbb{C}$

(b) $A=\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix},\ K=\mathbb{R}$

(c) $A=\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix},\ K=\mathbb{F}_5$

(d) $A=\begin{pmatrix}x+1 & 1 \\ x-1 & 2x-1\end{pmatrix}, \ K=\mathbb{R}(x)$

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To check if the matrix is diagonalizable we have to calculate the eigenvalues and if we have $n$ different eigenvalues in the givenfield, right?

I have done the following :

At (a) we have the characteristic polynomial $p(x)=(x-2)^3$. Over $\mathbb{C}$ there are $3$ different eigenvalues and so $A$ is diagonalizable, or not?

At (b) we have the characteristic polynomial $p(x)=x^2+1$. Over $\mathbb{R}$ there are no eigenvalues and so $A$ is not diagonalizable, or not?

At (c) we have the characteristic polynomial $p(x)=x^2+1$. Over $\mathbb{F}_5$ there are $2$ different eigenvalues, $2$ and $3$, and so $A$ is diagonalizable, or not?

At (d) we have the characteristic polynomial $p(\lambda )=(x+1-\lambda)(2x-1-\lambda)-(x-1)$. Over $\mathbb{R}(x)$ there are $2$ different eigenvalues, $x$ and $2x$, and so $A$ is diagonalizable, or not?

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    $\begingroup$ You seem to be misunderstanding some basic points. In (a) the "three" eigenvalues are all the same, $\lambda = 2$. In fact we don't have enough linearly independent eigenvectors for that eigenvalue to form a basis (and the matrix is not diagonalizable). $\endgroup$
    – hardmath
    Jul 23, 2021 at 5:25

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Everything is correct except for $(a)$.

There is only one eigenvalue and it's $2$. (It does have algebraic multiplicity of $3$ ) This doesn't mean it's not diagonalizable. You can try finding the dimension of the eigenspace of $A - 2I$ and see that it is not all of $\mathbb{C}^3$.

Or you just claim that $A$ is already in its Jordan Canonical Form so it's not diagonalizable.

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  • $\begingroup$ So in general : If we have a $n\times n$ matrix and there are $n$ different eigenvalues then it follows that the matrix is diagonalizable. If there are less than $n$ dinstinct eigenvalues then we cannot say anything, we have to check if the geometric multiplicity is equal to the algebraic multiplicity for each eigenvalue, right? $\endgroup$
    – Mary Star
    Jul 23, 2021 at 9:13
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    $\begingroup$ @MaryStar Precisely! $\endgroup$ Jul 23, 2021 at 17:37
  • $\begingroup$ Great!! Thank you very much!! :-) $\endgroup$
    – Mary Star
    Jul 26, 2021 at 18:52

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