# Are the matrices diagonalizable in the field $K$?

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



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?

• 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). Jul 23, 2021 at 5:25

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.
• 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? Jul 23, 2021 at 9:13