# Determining whether the matrix $A$ is diagonalizable depending on which field $K$ the matrix is over

Let $V$ be a finite-dimensional vector space over a field $K$ How do you go about deciding whether or not the matrix:

$A = \left({\begin{array}{cc} 1 & 6 \\ 3 & 5 \\ \end{array}}\right)$

can be diagonalised depending on the field for $K$. For instance if $K=\mathbb{R},\mathbb{C},\mathbb{Q}$?

The way I am approaching this is to assume that this is a matrix of some transformation $T:V\to V$, then find it's minimal polynomial, then if we can express the minimal polynomial as a product of distinct linear factors in the different field cases then that should tell us if $A$ is diagonalizable.

If this method is possible I would very much appreciate help with understanding how to make it work, as I am currently studying the relationship between minimal polynomials and characteristic polynomials.

EDIT: Ok, i'm saying $m_T(x)=(x-(3+\sqrt{22})^{m_1}(x-(3-\sqrt{22})^{m_2}$ where $m_1,m_2$ are positive integers. Therefore in the $\mathbb{R},\mathbb{C}$ this can be expressed this way, but in $\mathbb{Q}$ clearly it can't as $\sqrt{22}$ is irrational.

How about any field with char $K=2$?

EDIT#2

Also if $K=\{0,1,2,3,4,5,6\}$ with addition and multiplication modulo 7 it is not. As $X_A(x)=(1-x)(5-x)-3*6=x^2-6x+1$ which with the restriction on $K$ can't be factored?

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There's a typo in your title. – Rasmus Nov 7 '11 at 14:47

@LHS: In characteristic two you have $6=0$, $3=1=5$. - Your matrix is diagonalizable over $K$ iff $22$ is a nonzero square in $K$. – Pierre-Yves Gaillard Nov 7 '11 at 15:27