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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.2011-11-07

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You consider the minimal polynomial. You verify if the polynomial can be split into linear factors in the chosen field or not. When the polynomial can be split into linear factors, it means the matrix is diagonalisable in this field.

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    @Pierre-YvesGaillard: Ah ok, I see, thank you kindly for your help2011-11-07