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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 $K=2$?


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

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up vote 3 down vote accepted

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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Ok, excellent, so how do I go about finding the minimum polynomial? –  Freeman Nov 7 '11 at 14:51
@LHS: In your case, it's the characteristic polynomial. –  Pierre-Yves Gaillard Nov 7 '11 at 14:59
The minimal polynomial divides the characteristic polynomial. The degree of the minimal polynomial is one iff the matrix is a scalar times the identity. - Also in characteristic two your matrix in clearly not diagonalizable. –  Pierre-Yves Gaillard Nov 7 '11 at 15:12
The minimal polynomial always divides the characteristic polynomial, and it has the same set of roots as the characteristic polynomial (but with possibly lower multiplicities). In the above situation, this forces the two to coincide. –  Brad Nov 7 '11 at 15:13
@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

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