The matrix is diagonalizable over
i)$Q$
ii) any field where $1 + 1 = 0$
iii) $F_7$ = {0, 1, 2, 3, 4, 5, 6} with addition and multiplication modulo 7.
$\begin{bmatrix}1&6\\3&5\end{bmatrix}$
The eigenvalues are $\lambda_1=-\sqrt{22}+3 $ and $\lambda_2=\sqrt{22}+3 $
for $\lambda_1$, x=$x_2\begin{bmatrix}\frac{-\sqrt{22}-2}{3}&1\end{bmatrix}^T$
for $\lambda_2$, x=$x_2\begin{bmatrix}\frac{\sqrt{22}-2}{3}&1\end{bmatrix}^T$
Since the roots are not rational number then it is not diagonalizable over $Q$ and how to do the Field?
For field, I considered the characteristic polynomial over the field, for field 1+1=0, how to determined whether the roots are qualified or not?
For field 1+1=0, considering the multiple inverse, should it be field $F$={0,1}?
For $F_7$
Is it that for the characteristic polynomial: $\lambda^2-6\lambda-13=\lambda^2+6\lambda-6$?
Then there is no roots in $F_7$, so it is not diagonalizable in $F_7$?
And what is about field 1+1=0?
Best Answer
The characteristic polynomial over any field is $\;\chi(x)=x^2-6x-13\;$, and this quadratic's discriminant is $\;\Delta=36+52=88\;$
=== Over $\;\Bbb Q\;$ . Since $\;\sqrt{88}\notin\Bbb Q\;$ the roots $\;\chi(x)\;$ aren't in the base field and thus the matrix cannot be diagonalizable
=== Over a field $\;\Bbb F\;,\;\;\text{char}\,\Bbb F=2\;$ . In this case we have $\;\chi(x)=x^2-13=x^2+1=(x+1)^2\;$ and thus we have a double eigenvalue, so the matrix is diagonalizable iff the minimal polynomial of the matrix, $\;\mathcal M(x)\;$ , is $\;x+1=x-1\;$ , yet
$$\begin{pmatrix}1&6\\3&5\end{pmatrix}=\begin{pmatrix}1&0\\1&1\end{pmatrix}\pmod2\neq\begin{pmatrix}1&0\\0&1\end{pmatrix}$$
and thus again our matrix isn't diagonalizable.
=== Over $\;\Bbb F_7\;$ . Here we have $\;\chi(x)=x^2+x+1\;$ , and the discriminant is $\;88=2^2\pmod7\;$ , so we have two different eigenvalues here: $$\;\frac{-1\pm2}2=\begin{cases}-\cfrac32=-12=2\pmod7\\{}\\\cfrac12=4\pmod7\end{cases}\;$$
and then automatically the matrix is diagonalizable.