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Two matrices $A,B$ are said to be cogredient if there exists an invertible matrix $P$ such that $B = P^{t}AP$.

I know how to tell if two matrices are cogredient in algebraically closed fields, its as easy as looking the ranks of them. Also, I know how to tell if two matrices are cogredient in real closed fields, you can tell by looking at the signature (the difference between positive elements and negative elements in its diagonal).

Now, I'm facing the following exercise (from Nathan Jacobson's Basic Algebra 1):

Show that diag$\{1,1\}$ and diag$\{5,5\}$ are cogredient in $M_2(\mathbb{Q})$.

How should I go around proving such thing?, I'm working on neither an algebraically closed field nor a real closed field.

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    $\begingroup$ Since multiples of the identity commute with everything, you need to find a $P$ with $P^tP = 5 I$. It's an easy special case that can be done by brute force (a good idea helps, of course). $\endgroup$ – Daniel Fischer May 27 '15 at 23:37
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Letting $A = I$, then you want to find a matrix $P$ with rational entries such that

$$P^TP = \left(\begin{array}{cc} 5 & 0 \\ 0 & 5\end{array}\right).$$

That is to say that $P^T = 5 P^{-1}$. We know that $P^{-1}$ is given by

$$\frac{1}{ad-bc}\left(\begin{array}{rr} d & -b \\ -c & a\end{array}\right)$$

if $P = \left(\begin{array}{cc} a & b \\ c & d\end{array}\right).$ Moreover, $\det(P^TP) = (\det P)^2 = 25$ so $\det P = \pm 5$.

$$P^T = 5P^{-1} \Longrightarrow \left(\begin{array}{cc} a & c \\ b & d\end{array}\right) = \pm\left(\begin{array}{cc} d & -b \\ -c & a\end{array}\right).$$

So either $a=d$ and $c=-b$ or $a=-d$ and $b=c$. In the former case, we have $P = \left(\begin{array}{rr} a & b \\ -b & a\end{array}\right)$; in the latter case, we have $P = \left(\begin{array}{rr} a & b \\ b & -a\end{array}\right)$. Both are subject to the condition that $a^2 + b^2 = 5$. Try a solution with integers.

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  • $\begingroup$ Thanks for your answer, I think I'm unable to do such thing, any tips? $\endgroup$ – Miguelgondu May 27 '15 at 23:41
  • $\begingroup$ Supposing $b\neq 0$, $ad-bc = -5$? $\endgroup$ – Miguelgondu May 27 '15 at 23:43
  • $\begingroup$ I've edited my answer since I made a slight error. It's much clearer now. $\endgroup$ – Cameron Williams May 28 '15 at 0:07

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