# Conjugate elements of $GL_2(\mathbb{R})$

Decide whether or not the two matrices $A= \begin{pmatrix} 3 & 0 \\ 0 & 2 \end{pmatrix}$ and $B= \begin{pmatrix} 1 & 1 \\ -2 & 4 \end{pmatrix}$ are conjugate elements of the general linear group $GL_2(\mathbb{R})$

So I need to show that there exists a matrix $X= \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in GL_2(\mathbb{R})$ such that $a,b,c,d \in \mathbb{R}$ and such that $ad-bc \neq 0$

I solved the system $XAX^{-1}=B$ and found that the the matrix is of the form $X= \begin{pmatrix} a & 0 \\ b & 0 \end{pmatrix}$ which is not invertible hence A and B are not conjugate elements.

Is it correct? Is there an easier way to solve this type of question ?

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Have you ever taken a course in Linear Algebra? If so, consider the following hint. $$\text{Hint: what is the purpose of similar matrices?}$$ – user1729 Sep 10 '13 at 14:20
Diagonalize the matrix $B$ and you find that's similar to $A$. – user63181 Sep 10 '13 at 14:22
Group theory before basic linear algebra? Weird. – Git Gud Sep 10 '13 at 14:24

It can readily be seen that $2$ is an eigenvalue of $B$ because the sums of the lines is always $2$. Since the trace of $B$ is $5$ it follows that $3$ is also an eigenvalue of $B$, therefore $B$ is diagonalizable and similar to $A$, that is, there exists $X$ such that $X^{-1}BX=A$.

So you have made a miscalculation in solving the system. If you want to solve a system I suggest solving the equivalent, and in my opinion easier system:$XA=BX$.

Edit: Diagonalizability is a field-dependent concept. All that was said above is with respect to $\Bbb R$. It maybe the case that a real matrix isn't diagonalizable in $\Bbb R$ and it is $\Bbb C$.

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Do you perhaps mean $XA=BX$?... – user1729 Sep 10 '13 at 14:31
@user1729Yes, thanks. I had even thought of that and ended up writing the wrong thing for some reason. – Git Gud Sep 10 '13 at 14:33
@GitGud Could you include perhaps what are the various ways of showing that two matrices are similar ? Because I know how to show that they are not – Jean-Francois Rossignol Sep 10 '13 at 14:35
@Jean-FrancoisRossignol Oh, man. There's so much stuff on that related to SNF, JNF, invariant factors, etc. There's no easy, universal shortcut. However, if you're lucky enough that the matrices are diagonalizable, it's very easy to check. – Git Gud Sep 10 '13 at 14:41
@Jean-FrancoisRossignol Anyway, there are probably dozens of questions asking to find the JNF of specific matrices, maybe you can learn by example. – Git Gud Sep 10 '13 at 14:50

$XA=BX$ is equivalent to $c=2a$ and $d=b$, i.e., $$X=\begin{pmatrix} a & b \\ 2a & b \end{pmatrix}$$ We need $0\neq \det(X)=-ab$, which is possible with, say, $a=b=1$.

Remark: It is easy to see that $A$ and $B$ have the same characteristic polynomial, since they have the same trace and the same determinant. For $GL_2(\mathbb{Z})$ this is not enough to conclude that they are similar.

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Because the trace and the determinant coincide, so does the characteristic polynomial, in particular its roots=eigenvalues. Both eigenvalues are distinct, so both matrices similar to a diagonal matrices with entries $2$ and $3$.

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No, distinct eigenvalue must have algebraic multiplicity = geometric multiplicity, because the sum of all algebraic multiplicities is equal two and each alg. multiplicity at least one. – plusepsilon.de Sep 10 '13 at 16:10