How to determine two matrices are conjugate. Which of the following statements are true? 


*

*The matrices $
  A=\left[ {\begin{array}{cc}
   1 & 1 \\
   0 & 1\\
  \end{array} } \right]$ and $
  B=\left[ {\begin{array}{cc}
   1 & 0 \\
   1 & 1\\
  \end{array} } \right]$ are conjugate in $GL_2(\mathbb{R})$

*The matrices $
  A=\left[ {\begin{array}{cc}
   1 & 1 \\
   0 & 1\\
  \end{array} } \right]$ and $
  B=\left[ {\begin{array}{cc}
   1 & 0 \\
   1 & 1\\
  \end{array} } \right]$ are conjugate in $SL_2(\mathbb{R})$

*The matrices $
  C=\left[ {\begin{array}{cc}
   1 & 0 \\
   0 & 2\\
  \end{array} } \right]$ and $
  D=\left[ {\begin{array}{cc}
   1 & 3 \\
   0 & 2\\
  \end{array} } \right]$ are conjugate in $GL_2(\mathbb{R})$


I know the conjugate matrices have the same eigenvalues. But all of this matrices have the same eigenvalues. I am not sure how to determine that which matrices are conjugate to each other. 
 A: Hint

The definition of conjugacy is: if $A$ and $B$ are two matrices such that $$A= P^{-1}BP$$ then they are conjugate. In $GL(2,\mathbb{R})$ is the same as similarity so not only the eigenvalues have to be preserved: mainly two similar matrices in $GL(2,\mathbb{R})$ have the same Jordan normal form.

Notice that $A$ is in Jordan canonical form (just look at the eigenvalues). Notice that the matrix $D$ is diagonalisable.
Probabily this forum can help for the second question.
A: Quiver has already told you (and I am sure it is in you textbook) that two matrices, A and B, are conjugate if and only if there exist an invertible matrix, P, such that $A= P^{-1}BP$.  That is equivalent to $PA= BP$.
In the first problem, $A= \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}$ and $B= \begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix}$.  We can show they are conjugate by finding an appropriate P!
Since A and B are 2 by 2 P must be also and we can write it $\begin{bmatrix}a & b \\ c & d\end{bmatrix}$ so we must have
$\begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}= \begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix}\begin{bmatrix}a & b \\ c & d\end{bmatrix}$.
$\begin{bmatrix}a & a+ b\\ c & c+ d\end{bmatrix}= \begin{bmatrix}a * b \\ a+ c & c+ d\end{bmatrix}$.
We must have a= a, a+ b= b, c= a+ c, and c+ d= c+ d.
Both a+ b= b and c= a+ c reduce to a= 0 while c+ c= c+ d is always true.
Any matrix of the form $\begin{bmatrix}0 & b \\ c & d\end{bmatrix}$ will do.
