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Consider $\varphi:\mathfrak{sl}_2(\mathbb F)\to \mathfrak{gl}_3(\mathbb F)$ sending $\begin{pmatrix} a & b\\ c & -a \end{pmatrix}$ to $\begin{pmatrix} 2a-c & -2a-2b+3c & -2a+3b-2c\\ -c & 3c & -2a+b-2c\\ 0 & 2c & -2a-2c \end{pmatrix}$

I want to find a basis of $\mathfrak{sl}_2(\mathbb F)$ such that $\varphi(x)$ is the matrix of $\text{ad}(x)$ with respect to this basis. ($\text{ad}(x)(y)=[x,y]=xy-yx$)

I would appreciate a way to proceed because the bases I found don't work, and I can't seem to find matrices which satisfy the condition to extract a basis $\begin{pmatrix} a & b\\ c & -a \end{pmatrix}\begin{pmatrix} x_1 & y_1\\ z_1 & -x_1 \end{pmatrix}-\begin{pmatrix} x_1 & y_1\\ z_1 & -x_1 \end{pmatrix}\begin{pmatrix} a & b\\ c & -a \end{pmatrix}=(2a-c)\begin{pmatrix} x_1 & y_1\\ z_1 & -x_1 \end{pmatrix}-c\begin{pmatrix} x_2 & y_2\\ z_2 & -x_2 \end{pmatrix}$

where $\begin{pmatrix} x_1 & y_1\\ z_1 & -x_1 \end{pmatrix},\begin{pmatrix} x_2 & y_2\\ z_2 & -x_2 \end{pmatrix},\begin{pmatrix} x_3 & y_3\\ z_3 & -x_3 \end{pmatrix}$ are the elements of the basis we want to find.

And we do this for each column of the 3x3 matrix.

Is this the way to go ?

raisinsec
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    You wrote a commutator of $2\times 2$-matrices as a $3\times 3$-matrix. This is impossible. You can use the adjoint matrices $ad(x)$ here and compute a linear combination. – Dietrich Burde Sep 25 '22 at 15:41
  • Yes in fact I wanted to write the result of the commutator as a linear combination of elements of the basis, which correspond to the factors in the columns of the 3x3 matrix (since each column is supposed to be the image of an element of the basis). Note that what we want to do is precisely the inverse of your link. We don't want to calculate ad with a basis since we already have the 3x3 matrix. We want to find a basis whose image is the matrix. – raisinsec Sep 25 '22 at 19:37
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    But you can do the "inverse" process also the other way around, by using the adjoint matrices for the standard basis and rewriting them for another linear combination of this standard basis. So use $ad(e_i)$ to write down $r_1ad(e_1)+\cdots r_k ad(e_k)=ad(r_1e_1+\ldots r_ke_k)=A$, where $A$ is your given matrix. This gives you the scalars $r_i$. – Dietrich Burde Sep 25 '22 at 19:41
  • Ok then by calculating ad$(e_i)(e_j)$ and putting it into matrices I get $r_1\text{ad}(e_1)+r_2\text{ad}(e_2)+r_3\text{ad}(e_3)=\begin{pmatrix} 2r_3 & 0 & -2r_1\ 0 & -2r_3& 2r_2\ -r_2 & r_1 &0\end{pmatrix}=A$ which I find weird since it seems that this equality is only true for some well chosen $a,b,c$ – raisinsec Sep 26 '22 at 07:40
  • I'm not sure to understand. The matrix $A$ is $\varphi(x)$, $x$ written in the standard basis. It should be ad$(x)$ for some basis which we're going to find. I'm trying with your idea but I just get $r_1=r_2=r_3=0$ (I take $r_1=a$, $r_2=b$, $r_3=c$ and I try to solve). But even if we have a solution what would be the new basis using this coefficients ? – raisinsec Sep 26 '22 at 08:23
  • This time I'm sure not to understand. What link are you talking about ? Repeating the same question after I tell you I don't understand doesn't seem to work. – raisinsec Sep 26 '22 at 09:15
  • If your matrix is of the form $ad(x)$, and this is what you claim to be, for some basis, then $ad(x)(x)=0$. What exactly is not clear of my sentence? Again, where did you get the matrix $\begin{pmatrix} 2a-c & -2a-2b+3c & -2a+3b-2c\ -c & 3c & -2a+b-2c\ 0 & 2c & -2a-2c \end{pmatrix}$? – Dietrich Burde Sep 26 '22 at 09:24
  • Yes the first part was clear that's indeed unfortunate. I got the matrix from an exercise sheet on Lie algebras. It's a subquestion of some exercise. I checked several times and it's the same matrix as in the exercise. The condition $[x,x]=0$ gives an easy way to check if it can be an adjoint matrix, I'll keep it. So there is no chance of finding a basis which makes this matrix an adjoint matrix ? – raisinsec Sep 26 '22 at 10:06
  • I have to say this question seems kind of awkward. We have a standard basis of $\mathfrak{sl}_2$ (at least once you have written it as matrices like that) and with respect to that basis you get a nice matrix like the one you write in the comments. The other matrix $\varphi$ is then what you get if you write $x$ in terms of the standard basis but compute $\mathrm{ad}(x)$ with respect to some other basis instead which seems a weird thing to do. – Callum Sep 26 '22 at 12:41
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    So when you write $\mathrm{ad}(e_1)$ you need to write it with respect to the new basis $f_i=∑λ_{ij}e_j$ and find the $λ$'s that give you that matrix. You should be able to solve that in theory as this does appear to be a similar matrix to one written in the standard basis (they are both diagonalisable and have the same eigenvalues). I think however that this is a confusing and not that worthwhile exercise unless you have a specific reason for wanting to solve it. – Callum Sep 26 '22 at 12:41
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    I suppose that if you show the whole exercise, some of these remarks and questions might clarify. As it stands, the question doesn't appear to be reasonable. – Dietrich Burde Sep 26 '22 at 19:51
  • With the edits, this is a not implausible question. With the linked exercise, it boils down to finding a basis w.r.t. which the linear map given by $M_1=\pmatrix{2a&0&-2b\0&-2a&2c \-c&b&0 }$ (in standard basis) is expressed by OP's $M_2=\pmatrix{ 2a-c & -2a-2b+3c & -2a+3b-2c\ -c & 3c & -2a+b-2c\ 0 & 2c & -2a-2c}$. Existence of such basis is equivalent to the matrices being similar, i.e. having same Jordan form, which for $3\times3$-matrices just means they have same char. and minimal polynomial, which is easily checked. Now to find the $P$ such that $PM_1P^{-1}=M_2$ is computation ... – Torsten Schoeneberg Sep 27 '22 at 17:29
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    ... computation as in https://math.stackexchange.com/a/2171615/96384 – Torsten Schoeneberg Oct 06 '22 at 18:49

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