$f: SL_2(\mathbb{R}) \to GL_4(\mathbb{R})$ show that $Im(f)=SL_4(\mathbb{R})$ I was struggling with the following problem (from linear algebra):

Let $V$ be the vector space of the $2 \times 2$ matrices with real coefficients. Consider the action of the group $SL_2(\mathbb{R})$ on $V$, namely for any matrix $T \in SL_2(\mathbb{R})$ and for every matrx $A \in V$ we define $\Phi_T(A)=T\cdot A\cdot T^{-1}$. Prove that for every $T\in SL_2(\mathbb{R})$, $\Phi_T$ is an endomorphism of $V$. Then starting from a basis of $V$, write the corresponding homomorphism $SL_2(\mathbb{R}) \to GL_4(\mathbb{R})$, then show that the image lies inside $SL_4(\mathbb{R})$

First of all. Does anybody know the origin of this problem? Is there a book with standard solutions to this kind of problem? And is my solution any good? Apparently it is generalizable to many more dimensions, giving a nontrivial result on the determinant of such matrices.
And what does it happen when the trace is $0$? Can we show an homomorphism $SL_2(\mathbb{R}) \to SL_3(\mathbb{R})$ in the same way?
 A: I have tried to show the actual homomorphism. I found the matrix in $GL_4(\mathbb{R})$ associated to the general matrix in $SL_2(\mathbb{R})$, namely:
$$ \left( \begin{matrix} a & b \\ c & d\end{matrix} \right) \quad \text{with} \ ad-bc\neq0 $$
And it should be the following:
$$ \left( \begin{matrix}ad & -ac & bd &-bc\\ -ab & a^2 & -b^2 & ab \\ cd & -c^2 & d^2 & -cd\\ -bc & ac & -bd & ad\end{matrix}\right) $$
But honestly I find it a tedious task to look at its determinant to show that it is always $1$.
I found an alternative solution by looking at the abstract algebra behind it. We can compose the homomorphism $SL_2(\mathbb{R}) \to GL_4(\mathbb{R})$ with the homomorphism given by the determinant $GL_4(\mathbb{R}) \to (\mathbb{R^*}, \cdot)$. We get a new homomorphism $SL_2(\mathbb{R}) \to (\mathbb{R^*}, \cdot)$. But I can prove that this homomorphism is trivial! In this paper I found a proof that the only normal subgroups of $SL_2(\mathbb{R})$ are $\{\pm 1\}$ and the trivial ones. Given that the kernel must be a normal subgroup and that the matrix $$\left( \begin{matrix} 0 & -1 \\ 1 & 0\end{matrix} \right)$$ must map to the identity since it has order $4$ we have that the whole group maps to the identity giving the thesis!
A: Let $\Bbb k$ be a field, and let $V$ be a finite dimensional $\Bbbk$-vector space (of dimension $n$). The special linear group $SL(V)$ is generated shear mappings. All shear mappings are conjugate in $GL(V)$, so that the linear maps "conjugation by a shear map" are conjugate as endomorphisms of $\mathrm{End}(V)$, and thus have the same determinant. If we prove tha conjugation by a shear mapping induces a linear map of determinant one on $\mathrm{End}(V)$, then from what precedes we see that conjugation by a map of determinant one has determinant one aswell.
I'll describe two paths to the result. The first is a direct computation in an appropriately chosen basis of $\mathrm{End}(V)$. The second approach is simpler and more formal : it uses the determinant and the fact that the square of a shear mapping is a shear mapping (or the identity in characteristic $2$).
First proof
Let $u$ be a shear mapping. By definition there exists nonzero linear form $f:V\to \Bbbk$ and a nonzero vector $h\in H=\ker(f)$ such that for all $v\in V$
$$u(v)=v+f(v)h$$
Consider a basis $\mathcal{B}$ of $V$ of the form $(e_1,\dots,e_n)=(h_1,\dots,h_{n-2},h,x)$ where $h$ is as above, $f(x)=1$ and $(h_1,\dots,h_{n-2},h)$ is a basis of $H$. Then
$$\mathrm{Mat}(u;\mathcal{B})=\begin{pmatrix}1\\&1\\&&\ddots\\&&&1\\&&&&1&1\\&&&&&1\end{pmatrix}$$
Consider the basis $(E_{i,j})_{1\leq i,j\leq n}$ of $\mathrm{End}(V)$ associated to the basis $\mathcal{B}$ defined on basis elements by 
$$E_{i,j}(e_k)=\begin{cases}k=j:& e_i\\ k\neq j:&0\end{cases}$$
Then $u=\mathrm{id}+E_{n-1,n}$ and $u^{-1}=\mathrm{id}-E_{n-1,n}$; for any $i,j$
$$\begin{array}{rcl}
u\circ E_{i,j}\circ u^{-1} &=&(\mathrm{id}+E_{n-1,n})E_{i,j}(\mathrm{id}-E_{n-1,n})\\
&=&E_{i,j}+E_{n-1,n}E_{i,j}-E_{i,j}E_{n-1,n}-E_{n-1,n}E_{i,j}E_{n-1,n}\\
&=&
\begin{cases}
i\neq n,j\neq n-1:&E_{i,j}\\
i= n,j\neq n-1:&E_{n,j}+E_{n-1,j}\\
i\neq n,j= n-1:&E_{i,n-1}-E_{i,n}\\
i=n,j=n-1:&E_{n,n-1}+E_{n-1,n-1}-E_{n-1,n}-E_{n,n}
\end{cases}\\
&=&
\begin{cases}
i\neq n,j\neq n-1:&E_{i,j}\\
i= n,j\neq n-1:&E_{i,j}+\cdots\in X\\
i\neq n,j= n-1:&E_{i,j}+\cdots\in X\\
i=n,j=n-1:&E_{i,j}+\cdots\in Y
\end{cases}
\end{array}$$
where $X=\mathrm{Vect}(E_{i,j})_{i\neq n,j\neq n-1}$ and $Y=\mathrm{Vect}(E_{i,j})_{(i,j)\neq(n,n-1)}$, so that the matrix of conjugation by $u$, when represented in some basis obtained by ordering the vectors $(E_{i,j})_{i\neq n,j\neq n-1}\cup(E_{n,j})_{j\neq n-1}\cup(E_{i,n-1})_{i\neq n}\cup(E_{n,n-1})$ inside each set of parantheses is upper triangular with a diagonal made of ones.
Second proof
Another line of reasoning goes as follows: we have a linear group action $\rho$ of $SL(V)$ on $\mathrm{End}(V)$
$$\rho:SL(V)\to GL(\mathrm{End}(V)),\quad u\mapsto u\circ-\circ u^{-1}$$
Which we may compose with the determinant
$$\det:GL(\mathrm{End}(V))\to\Bbbk^{\times}$$
Notice that the square of a shear mapping is either a shear mapping (if $\mathrm{char}(\Bbbk)\neq 2$) or the identity (if $\mathrm{char}(\Bbbk)=2$) so that for any shear mapping $u$
$$\det(\rho(u))^2=
\begin{cases}
\det(\rho(u))&\text{if }\mathrm{char}(\Bbbk)\neq 2\\
1 &\text{if }\mathrm{char}(\Bbbk)=2
\end{cases}$$
So that, in both cases (since $1=-1$ in characteristic $2$) $\det(\rho(u))=1$. The generation and conjugation argument from above then tells us that $\rho$ maps $SL(V)$ into 
$SL(\mathrm{End}(V))$.
A: The question setter probably mean this. Consider the following ordered basis $\mathcal B$ of $V=M_2(\mathbb R)$.
$$
\mathcal B=\left\{
B_1=\pmatrix{1&0\\ 0&0},
\ B_2=\pmatrix{0&0\\ 1&0},
\ B_3=\pmatrix{0&1\\ 0&0},
\ B_4=\pmatrix{0&0\\ 0&1}
\right\}.
$$
We can identify every $B=\pmatrix{a&c\\ b&d}\in V$ with the vector $\renewcommand{\vec}{\operatorname{vec}}\vec(B)=(a,b,c,d)^\top\in\mathbb R^4$ (this is known as the vectorisation of a matrix). Now, for any $T\in SL_2(\mathbb R)$, the group action $B\mapsto TBT^{-1}$ is actually an automorphism. It follows that the image of $\mathcal B$ under the group action is still a basis of $V$. Consequently, the matrix
$$
f(T) = \left[\vec(TB_1T^{-1}),\vec(TB_2T^{-1}),\vec(TB_3T^{-1}),\vec(TB_4T^{-1})\right]
$$
lies inside $GL_4(\mathbb R)$. If one knows what a Kronecker product is, one immediately sees that
$$
f(T)=(T^{-\top}\otimes T)
\underbrace{\left[\vec(B_1),\vec(B_2),\vec(B_3),\vec(B_4)\right]}_{=I_4}
=T^{-\top}\otimes T.
$$
By the properties of Kronecker product (see the above-linked Wikipedia entry), it is also obvious that $f$ is a group homomorphism from $SL_2(\mathbb R)$ to $GL_4(\mathbb R)$ and $\det f(T)=1$, i.e. $\operatorname{Im}(f)\subset SL_4(\mathbb R)$. Also, $f$ is injective. Therefore $\operatorname{Im}(f)$ is isomorphic to $SL_2(\mathbb R)$. This is probably why the last sentence in the question says that "the image lies inside $SL_2(\mathbb R)$", if the $SL_2(\mathbb R)$ here is not a typo to $SL_4(\mathbb R)$.
The 4-by-4 matrix that you have shown is $T\otimes T^{-\top}$. So, it also works.
