Is it true that $\mathrm{GL}_n(\mathbb{R})=\mathrm{SL}_n(\mathbb{R}) \oplus \{\lambda E : \lambda \in \mathbb{R}^{*}\}$ Does the equality $\mathrm{GL}_n(\mathbb{R})=\mathrm{SL}_n(\mathbb{R}) \oplus \{\lambda E : \lambda \in \mathbb{R}^{*}\}$ hold?
I think yes! That's the reason: we have the following exact sequence:
$$1 \longrightarrow \mathrm{SL}_n(\mathbb{R}) \longrightarrow \mathrm{GL}_n(\mathbb{R}) \longrightarrow \{\lambda E : \lambda \in \mathbb{R}^{*}\} \longrightarrow 1$$
To be split, there should be a homomorphism $\varphi:\mathrm{GL}_n(\mathbb{R}) \to \mathrm{SL}_n(\mathbb{R})$ such that its composition with the first homomorphism gives the identity. So I simply take
$$\varphi: \mathrm{GL}_n(\mathbb{R}) \ni A \longmapsto \frac{1}{\det A} A\in \mathrm{SL}_n(\mathbb{R}) $$
Am I correct? Thanks in advance.
 A: No, that is not correct. However, the group $GL(n,\mathbb{R})^+$ of all $n\times n$ matrices whose determinant is greater than $0$ is the direct sum of $SL(n,\mathbb{R})$ and the group of all scalar matrices (with real entries). Just consider the map\begin{array}{ccc}GL(n,\mathbb{R})^+&\longrightarrow&SL(n,\mathbb{R})\\M&\mapsto&\displaystyle\frac1{\sqrt[n]{\det A}}A.\end{array}
A: In the general situation, the sequence splits in the sense that there is a section $\mathbb{R}^\times \to GL_n(\mathbb{R})$ of $\det$. Such a section is for instance $\lambda\mapsto (\lambda-1)E_{1,1}+I_n$ (this is the diagonal matrix with a $\lambda$ in the top-right-hand corner, and $1$'s on the rest of the diagonal).
This section means that $GL_n(\mathbb{R})\simeq SL_n(\mathbb{R})\rtimes \mathbb{R}^\times$, the semi-direct product being induced by the section of $\det$.
When $n$ is odd, one may choose another section such as $\lambda \mapsto \lambda^{\frac{1}{n}}I_n$, which induces a trivial action: hence in this case the semi-direct product is a direct product : $GL_n(\mathbb{R})\simeq SL_n(\mathbb{R})\times \mathbb{R}^\times$ and one may check that this coincides with your decomposition.
When $n$ is even, $SL_n(\mathbb{R})\cap \{\lambda I_n, \lambda \in \mathbb{R}^\times\} \neq \{I_n\}$, indeed it contains $-I_n$. But we may still wonder if there is some isomorphism $SL_n(\mathbb{R})\times \mathbb{R}^\times \to GL_n(\mathbb{R})$ sending $SL_n(\mathbb{R})\times \{1\}$ to $SL_n$: the answer is still no. Indeed if $f$ is such an isomorphism then $f(I_n, -1)$ is in the center of $GL_n(\mathbb{R})$, because $(I_n,-1)$ is in the center of $SL_n(\mathbb{R})\times \mathbb{R}^\times$, but also $f(I_n, -1)^2 = I_n (*)$. Being in the center implies that it's of the form $\lambda I_n$ and $(*)$ implies $\lambda^2= 1$.
Now $-1\neq 1$ so $\lambda$ cannot be $1$ ($f$ is injective). Therefore $-I_n = f(I_n,-1)$, therefore $-I_n \notin SL_n(\mathbb{R})$, and so $n$ is odd.
Addendum : We may also wonder about general isomorphisms $SL_n(\mathbb{R})\times \mathbb{R}^\times \to GL_n(\mathbb{R})$, without requiring a priori that $SL_n(\mathbb{R})$ be sent to itself. Well it turns out that they don't exist either. In fact, for a very simple reason, every morphism $SL_n(\mathbb{R})\to GL_n(\mathbb{R})$ has its image contained in $SL_n(\mathbb{R})$. This is because $SL_n(\mathbb{R})$ is perfect, that is, it is its own derived subgroup: $SL_n(\mathbb{R}) = [SL_n(\mathbb{R}), SL_n(\mathbb{R})]$, therefore it is sent by every morphism to the derived subgroup, and the one of $GL_n(\mathbb{R})$ is $SL_n(\mathbb{R})$. Therefore the conclusion is as follows :

For $n$ odd, there is an isomorphism $SL_n(\mathbb{R})\times \mathbb{R}^\times \to GL_n(\mathbb{R})$. This sends $SL_n$ to itself, and on $\mathbb{R}^\times$ it acts as $x\mapsto \lambda(x)I_n$ where $\lambda:\mathbb{R}^\times \to \mathbb{R}^\times$ is an isomorphism.
For $n$ even, there is an isomorphism $SL_n(\mathbb{R})\rtimes \mathbb{R}^\times \to GL_n(\mathbb{R})$ induced by a section of the evident short exact sequence, and there is no isomorphism $SL_n(\mathbb{R})\times \mathbb{R}^\times \to GL_n(\mathbb{R})$

