Let $K/k$ be a finite separable extension. If necessary, we can assume $[K : k] = 2$. Let $H$ be a $K$-closed subgroup of $\operatorname{GL}\_n$, and let $\tilde{H} = \operatorname{Res}\_{K/k}H$. Since $\tilde{H}$ is a linear algebraic group over $k$, it should embed into some $\operatorname{GL}\_m$. I was wondering if there is a nice way to do this in general if, say, we fix a basis of $K/k$.


Of course since restriction of scalars respects closed subgroup immersions, it suffices to do this for $H = \operatorname{GL}\_n$. 


**Example**: $K = \mathbb C, k = \mathbb R$, $H = \operatorname{GL}\_{1,\mathbb C}$. If $Z$ is the center of $\operatorname{GL}\_{2,\mathbb R}$, we can realize $\tilde{H}$ inside $\operatorname{GL}\_{2,\mathbb R}$ as the product $Z.\operatorname{SO}\_2$, since a nonzero complex number $re^{i\theta}, e^{i\theta} = a+bi$ identifies with


$$\begin{pmatrix} r \\ & r \end{pmatrix} \begin{pmatrix} a & b \\ -b & a \end{pmatrix}$$


**(Non)-example**: Again $K = \mathbb C, k = \mathbb R$, $H = \operatorname{GL}\_{n,\mathbb C}$. Then $\tilde{H}$ is $\mathbb R$-isomorphic to a Levi subgroup of the unitary group $U(n,n)$. But this isn't quite what I want, because $U(n,n)$ is an outer form of $\operatorname{GL}\_{2n}$. In other words, this embedding $\tilde{H} \rightarrow \operatorname{GL}\_{2n}$ is not defined over $\mathbb R$. 


**(Non)-example 2**: If $K/k$ is quadratic with nontrivial automorphism $\sigma$, then we have an embedding $\operatorname{GL}\_n(K) \rightarrow \operatorname{GL}\_{2n}(k)$ on points by


$$g \mapsto \begin{pmatrix} g \\ & \sigma(g) \end{pmatrix}$$


This embedding respects the Galois action. However, it is not a morphism of group schemes. 



If $H \subset \operatorname{GL}(n)$, can we realize $\operatorname{Res}_{K/k} H$ inside $\operatorname{GL}([K : k]n)$?
Thanks to LSpice for answering. In the quadratic case, we have $K = k(\sqrt{d})$ for $d \in k$. A basis $e\_1, ... , e\_n$ of an $n$-dimensional $K$-vector space has basis $e\_1, ... , e\_n, \sqrt{d} e\_1, ... , \sqrt{d}e\_n$ over $k$. For $g = (\alpha\_{ij}) \in \operatorname{GL}\_n(K)$ with $\alpha\_{ij} = a\_{ij} + \sqrt{d}b\_{ij}$, we have


$$g.e\_k = a\_{1k}e\_1 + \cdots + a\_{nk}e\_n + b\_{1k} \sqrt{d}e\_1 + \cdots + b\_{nk} \sqrt{d}e\_n$$


$$g.\sqrt{d}e\_k = b\_{1k}d e\_1 + \cdots + b\_{nk}d e\_n + a\_{1k} \sqrt{d}e\_1 + \cdots + a\_{nk}\sqrt{d}e\_n$$


Writing $g$ uniquely as $A + \sqrt{d}B$ for $A, B \in \operatorname{Mat}\_n(k)$, the map


$$g \mapsto \begin{pmatrix} A & dB \\ B & A \end{pmatrix}$$
embeds $\operatorname{GL}\_n(K)$ into $\operatorname{GL}\_{2n}(k)$. It's easy to check that the corresponding subgroup variety of $\operatorname{GL}\_{2n}$ is defined over $k$ and isomorphic to $\operatorname{Res}\_{K/k}\operatorname{GL}\_n$. 


