Let $L/K$ be field extension and let $V$ be a $K$-vector space. Then do we have an isomorphism $$L\otimes_K \text {End}_K(V)\cong \text{End}_L(L\otimes_K V)$$ as $L$-algebras?
My attempt: for $\lambda \in L$ and $f\in \text{End}_K(V)$ let $\varphi_{\lambda, f}:L\otimes_K V \to L\otimes_K V$ be the $L$-linear map given by $\alpha \otimes v \mapsto \lambda\alpha\otimes f(v)$.
Then the mapping $\varphi:L\times \text{End}_K(V)\to \text{End}_L(L\otimes V)$ given by $\varphi(\alpha, f)=\varphi_{\alpha,f}$ is $K$-bilinear, so it gives rise to a unique $K$-linear map $\bar{\varphi}:L\otimes_K \text {End}_K(V)\cong \text{End}_L(L\otimes_K V)$ such that $\bar{\varphi}(\lambda\otimes f)=\varphi_{\lambda,f}$. It is also a homomorphism of $L$-algebras.
My problem is showing that $\bar{\varphi}$ has an inverse.
Many thanks.
It is a general fact of life that there is an isomorphism of $L$-vector spaces: $$\hom_L(L \otimes_K X, Y) \cong \hom_K(X, Y),$$ where $X$ is a $K$-vector space and $Y$ is an $L$-vector space. To see this, if $f \in \hom_K(X, Y)$, let $\bar{f} \in \hom_L(L \otimes_K X, Y)$ be defined by $\bar{f}(\lambda \otimes x) = \lambda f(x)$. The inverse morphism maps $g \in \hom_L(L \otimes_K X, Y)$ to $\tilde{g} \in \hom_K(X,Y)$, where $\tilde{g}(x) = g(1 \otimes x)$.
It follows that: $$\operatorname{End}_L(L \otimes_K V) = \hom_L(L \otimes_K V, L \otimes_K V) \cong \hom_K(V, L \otimes_K V).$$ You want to show that this is isomorphic to $L \otimes_K \hom_K(V,V)$ $(= L \otimes_K \operatorname{End}_K(V))$.
In general there is a canonical linear map: $$\Psi : Z \otimes_K \hom_K(X, Y) \to \hom_K(X, Z \otimes_K Y)$$ that maps $z \otimes f$ ($z \in Z$, $f : X \to Y$) to $\psi_{z,f} : X \to Z \otimes_K Y$, where $\psi_{z,f}(x) = z \otimes f(x)$ (this corresponds to your $\varphi$.) It's also easy to see that when $Z$ is a $K$-algebra (eg. $L$) then this is a morphism of algebras).
Case 1: $L$ is finite dimensional
When $Z$ is finite dimensional over $K$, this map is an isomorphism. Indeed let $(z_k)_{k=1}^n$ be a basis of $Z$ in that case. For $f \in \hom_K(X, Z \otimes_K Y)$, every $f(x)$ can be written as a finite sum $\sum_{k=1}^n z_k \otimes f'_k(x)$, where $f'_k(x) \in Y$. Using the linearity of $f$, you can check that in fact, all the $f'_k$ are $K$-linear maps $X \to Y$, and so you have a well-defined element $$f' := \sum_{k = 1}^n z_k \otimes f'_k \in Z \otimes_K \hom_K(X, Y).$$
Then $f \mapsto f'$ is the inverse of the map $\Psi$ we wrote earlier, which is thus an isomorphism. So in the special case $X = Y = V$, $Z = L$, you get the isomorphism you want (and as I said it is an isomorphism of $K$-algebras), in the case that $L$ is finite-dimensional over $K$.
Case 2: $V$ is finite dimensional
When $X$ is finite-dimensional with basis $(x_k)_{k=1}^n$, $\Psi$ is also an isomorphism. The inverse is even more straightforward: if $f \in \hom_K(X, Z \otimes_K Y)$, let $$f(x_k) = \sum_{i = 1}^{m_k} z_i \otimes y_i.$$
For $y \in Y$ and $1 \le j \le n$, let $\delta_{j,y} : X \to Y$ be the map defined by $$\delta_{j,y}(x_i) = \begin{cases} y & i = j \\ 0 & i \neq j \end{cases}$$
And let $$\theta(f) = \sum_{k=1}^n \sum_{i=1}^{m_k} z_i \otimes \delta_{k, y_i}.$$
Then $\theta$ is the inverse of $\Psi$, so again you get an isomorphism (in the special case $X = Y = V$, $Z = L$) when $V$ is finite dimensional. (This proof essentially comes from the fact $\hom(X,Y) \cong X^* \otimes Y$ when $X$ is finite-dimensional, or "equivalently" $\hom(K^n, Y) \cong Y \oplus \dots \oplus Y$ and $Z \otimes (U \oplus V) \cong Z \otimes U \oplus Z \otimes V$).