Basis of extension of scalars Let $A$ be a $k$-algebra of finite dimension, where $k$ is a field, with $k$-basis $e_1,\cdots,e_n$. Let $K$ be some field extension of $k$. The extension of scalars is defined as $A\otimes_kK$.
What's a $K$-basis of $A\otimes_kK$? Would $e_1\otimes1,\cdots,e_n\otimes1$ work?
It clearly generates $A\otimes_kK$. However, I am having trouble showing linear independence.
Let $\lambda_1,\cdots,\lambda_n \in K$, suppose $\sum_{i=1}^ne_i\otimes\lambda_i=0$, why does this imply $\lambda_1=\cdots=\lambda_n=0$?
If $\lambda_i\in k$ then I can exploit bilinearity to get $\sum_{i=1}^ne_i\otimes\lambda_i=0=(\sum_{i=1}^ne_i\lambda_i)\otimes1.$
So $(\sum_{i=1}^ne_i\lambda_i)=0$ and $\lambda_1=\cdots=\lambda_n=0.$
But I don't know what to do in the case where $\lambda_i\in K.$
Any hints are appreciated.
 A: I believe that Lord Shark's answer is the right way to approach the problem. I will try to explain their answer.
Suppose $\sum_{i=1}^ne_i\otimes_k\lambda_i=0$ for some $\lambda_1,\ldots,\lambda_n\in K$. If we define a $k$-bilinear map $\beta_i:A\times K\to K$ as in Lord shark's answer, that is, $$\beta_i\left(\sum_{j=1}^n a_j e_j,b\right)=a_ib,$$ we have a $k$-linear map $f_i: A\otimes_k K\to K$ which takes $\left(\sum_{j=1}^n a_j e_j\right)\otimes_k b$ to $a_ib$. 
Note that $f_i(e_i\otimes_k b)=b$ and $f_i(e_j\otimes_k b)=0$ if $i\neq j$.
So $$f_i\left(\sum_{j=1}^ne_j\otimes_k\lambda_j\right)=\lambda_i.$$ But $$f_i\left(\sum_{j=1}^ne_j\otimes_k\lambda_j\right)=f_i(0)=0.$$ So $\lambda_i=0$. Since $i$ is arbitrary, the claim follows.
A: For each $i$, there's a $k$-bilinear map $\beta_i:A\otimes_k K\to K$
defined by
$$\beta_i\left(\sum_{j=1}^n a_j e_j,b\right)=a_ib.$$
This corresponds to a $k$-linear map taking
$$\sum_{j=1}^n e_j\otimes b_j$$
to $b_i$. So if $\sum_{j=1}^n e_j\otimes b_j=0$ then $b_i=0$ for all $i$.
