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Let $\mathbb F$ be a subfield of a field $\mathbb K$ satisfying the condition that the dimension of $\mathbb K$ as a vector space over $\mathbb F$ is finite and equal to $r$. Let $V$ be a vector space of finite dimension $n>0$ over $\mathbb K$. Find the dimension of $V$ as a vector space over $\mathbb F$.

I am not sure how to even approach this problem. Looking around online I found $\dim_{\mathbb F}(V)=\dim_{\mathbb F}(\mathbb K)\dim_{\mathbb K}(V)$ for a field extension $\mathbb K/\mathbb F$. Where does this come from?

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Well you gave the answer yourself. The reason that equation is true is as follows. Suppose $\{k_1,\dots,k_r\}$ is a basis for $K$ over $F$, and $\{v_1,\dots,v_n\}$ is a basis for $V$ over $K$. Then it suffices to show $\{k_iv_j : 1 \le i \le r, 1 \le j \le n\}$ is a basis for $V$ over $F$. But this is just unfolding definitions and is left as an exercise to you.

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Let $\dim_K V = m$ and write $v\in V$ as $\displaystyle\sum_{i=1}^m a_i v_i$. Then write $a_i\in K$ as $\displaystyle\sum_{j=1}^nb_{ij}a_j$. Put them together to get

$$v = \sum_{i=1}^m\sum_{j=1}^nb_{ij}v_{ij}$$

where $\displaystyle a_iv_i = \sum_{i=1}^n b_{ij}v_{ij}$. The minimality of $mn$ is given by minimality of each of $m$ and $n$ in their respective settings.

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