Proving rank of linear operator equals the rank of the B matrix of linear operator

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I am new to linear algebra, and I am struggling with this question. Any help would be really appreciated.

Let $B$ and $C$ both be bases for $\Bbb R^n$ and let $L: \Bbb R^n → \Bbb R^n$ be a linear operator. Prove that $\text{rank}([L]_B) = \dim(\text{Range}(L))$

What I have concluded so far is that we are trying to prove that $\text{rank}([L]_B) = \text{rank}(L)$, but I also do not know how to prove this.

I would be grateful for any help!

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Hint: using the definition of the matrix of a linear map, you can show first that the range of $L$ is the vector subspace of $\Bbb R^n$ generated by the columns of $[L]_B$.