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Could anyone let me know how I go about this question? I have no idea what to do with the permutations. I don't even understand what I'm being asked here. Thanks.

Suppose that $(u_1,\dots,u_n)$ and $(v_1,\dots,v_n)$ are bases of a vector space $V$.

Show that there is a permutation $\pi$ such that $(u_1,\dots,u_{i-1},v_{π(i)},\dots,v_{π(n)})$ is a basis of $V$, for each $i = 1,2,\dots,n$.

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So you have $n$ basis-vectors $u_j$ for your $n$-dimensional vector space. Suppose we remove just one one $u$ basisvector, say $u_n$. What remains ($u_1,\ldots,u_{n-1}$) is a basis for an $n-1$ dimensional subspace of your original vector space.

What do you have to add to ($u_1,\ldots,u_{n-1}$) to get a basis for your original vector space? It suffices to add a vector which is in $V$ but not in span($u_1,\ldots,u_{n-1}$). Now ($v_1,\ldots,v_n$) is guaranteed to contain a vector not in this span - otherwise it wouldn't be a basis of $V$. Suppose $v_k$ is (one of) the $v$ that are not in span($u_1,\ldots,u_{n-1}$). Define your permutation such that $\pi(n)=k$, and you have your answer.

A similar argument will work if you remove more than 1 of the $u$ vectors.

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