Let $V$ be a finitely generated vector space with $dimV=n$. Let $\mathfrak{B}$ be the set of all bases of $V$, and let $B_1\in\mathfrak{B}$ be the basis $\{\alpha_1, ....,\alpha_n\}$. Prove that there exists a bijection $S$ from $\mathfrak{B}$ to the set of all invertible matrices of order $n\times n$ (we'll denote it by $A$) such that for all $B\in\mathfrak{B}\rightarrow S(B)$ is a change of basis matrix from $B_1$ to some other basis.

Hello everyone. I tried defining the function $S$ as $S(B)=$ $$\begin{pmatrix}|&|&&|\\ [\alpha_1]_{B}&[\alpha_2]_B&\cdots&[\alpha_n]_B\\|&|&&|\end{pmatrix}$$ i.e the $i$th column of $S(B)$ is the coordinate vector of the $i$th element of $B_1$ with respect to the basis $B$. It is an invertible matrix since the coordinate mapping is an isomorphism of vector spaces, for each set $S_0$ and basis $C$ ($S_0\subseteq SpC)$ $\rightarrow$ the set $[S_0]_C$ is linearly independent iff $S_0$ is linearly independent. So the columns of $S(B)$ form an invertible matrix in $A$ since $B$ is a basis and a linearly independent set in particular, hence $S$ is well defined.

Now, I've tried proving that for every $v\in V$, $S(B)[v]_{B_1}=[v]_B$ but had no success. I guess I'm doing it all wrong. I would be happy to hear your thoughts. Thank you in advance :)


1 Answer 1


Hint: You're on the right track.
We have $$\pmatrix{{\bf a_1}&\dots&{\bf a_n}}\pmatrix{v_1\\ 0\\ \vdots\\ 0}\ =\ v_1\cdot{\bf a_1}$$

  • $\begingroup$ Thank you, I understand $S(B)[v]_{B_1}=\sum^{n}_{i=1} [\alpha_i]_B v_i$ but I still can’t see the outcome, does this complete the proof? $\endgroup$
    – Noy
    Mar 14, 2018 at 13:06
  • $\begingroup$ Well, $v= v_1\alpha_1+\dots+v_n\alpha_n$... $\endgroup$
    – Berci
    Mar 14, 2018 at 13:58

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