I've been looking all over, but I haven't found anything satisfactory.

We've been shown in class by a commutative diagram that, given an $n$-dimensional vector space $V$ over a field, $\mathbb{F}$, and bases, $\mathcal{B}=\{v_1,...,v_n\}$ and $\mathcal{C}=\{u_1,...,u_n\}$, the coordinate maps $[]_{\mathcal{B}}:V\rightarrow \mathbb{F}^n$ and $[]_{\mathcal{C}}:V\rightarrow \mathbb{F}^n$ give rise to a unique map $P=[]_{\mathcal{B}}\circ []^{-1}_{\mathcal{C}}:\mathbb{F}^n\rightarrow \mathbb{F}^n$, which is our change of basis matrix.

But I am having a lot of trouble proving that $P$ is unique. Can anyone enlighten me as to why this is necessarily true?

  • 5
    $\begingroup$ The definition in your post explicitly constructs $P$, hence gives a single well-defined map. I do not see where problems with uniqueness should come from. It's like asking "why is $\cos\circ\sin^{-1}$ unique?". $\endgroup$ – M. Winter Feb 7 at 9:22
  • $\begingroup$ Because if it was not unique, there would be at least a vector with several possible coordinates in $\mathcal C$ $\endgroup$ – Evpok Feb 8 at 12:42

Remember that any linear map on any linear space $\;V\;$ is uniquely and completely determined once we know its action on any basis of $\;V\;$ ...and that's all.

If you want to do this proof, suppose there's another map $\;Q:V\to V\;$ s.t. it coincides on "the old basis" $\;\mathcal B\;$ with $\;P:\;\; Qv_i=Pv_i\;\;\forall\,i=1,2,...,n\;$ , then (using linearity of the maps), for any

$$v=\sum_{k=1}^n a_iv_i\in V\;,\;\;Qv=\sum_{k=1}^na_iQv_i=\sum_{k=1}^n a_iPv_i=Pv$$

so $\;Q\equiv P\;$.


A basis is an ordered set of vectors that are independent and generates the whole vector space. If you have two basis $\mathcal B$ and $\mathcal C$ as in your post, then a change of basis $f$ from $\mathcal B$ to $\mathcal C$ must satisfy $f(v_i)=u_i$ for $i=1,\ldots,n$ (in this precise order). There is a unique map satisfying this requirement by a theorem stating that if a linear map is defined over a basis then it is uniquely defined over the whole space. Since the requirement fix the images of the elements of the basis $\mathcal B$ then there is only a unique map that satisfies those conditions.


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