# Why is the Change of Basis map unique?

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?

• 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?". – M. Winter Feb 7 at 9:22
• Because if it was not unique, there would be at least a vector with several possible coordinates in $\mathcal C$ – Evpok Feb 8 at 12:42

## 2 Answers

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.