I'm reading "Linear Algebra via Exterior Products" by Winitzki. The part leading up to the canonical isomorphism definition firstly proves the result that an $n$-dimensional vector space $V$ over $\mathbb{R}$ is isomorphic to $\mathbb{R}^n$. Suppose the isomorphism is defined as $\hat{M}v = (x_1, \ldots, x_n)$, where the coordinates $x_i$ are given in the standard basis $\{e_i\}$.

Then it says:

Note that the isomorphism $\hat{M}$ will depend on the choice of the basis: a different basis $\{e′_i\}$ yields a different map $\hat{M_1}$. For this reason, the isomorphism $\hat{M}$ is not canonical.


A linear map between two vector spaces $V$ and $W$ is canonically defined or canonical if it is defined independently of a choice of bases in $V$ and $W$.

I can't understand why changing the basis of $\mathbb{R}^n$ would yield a different linear map. Shouldn't the map remain the same? Only the representation of the output vectors would change because of the basis change.

And by that logic, for any linear map, changing the basis would change the coordinates, and hence the representation, of the output vectors. So what exactly is meant by a linear map "defined independently of the choice of bases"?

  • $\begingroup$ Two maps are different if their image on at least one element of hte domain is different...and this precisely happens in this case, of course, as linear maps are completely and uniquely determined by their values on any basis of the domain. $\endgroup$ – DonAntonio Jan 11 '18 at 7:19
  • $\begingroup$ The map $\hat{M}$ is the representation map $\endgroup$ – eepperly16 Jan 11 '18 at 7:23

Let me try and address what I see as the root of your question. Suppose I have two vector space $V$ and $W$ and a linear map $T : V \to W$ between them. Clearly, the action of $T$ on a vector $v \in V$ is independent of any bases $\beta,\gamma$ I may have chosen for $V$ and $W$. $T(v)$ is $T(v)$ no matter what bases I choose for $V$ and $W$.

Our situation here is much different. We haven't been given a linear transformation $T : V \to \Bbb R^n$ but are asked to build a linear transformation $\hat{M} : V \to \Bbb R^n$ which is an isomorphism. Here we first choose a basis $\beta = \{v_1,\ldots,v_n\}$ for $V$ and then define

$$ \hat{M}(v) = \hat{M}(x_1v_1 + x_2v_2 + \cdots + x_nv_n) = (x_1,x_2,\ldots,x_n). $$

Because every vector in $V$ can be expanded uniquely as a linear combination of $\beta$, $\hat{M}$ is unique and well-defined. Such a map is sometimes called a representation map $\hat{M}$ as it "represents" elements of the abstract vector space $V$ in the concrete vector space $\Bbb R^n$.

However, if we pick a different basis $\beta'$ and compute a new representation map $\hat{M}'$, we have no guarantees that $\hat{M} = \hat{M}'$. In fact, if $\beta \ne \beta'$ then $\hat{M} \ne \hat{M}'$. Alejandro's answer gives a great example of how picking $\beta \ne \beta'$ does indeed give $\hat{M} \ne \hat{M'}$.

There is no contradiction here. We have simply built two different linear transformations $\hat{M}$ and $\hat{M}'$ using different bases. While $\hat{M}$ and $\hat{M}'$ are both ways of representing vectors in $V$ as vectors in $\Bbb R^n$ they are different transformations.

You might ask if its possible to build a map from $V$ to $\Bbb R^n$ without picking a basis first for $V$. This turns out to be impossible in general. Thus we may say there is no canonical isomorphism from $V$ to $\Bbb R^n$: any isomorphism from $V$ to $\Bbb R^n$ will depend on which basis you use to make the representation map.

For example, if I say to you that $V$ has dimension three, there is no way for me you can construct for me a linear map $T : V \to \Bbb R^3$ without picking a basis (or something equivalent).


Consider the vector space of linear and constant polynomials with real coefficients: $$V=\{a+bx: a,b \in \mathbb R\}.$$

Let's also consider the bases $B_1=\{1,x\}$ and $B_2=\{x,x+1\}$. If you express in each basis the polynomial $f(x)=2-3x$ you get $$f(x)=2\cdot 1+(-3)\cdot x$$ and $$f(x)=(-5)\cdot x+2\cdot (x+1).$$

Then, if we consider the first basis the isomorphism $\hat M$ is such that $$\hat M(2-3x)=(2,-3)$$ but if we take the second basis we get $$\hat M(2-3x)=(-5,2).$$

So that when you say "let $B$ be a basis...", the choice of that basis determines which homeomorphism you're defining.


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