Isomorphism of vector spaces

Question

There is an example given in my lecture notes, which I feel a bit uncertain about

$ \{ $functions$ \{1,2,...,n \} \to \mathbf F \} \cong \mathbf F^n $ by the map $ f \mapsto (f(1),f(2),...f(n))$ where $\cong$ means isomorphic

My question is what if $f(k)=0$ for some k ? Then the mapping would not be surjective. Could somebody help me on this please?

Answer 1

Seems that you're considerating $f$ and the map you have as one thing, which isn't the case. Let us first set some notations: $V=\{\text{functions}\,\{1,\cdots,n\}\to\mathbf{F}\}$ and let $$\varphi:V\to\mathbf{F}^n\\f\mapsto (f(1),\cdots,f(n))$$

$\varphi$ is surjective because the image of a basis of $V$ spans $\mathbf{F}^n$. A basis of $V$ is $\mathcal{B}=\{f_1,\cdots,f_n\}$ where for each $i\in\{1,\cdots,n\}$, $f_i$ is the function defined by:

1) $\forall k\in\{1,\cdots, n\}\backslash\{i\},\,f_i(k)=0$

2) $f_i(i)=1$

Then it's easy to see that $\varphi(\mathcal{B})=\mathcal{B}_c$ where $\mathcal{B}_c$ is the canonical base of $\mathbf{F}^n$.

$\varphi$ is also injective because you can easly prove that $\operatorname{null}\,\varphi=\ker\varphi=\{0_V\}$ where $0_V$ is the map that maps all the elements of $\{1,\cdots, n\}$ to $0$.

Answer 2

You need to write this in full.

Let $V$ be the set of functions $\{1,2,...,n \} \to \mathbf F$. Then $V$ is a vector space over $ \mathbf F$. (Check!)

Now let $T: V \to \mathbf F^n$ be given by $T(f)=(f(1),f(2),...f(n))$.

Then $T$ is a linear transformation which is an isomorphism. (Check!)

You need to check that $T$ is linear, injective, and surjective.

Note that you need to reason about $T$, not really about the individual $f$ in $V$.

Answer 3

The map is surjective because given any $(x_1,x_2,\dots,x_n) \in \mathbf F^n$ there exists a function from $\{1,2,\dots, n\}$ to $\mathbf F$ such that $f(i)=x_i$ for all $i$.

In fact, that function $f$ from the vector space $V=\{$functions $\{1,2,\dots,n\}\to\mathbf F\}$ is defined by $$f(1)=x_1,f(2)=x_2,\dots,f(n)=x_n,$$ because to define such a function all you have to do is tell it how to act on each element of its domain, which is $\{1,2,\dots,n\}$.

There may be functions in $V$ that map every $i$ to a constant, but it doesn't matter, because there are other functions that don't.

Something you can do to convince yourself it is an isomorphism is try to figure out the preimage of every basis vector of $\mathbf F^n$.

For example, let's look at the standard basis $(e_1,e_2,\dots,e_n)$, where $e_i$ has a 1 in the $i$th coordinate and 0 everywhere else.

The vector $e_1$ corresponds under the isomorphism to the function $f_1$ that sends 1 to 1 and everything else to 0, or $f_1(1)=1,f_1(2)=0, \dots, f_1(n)=0$.

The vector $e_2$ corresponds under the isomorphism to the function that sends 2 to 1 and everything else to 0, or $f_2(1)=0,f_2(2)=1, f_2(3)=0, \dots, f(n)=0$.

And so on. Note that the function is different in each case. Try to check that $f_1,f_2,\dots,f_n$ are a basis for $V$.