Let $E$ be a vector space with countable basis.Construct a linear isomorphism between $E$ and $E \oplus E.$

Question

In the book of linear algebra by Werner Greub, at page $63$, it is asked that

Let $E$ be a vector space with countable basis.Construct a linear isomorphism between $E$ and $E \oplus E.$

I first thought that the direct sum is an internal sum, and defined a map $\phi: E \to E \oplus E$ by $$x_v \to x_v + x_v,$$ where $\{x_v\}$ is a basis for $E$, and I argued that for any $u,v$ the vectors $x_v + x_v$ and $x_u + x_u$ are linearly independent.Moreover, since this set is a maximal linearly independent set, it is a basis for $E$.

So my questions are;

-Is my proof correct ?

-why do we need $E$ to have a countable basis ? Is it because, as I do, we need to be able to give indices to each member of the basis ? I mean if the basis wasn't countable, we couldn't give them indices, right ?

-Is there any alternative mapping than that satisfying the same conditions ?

Answer 1

It is well-known that

If $V$ and $W$ are vector spaces over the same field $K$ such that $V$ has a base $\{v_i; i \in I\}$ and $W$ has a base $\{w_j: j \in J\}$ such that $|I| = |J|$ then $V$ and $W$ are isomorphic as vector spaces.

The proof is quite easy: let $f: I \to J$ be a bijection, promised by $|I| = |J|$, then define $I(v_i) = w_{f(j)}$ and extend by linearity. Then $I: V \to W$ is an isomorphism.

If $I$ is an infinite set and $E$ has a base $\{b_i : i \in I\}$, then it is easy to see that $\{(b_i,0), (0, b_i): i \in I\}$ is a base for $E \oplus E$ (as an external direct sum; an internal sum makes no sense here) and its size is $|I| + |I| = |I|$ as $I$ is infinite. So $E$ and $E \oplus E$ have the same size basis,so are isomorphic.

Answer 2

Your map is $x\mapsto (x,x)$. This is injective, but not surjective.

Let your basis be $e_1,e_2,e_3,\ldots$. Consider the subspace spanned by $e_1,e_3,e_5,\ldots$ and the subspace spanned by $e_2,e_4,e_6,\ldots$.

Countability isn't crucial here, at least if you believe the Axiom of Choice.

Answer 3

Actually, the definition of direct sum $\oplus$ is slightly different than that:

If $A$ and $B$ are vector spaces with corresponding bases $\{a_1,\ldots, a_k\}$ and $\{b_1,\ldots, b_\ell\}$, then the direct sum $A\oplus B$ can be defined as the vector space whose basis is the disjoint union of the two bases:

$A\oplus B$ is a vector space with basis $\{a_1, \ldots, a_k, b_1, \ldots, b_\ell\}$.

If $A=B=E$, and $E$ has a countable basis $\{e_k : k\in \mathbb{N}\}$, then $E\oplus E$ also has a countable basis, because it is the disjoint union of two countably infinite sets.

To define a linear map between any two spaces $A\rightarrow B$, it is enough to define what it does to basis vectors. In the case of a linear isomorphism, the mapping must be a bijection between the basis vectors.

In particular, if you are sending $\phi: E \rightarrow E\oplus E$, how will you establish a bijection between the basis of $E$ and the basis of $E\oplus E$?

Call the basis of $E$ $\{e_i\}$ and refer to the basis of $E\oplus E$ as $\{\color{maroon}{e_i}\}\cup \{\color{green}{e_i}\}$. Here, the colors are used to indicate the direct sum.

Your proposed map is something like $\phi : e_i \mapsto \color{maroon}{e_i} + \color{green}{e_i}$. This does not span the whole space, because for example there is no vector $v$ such that $\phi(v) = \color{green}{e_i}$ only. (The proof of this comes from the fact that the $e_i$ are linearly independent, so there is no way to make a linear sum of them vanish without all of the coefficients being equal to zero.)

Instead, consider the following map on basis elements:

$$\phi(e_k) = \begin{cases}\color{maroon}{e_{k/2}}& \text{ if }k \text{ is even}\\\color{green}{e_{(k+1)/2}}& \text{ if }k \text{ is odd}\\\end{cases}$$

In other words,

$$\begin{align*} e_1 & \mapsto \color{green}{e_1}\\ e_2 & \mapsto \color{maroon}{e_1}\\ e_3 & \mapsto \color{green}{e_2}\\ e_4 & \mapsto \color{maroon}{e_2}\\ e_5 & \mapsto \color{green}{e_3}\\ e_6 & \mapsto \color{maroon}{e_3}\\ \vdots & \\ \end{align*}$$

Then $\phi$ is a bijection between the basis of $E$ and the basis of $E\oplus E$, which means that it defines a linear isomorphism between them.