Given two vector spaces $V,W$ with finite dimension,show that they are isomorphic if and only if $\text{dim(V)}=\text{dim(W)}$.

Question

Given two vector spaces $V,W$ over the field $\mathbb F$ with finite dimension,show that they are isomorphic if and only if $\text{dim(V)}=\text{dim(W)}$.

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$\Longrightarrow$

This this direction looks pretty straightforward to me,since if the two vector spaces are isomorphic then it means that an isomorphic exists between them,which is a bijective homomorphism,and we know that if there exist a bijection between two sets then their cardinality is the same and if they are finite then $$\text{dim(V)}=\left|V\right|=\left|W\right|=\text{dim(W)}$$

$\Longleftarrow$

One can use the fact that every finite dimension vector space $V$ with $\text{dim (V)}=n$ is isomorphic to $\mathbb F^n$ and use the fact that the composition of isomorphism is another isomorphism.

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I'm not sure if my proof is right,it would be nice if someone checks that,thanks.

Answer 1

In $\Longrightarrow$, $\text{dim(V)}=\left|V\right|=\left|W\right|=\text{dim(W)}$ is false since dimension of a finite dimensional vector space is the size of a basis for that vector space, not the cardinality (or size) of the vector space itself.

To prove this part, first define $B_V = \{v_1,...,v_n\}$ as a basis for $V$ with $\dim(V) = n$. Then, since $V$ and $W$ are isomorphic, there exists a linear transformation $T:V \to W$ such that $T$ is a bijection. Then, since $T$ is a linear transformation, it is uniquely determined by where it sends $v_1,...,v_n$ to. Now, you can show that $B_W = \{T(v_1),...,T(v_n)\}$ is a basis for $W$ by using the assumption that $T$ is bijective (injectivity will be useful for showing linear independence of $B_W$ and surjectivity will be useful for showing $\text{Span}(B_W) = W$).

For $\Longleftarrow$, I think your second argument is fine.

Answer 2

Hints:

$\implies:$ Show that an injective linear map $\;T:V\to W\;$ sends linearly independent (=l.i.) sets to l.i. sets, and thus an isomorphism $\;T\;$ as above maps a basis $\;A\;$ of $\;V\;$ to a l.i. set $\;B\;$ in $\;W\;$ . That $\;B\;$ is in fact a basis of $\;W\;$ follows from the fact that $\;T\;$ is also surjective and thus maps generating sets to generating sets, and thus $\;B\;$ is in fact l.i. generating set = a basis, with the very same of elements as $\;A\;$ , and thus $\;\dim V=\dim V\;$

$\;\Longleftarrow\;:$ If $\;\dim V=\dim W\;$ then choosing basis $\;A=\{a_i\}_{i\in I}, B=\{b_j\}_{j\in J}\;$ in $\;V, W\;'$ resp., define $\;T:V\to W\;$ by $\;Ta_i:=b_{\phi(i)}\;$, where $\;\phi:A\to B\;$ is a bijective map (which exists by assumption). Thus...continue and finish the proof.

Observe that the above proof works for any cardinality of $\;\dim V=|A|\,,\;\;\dim W=|B|\;$, and it can be slightly less messy if we assume finite dimensionality of all the vector spaces involved.

Answer 3

Both implications of your proof seem to misuse the definition of vector space dimension. Vector space dimension is not the same as cardinality of the vector space. Rather, the dimension of a vector space is the number of elements of a basis in that vector space.

$\Leftarrow$:

Let $n =\operatorname{Dim} V$. This means that there is a basis of $V$ consisting of $n$ elements. Let $v_1, ... , v_n$ be such a basis. Then every element of $V$ is equal to $c_1v_1 + \cdots +c_nv_n$ for unique scalars $c_i$.

If $\operatorname{Dim}V = \operatorname{Dim}W$, then there is a basis $w_1, ... , w_n$ of $W$ also consisting of $n$ elements. Define a function

$$T: V \rightarrow W$$

by $T(c_1v_1 + \cdots + c_nv_n) = c_1 w_1 + \cdots + c_nw_n$.

You are done if you can show that $T$ is an isomorphism. In other words, if you can show that $T$ is a linear transformation, and that it is a bijection from $V$ onto $W$.