Proving $V$ is isomorphic to $W$ iff $\dim V=\dim W$

Question

Let $V$ and $W$ be two finite vector spaces over $F$.

Prove that $V$ is isomorphic to $W$ iff $\dim V=\dim W$

I think I got the general approach but I don't think it's rigorous enough.

$\Rightarrow$

Suppose $V$ is isomorphic to $W$ then there's a linear map $T$ that is a bijection. Let $\mathbb B=\{v_1,...,v_n\}$ be a basis for $V$. We know that $V=\displaystyle\sum^{n}_{i=1}\alpha_iv_i$. Since $T$ is a bijection there n elements in $W$ such that $Tv_i=w_i$. So for each $w\in W$ there's a single representation of $\displaystyle\sum^{n}_{i=1}\alpha_iw_i$. So we get that the set $\mathbb K=\{w_1,...,w_n\}$ is linearly independent thus it's basis so $\dim V=\dim W$.

$\Leftarrow$

I need to show that if there's a linear map between the two spaces then it's a bijection thus it's isomorphic but I'm not sure to word it right.

Answer 1

The left to right implication is basically correct, but I would do it differently, as a consequence of two more general facts.

If $f\colon V\to W$ is a linear map and $\{v_1,\dots,v_n\}$ is a spanning set for $V$, then $\{f(v_1),\dots,f(v_n)\}$ is a spanning set for the image (or range) of $f$. Just observe that, if $v\in V$, then $v=\alpha_1v_1+\dots+\alpha_nv_n$, so $$ f(v)=\alpha_1 f(v_1)+\dots+\alpha_n f(v_n). $$

If $f\colon V\to W$ is an injective linear map and $\{v_1,\dots,v_n\}$ is a linearly independent set, then $\{f(v_1),\dots,f(v_n)\}$ is a linearly independent subset of $W$; indeed, if $$ 0=\alpha_1 f(v_1)+\dots+\alpha_n f(v_n) $$ you can write $$ 0=f(0)=f(\alpha_1v_1+\dots+\alpha_nv_n) $$ and, by injectivity, $\alpha_1v_1+\dots+\alpha_nv_n=0$, so $\alpha_1=\dots=\alpha_n=0$.

Now, let $f\colon V\to W$ be linear and bijective. If $\{v_1,\dots,v_n\}$ is a basis of $V$, it is both a spanning set of $V$ and linearly independent. Then $\{f(v_1),\dots,f(v_n)\}$ is a spanning set of $\operatorname{im}(f)=W$ ($f$ is surjective) and linearly independent ($f$ is injective).

The converse implication follows from the fact that assigning the images for the vectors in a basis of $V$ uniquely defines a linear map. So, if $\{v_1,\dots,v_n\}$ and $\{w_1,\dots,w_n\}$ are bases of $V$ and $W$, consider the unique linear map $f\colon V\to W$ such that $$ f(v_k)=w_k,\quad k=1,2,\dots,n. $$ Then $f$ is surjective, because a spanning set for the image of $f$ is a basis of $W$ by construction. The rank-nullity theorem says that the null space (kernel) of $f$ has dimension $0$, so $f$ is injective.

Answer 2

Your proof of $\Rightarrow$ is technically correct, but if i was your teacher, I would demand that you explain more clearly why $\{w_1,\dots,w_n\}$. I think you got the idea, but just spend one more sentence on it.

In the other direction, you know that $\dim V=\dim W$. To prove that $V$ and $W$ are isomorphic, construct two bases, one for the space $V$ and one for the space $W$. Then build a linear mapping using the two bases and prove that the linear mapping is bijective.

Answer 3

The most important point is that you must first know that dimension is well defined (for finitely generated spaces): it that same space has two finite bases, they must contain the same number of vectors. This requires soom work to prove, but is assumed to be known in the problem statement.

Now an the image under an isomorphism of a basis is a basis; this can be immediately deduced from the properties defining a basis. That's your $\Rightarrow$ direction.

Conversely, if both spaces have bases with the same number of elements, there is a bijection $f$ between those bases. Now any map from a basis of a vector space $V$ to a vector space $W$ can be uniquely extended to a linear map $\phi: V\to W$. Applying that to $f$ and then to the inverse bijection $f^{-1}$ between the bases, one gets linear maps $\phi: V\to W$ and $\psi: W\to V$ that are inverses of each other (both compositions linearly extend the identity map on one of the bases, giving the identity operator on its vector space) so $\phi$ and $\psi$ are (mutually inverse) isomorphisms.

Answer 4

How to do it (the logic):

Take the one-one and onto linear isomorphism. One-one ness implies linearly independent elements go to linearly independent elements under the map. So image of basis is linearly independent and spans whole range space due to ontoness. So dimensions are equal rank-nullity.

Now for the other part : Take a basis of V and of W and take a bijective map, a linear transformation is defined by its action on the basis, and you get a linear transformation. Its clearly an isomorphism.

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Ok then. $V = \langle v_1, \dots, v_n \rangle$. $V$ and $W$ are isomorphic, means there exists $f : V \rightarrow W$ linear, 1-1 and onto.

Claim : {f(v_i)}_{1 \leq i \leq n} is linearly independent.

Consider \begin{eqnarray*} & \sum_{i = 1}^{n} \alpha_i f(v_i) = 0 \\ & \sum_{i = 1}^{n} f_i(\alpha_i v_i) = 0 \\ & \sum_{i = 1}^{n} \alpha_i v_i = 0 \; (f \mbox{ is } 1-1) \\ & \Rightarrow \alpha_i = 0 \; \forall \; 1 \leq i \leq n. \end{eqnarray*}

$\langle f(v_1), \dots, f(v_n) \rangle = f(V) = W$ as $f$ is onto. So dim W = n.

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Other part : $V = \langle v_1, \dots, v_n \rangle$ and $W = \langle w_1, \dots, w_n \rangle$. Consider $f$ s.t. $f(v_i) \mapsto w_i$. This is clearly linear and onto.

For one-one, we will just prove $Ker f = \{0\}$ (as $f(v) = f(v')$ is same as $v-v' \in Ker f$). We do it by contradiction. Suppose $v \neq 0 \in Ker f$. Let $v = a_1 v_1 + \dots + a_n v_n$ not all $a_i$s zero.

$f(v) = f(a_1 v_1 + \dots + a_n v_n) = a_1 f(v_1) + \dots + a_n f(v_n) = a_1 w_1 + \dots + a_n w_n = 0$ which implies $a_i = 0$ for all $1 \leq i \leq n$ because elements of a basis are linearly independent by definition.

That does it !