Yesterday I was studying about isomorphisms between vecto spaces, and basically the fundamental note I highlighted stated that "two vector spaces whose dimensions are different cannot be isomorphic, ever."
One of the canonical examples was like $\mathbb{R}^4 \not\cong \mathbb{R}^3$. Now the notes did not prove this, relying just on the above statement.
But then I thoguht to prove it, by using the definition of isomorphism that is, a linear map $1:1$ and onto. So I thought about this example: suppose there exist a map $h$ such that $h:\mathbb{R}^4 \to \mathbb{R}^3$ defined by the following action
$$\begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix} \mapsto \begin{pmatrix} a+b \\ c \\ d \end{pmatrix}$$
So then I observed that this map cannot be injective ($1:1$) since it would mean that given two vectors $v_1, v_2$
$$v_1 = \begin{pmatrix} a_1 \\ b_1 \\ c_1 \\ d_1 \end{pmatrix} \qquad v_2 = \begin{pmatrix} a_2 \\ b_2 \\ c_2 \\ d_2 \end{pmatrix}$$
then $h(v_1) = h(v_2)$ means
$$\begin{pmatrix} a_1+b_1 \\ c_1 \\ d_1 \end{pmatrix} = \begin{pmatrix} a_2+b_2 \\ c_2 \\ d_2 \end{pmatrix}$$
And as far $c_1 = c_2$, $d_1 = d_2$, we are left with infinite possibilities about $a_1 + b_1 = a_2 + c_2$
Since it is not injective, it cannot be an isomorphism.
- Question(s): is this all right? Or did I just wrote a load of malarkey? :)
- How could I prove that this map is not onto? I tried but I cannot find anything good. Or maybe it is onto...
- If my reasoning is bugged, please can you fix it where necessary? Thank you so much!
The idea behind your developments is correct, but your proof needs to be systematized to general cases. The easiest formulation turns out to be a proof by contradiction.
Let $U,V$ be $n$- and $m$-dimensional vector spaces respectively, whose bases are denoted by $\{u_1,\ldots,u_n\} \subset U$ and $\{v_1,\ldots,v_m\} \subset V$. Without loss of generality, we can suppose that $n > m$. Now, let's assume that $U \cong V$. In consequence, there exists an isomorphism $\phi : U \rightarrow V$. Without loss of generality again, we can choose $\phi(u_1) = v_1$, $\phi(u_2) = v_2$, etc., until $\phi(u_m) = v_m$. Then, $\phi(u_{m+1}) := \lambda_1v_1 + \ldots + \lambda_mv_m$. Yet, one has $\phi(\lambda_1v_1 + \ldots + \lambda_mv_m) = \lambda_1v_1 + \ldots + \lambda_mv_m$ by linearity, meaning that $\phi$ is not injective and a fortiori not isomorphic, hence finally $U \not\cong V$ by contradiction.