Theorem: Two finite-dimensional vector spaces are isomorphic if and only if they have the same dimension.
I can understand how to prove that if they are isomorphic then they have the same dimension. Yet for the other direction I cannot totally understand.
To quote Axler's Linear Algebra Done Right, 2nd edition page 55:
To prove the other direction, suppose $V$ and $W$ are finite-dimensional vector spaces with the same dimension. Let $(v_1,…,v_n)$ be a basis of $V$ and $(w_1,…w_n)$ be a basis of $W$. Let $T$ be the linear map from $V$ to $W$ defined by $$T(a_1v_1+…+a_nv_n)=a_1w_1+…+a_nw_n \ (*)$$ Then $T$ is surjective because $(w_1,…,w_n)$ spans $W$, and $T$ is injective because $(w_1,…,w_n)$ is linearly independent. Because $T$ is injective and surjective, $T$ is invertible.
I cannot understand how on earth can we define $T$ that satisfy (*) above. I don't think Axler has given a proof that this can be defined. Could somebody help me on this please?
Best Answer
You could also say that if $v \in V$, then $T(v)$ is defined to be the vector $a_1 w_1 + \cdots + a_n w_n$, where $(a_1,\ldots,a_n)$ is the unique $n$-tuple of scalars such that \begin{equation} v = a_1 v_1 + \cdots + a_n v_n. \end{equation}
The key point is that there is only one way to write $v$ as a linear combination of the vectors $v_1,\ldots, v_n$. The coefficients in that linear combination are unique. So, given $v$, the output vector $a_1 w_1 + \cdots + a_n w_n$ is perfectly well-defined.