I've stumbled upon this sentence in my Linear Algebra notes:
An isomorphism is a bijective linear tranformation between two $K$-vector spaces, that preserves structure.
Let's consider an isomorphism $f:V\to W$, with $\mathcal{B}$ a basis for $V$.
Since isomorphisms preserve structure, can I assume that all vectors in the basis $\mathcal{B}$ will be mapped to a basis vector in $W$? In other words: are basis elements mapped to basis elements, under isomorphisms?
Best Answer
Suppose $v_1,\ldots,v_n$ is a basis for $V$. We will show $f(v_1),\ldots,f(v_n)$ is a basis for $W$.
First let's show that these vectors are linearly independent. Suppose that $\sum \lambda_i f(v_i) = 0$. Then indeed $f\left( \sum \lambda_i v_i\right) = 0$ and since $f$ is an isomorphism, it is injective and hence this implies that $\sum \lambda_i v_i = 0$. Since $v_i$ is a basis for $V$, this implies $\lambda_i = 0$ for all $i$ and hence that the $f(v_i)$ are linearly independent.
Next we show they are spanning. Let $w \in W$. Since $f$ is surjective, there exists $v \in V$ such that $f(v) = w$. Write $v = \sum \lambda_i v_i$ for some $\lambda_i \in K$. Then $$w = f\left(\sum \lambda_i v_i\right) = \sum \lambda_i f(v_i)$$ Hence the $f(v_i)$ are also spanning, and since they are a linearly independent spanning set they are a basis for $W$.