TITLE: prove mapping is basis of $Y$ iff vectors being mapped are a basis of $X$
QUESTION [2 upvotes]: Let $X$ and $Y$ be linear spaces of dimension n over a field K. Let $\phi: X \rightarrow Y$ be an isomorphism. Prove that $\{\phi(x_1),\ldots,\phi(x_n)\}$ is a basis of Y if and only if $\{x_1,\ldots,x_n\}$ is a basis of $X$.
So,
Suppose $\{x_1,\ldots,x_n\}$ is not a basis of $X$, then we can write $k_1x_1 + k_2x_2 = 0$ with $k_1, k_2 \neq 0$.
Applying the isomorphism, $\phi(k_1x_1 + k_2x_2) = k_1\phi(x_1) + k_2\phi(x_2) = 0$, then with $k_1, k_2 \neq 0$, we have a contradiction that $\{\phi(x_1),\ldots,\phi(x_n)\}$ is a basis of $Y$.
Thoughts?

REPLY [2 votes]: I'd do this directly, rather than through contradiction, though contradiction works well with Bill Cook's suggestions! Let $\beta = \{x_{1}, ..., x_{n}\}$ be a basis of $X$ and let $\phi : X \to Y$ be a linear isomorphism. Then by homomorphism (preservation of operation), $\phi(0_{X}) = 0_{Y}$. And so:
$$ \phi(0) = \phi(\sum_{i=1}^{n} 0 x_{i}) = 0 \sum_{i=1}^{n} \phi(x_{i}) = 0$$
By $\phi$ being an isomorphism, it is injective. So this is the only mapping to $0_{Y}$, and so linear independence is preserved. Hence, $\phi(\beta)$ is a basis of $Y$.
And since $\phi$ is an isomorophism, it is onto as well and invertible, so the other direction is just working backwards.
Just a second way to look at the problem. :-)
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