prove mapping is basis of $Y$ iff vectors being mapped are a basis of $X$ 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?
 A: Your proof is more or less correct if you assume $n=2$. To repair it...
Suppose $\{x_1,\dots,x_n\}$ is not a basis for $X$, so this set either fails to be linearly independent or fails to span. Suppose it is linearly independent but fails to span, then we have a spanning set of size $n$ in an $n$-dimensional vector space. Thus it is a basis and is linearly independent (contradiction). Therefore, if $\{x_1,\dots,x_n\}$ is not a basis for $X$, it must be linearly dependent.
Thus there exists $k_1,\dots,k_n$ not all $0$ such that $k_1x_1+\cdots+k_nx_n={\bf 0}$. This implies that ${\bf 0} = \phi({\bf 0}) = \phi(k_1x_1+\cdots+k_nx_n)=k_1\phi(x_1)+\cdots+k_n\phi(x_n)$ (where we recall that not all of the $k_i$'s are $0$). So $\{\phi(x_1),\dots,\phi(x_n)\}$ is linearly dependent. So it is not a basis.
To get the converse. Suppose $\{\phi(x_1),\dots,\phi(x_n)\}$ is not a basis for $Y$ and then run the same argument using $\phi^{-1}$ (which is an isomorphism since $\phi$ is an isomorphism).
This certainly isn't the only way to prove this statement, but it works well enough. :)
A: 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. :-)
