Proof: $V$ and $W$ Vector Spaces, with finite $\dim (n)\ge1$ and $\gamma:V \to W$ an isomorphism

Question

Proof: $V$ and $W$ Vector Spaces, with finite $\dim (n)\ge1$ and $\gamma:V \to W$ an isomorphism, prove that $(\alpha_1,\dots,\alpha_n)$ is base of $V$ if and only if $(\alpha_1,...,\alpha_n)$ is base of $W$.

Can someone help me writing the proof?, on the two ways, i really don't know how can i do it.

Answer 1

In order to prove that if $(\alpha_1, \dots, \alpha_n)$ is a basis of $V$ then $B = (\gamma(\alpha_1),...,\gamma(\alpha_n))$ is a basis of $W$, we have to prove that (sketch):

1. $B$ spans $W$: let $w \in W$. Then, since $\gamma$ is surjective, $w = \gamma(v)$ for some $v \in V$. Now write $v$ as a linear combination of $(\alpha_1, \dots ,\alpha_n)$ and compute $\gamma(v)$. (Use the fact that $\gamma$ is linear!)

2. $B$ is linearly indipendent: if $a_1 \gamma(\alpha_1) + \dots + a_n \gamma(\alpha_n) = 0$, then, since $\gamma$ is linear, $\gamma(a_1 \alpha_1 + \dots + a_n \alpha_n) = 0$. Now try to deduce that $a_i = 0$. (What does the kernel of an isomorphism look like?)

Conversely, to prove that if $(\gamma(\alpha_1),...,\gamma(\alpha_n))$ is a basis of $W$ then $(\alpha_1, \dots, \alpha_n)$ is a basis of $V$, switch the roles of $V$ and $W$ in the above proof and remember that $\gamma^{-1}$ is an isomorphism.

Answer 2

If gamma is an isomorphism then only for the zero vector do you have 0v -> 0w.

If alpha is a basis in V then any linear combination is non-zero so the mapping of this combination is also non-zero.

Then use the properties of isomorphism to claim that any linear combination of the mapped basis is non-zero and hence that the mapped basis is linearly independant and being of n vectors is a basis in W.

Answer 3

If B = {a(1), a(2),...,a(n)} is a basis of V, then we can show B' = { b(1), b(2),..., b(n)} is a basis of W with b(i) = f(a(i)) and f: V --> W is an isomorphism. Observe that:for scalars r(i)'s: r(1)*b(1) +..+ r(n)*b(n) = 0 <==> r(1)*f(a(1)) +...+ r(n)*f(a(n)) = 0 <==> f(r(1)*a(1) +..+ r(n)*a(n)) = 0 <==> r(1)*a(1) + ...+ r(n)*a(n) = 0 since f is one-to-one and f(0) = 0. This means the a(i)'s are linearly independent iff the b(i)'s are linearly independent. Next if B spans V .Let v' in W ==> v' = f(v) for some v in V ==> v' = f(r(1)*a(1) +...+ r(n)*a(n)) = r(1)*f(a(1)) +..+ r(n)*f(a(n)) = r(1)*b(1) +...+ r(n)*b(n) ==> B' spans W. The other way is done similarly. So B is a basis of V <==> B' is a basis of W.