Let $V, W$ be finite-dimensional real vector spaces of equal dimension and $T: V \to W$ be an isomorphism. If $\{v_1, v_2, \ldots, v_n\}$ is a basis for $V$ then $\{T(v_1), T(v_2), \ldots, T(v_n)\}$ is a basis for $W.$


Let $V, W$ be finite-dimensional real vector spaces. If $\dim(V) = \dim(W),$ then $V, W$ are isomorphic

Proof of the Theorem:

Let $\{v_1, v_2, \ldots, v_n\}$ be a basis for $V.$ By the lemma above we have $\{T(v_1), T(v_2), \ldots, T(v_n)\}$ is a basis for $W$.

Let $T: V \to W$ be a transformation as $T(k_1v_1 + k_2v_2 + \ldots + k_nv_n) = k_1T(v_1) + k_2T(v_2) + \ldots + k_nT(v_n)$.

Let $u \in V$ such that $T(u) = 0.$ Suppose $u = k_1v_1 + k_2v_2 + \ldots + k_nv_n.$ Then $$T(u) = T(k_1v_1 + k_2v_2 + \ldots + k_nv_n) = k_1T(v_1) + k_2T(v_2) + \ldots + k_nT(v_n) = 0$$ and so all $k_i = 0$ because $T(v_i)$ are basis vectors which means $u = 0$. This implies $\ker(T) = \{0\}.$

I am confused about the way the given lemma is used in the proof. The lemma says that $T$ must be an isomorphism. But isn't that what we are trying to prove?

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    $\begingroup$ The proof doesn't make any sense. They need to tell you what $T$ is. $\endgroup$ – D_S Jan 15 '17 at 20:37
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    $\begingroup$ What you should do to prove the theorem is to choose a basis $w_1, ... , w_n$ of $W$. Every element of $V$ can be written as $c_1v_1 + \cdots + c_nv_n$ for unique scalars $c_i$. Define $T: V \rightarrow W$ to be the function $$T(c_1v_1 + \cdots + c_nv_n) = c_1 w_1 + \cdots + c_nw_n$$ Then you can show that $T$ is a linear transformation which is an isomorphism. $\endgroup$ – D_S Jan 15 '17 at 20:39
  • $\begingroup$ So if I choose a basis for $W$ I can simply ignore the lemma, correct? $\endgroup$ – user406949 Jan 15 '17 at 20:44
  • $\begingroup$ Yes $\space \space$ $\endgroup$ – D_S Jan 15 '17 at 20:45

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