With $V$ and $W$ being vector spaces, and $T: V \rightarrow W$ being a linear transformation: c) Suppose $B: (v_1, v_2, \cdots, v_n)$ is a basis for $V$ and $T$ is one-to-one and onto. Prove that $T(B)$ = $\left\{T(v_1), T(v_2), \cdots, T(v_n)\right\}$ is a basis for $W$.

In the 2 previous sub-questions, it has been proved that $T$ is one-to-one if and only if it carries linearly independent subsets of $V$ into linearly independent subsets of $W$.

So I started by stating that because of the previous sub-question, since $B$ is linearly independent, then $T(B)$ is L.I.

Then I used $\mathrm{nullity}(T) + \mathrm{rank}(T) = \dim(V)$

I said that since $T$ is one-to-one, then $\mathrm{nullity}(T) = 0$, and since we know $\dim(V) = n$, then $\mathrm{rank}(T)$ is also $n$.

Then, I finished by saying that since $T(B)$ is L.I. with $n$ elements and that $\mathrm{rank}(T) = n$, $T(B)$ spans $W$.

Is this correct and sufficient?

  • 1
    $\begingroup$ Use the fact that $T$ is onto to conclude that $\dim(W)=n$ and you should be good. $\endgroup$ Feb 17, 2014 at 19:44


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