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Question from Serge Lang's Introduction to Linear Algebra:

Let $F: V \to W$ be a linear map, whose kernel is $\{0\}$. Assume that $V$ and $W$ have both the same dimension $n$. Show that the image of $F$ is all of $W$.

Thoughts & attempt:

  • We can express any element $w \in W$ as $\sum_{i}^n c_iw_i$, where $w_1,...,w_n$ are linearly independent.

I'm trying to reach a point where I can coherently say $\sum_i^nF(c_iv_i) = \sum_{i}^n c_iw_i$ by assuming that $v_1,..., v_n$ form a basis of $V$.

Is it enough to say that since the ker = $\{0\}$, linear independence is preserved, and therefore $W$ also has $n$ dimension?

EDIT: I meant to consider $c_i$ on each side of the equation as not necessarily equivalent scalars.

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    Note: your "point" cannot be reached, since it implies for instance $F(v_1) = w_1$, and that need not be true of $F$. – 2017-01-02
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    @MeesdeVries I don't see how. See my edit. The scalars need not be equal. $F(v_i) = w_i$ but that says nothing of what they actually are. – 2017-01-02
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    Rank-Nullity theorem – 2017-01-02

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Let $\{v_1,\ldots,v_n\}$ be a basis for $V$. We claim that $\{T(v_1),\ldots,T(v_n)\}$, is a linearly independent set in $W$ and since $\dim W=n$ it is also a basis for $W$. Indeed suppose that $$ c_1T(v_1)+c_1T(v_2)+\dotsb+c_nT(v_n)=0 $$ Use linearity, the fact that the kernel is trivial and the fact that the $v_i$ are linearly independent to deduce that $c_i=0$ for all $i$. It follows that $W=\text{im} T$