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Let $E$ and $F$ be two finite dimensional vector spaces and $f: E \to F$ a linear transformation. If $v_1, ... , v_r \in E$ are linearly independent, prove that $\operatorname*{Im} f = \left \langle f(v_1), ... , f(v_r) \right \rangle \Leftrightarrow E = \operatorname*{Ker} f + \left \langle v_1,... ,v_r \right \rangle$.

I have tried both using Grassman's formula and the rank-nullity theorem with the corresponding inequalities due to the fact that $v_1, ... , v_r \in E$ are linearly independent, but I don't get a consistent proof for any of the implications.

NOTE: If $u_1, ... , u_n$ are vectors, $\left \langle u_1,...,u_n \right \rangle$ denotes the LINEAR SPAN of the vectors.

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I have tried both using Grassman's formula and the relationship between the dimension of the kernel, image, and domain of a linear map (dimE=dimImf+dimKerf) with the corresponding inequalities due to the fact that v1,...,vr∈E are linearly independent, but I don't get a consistent proof for any of the implications.

None of those are going to be sufficient, because they're all results about dimensions, and we don't want a result about dimensions. There's a possibility that you could use it, combined with other results, but it's not actually needed here, and probably shouldn't be the first approach you think of.


Instead, first note that, if $E = \mathrm{Ker}(f) + \langle v_1,\ldots,v_r\rangle$, then we can choose some $w_1,\ldots,w_{n-r}\in\mathrm{Ker}(f)$ to extend $\langle v_1,\ldots,v_r\rangle$ to a basis of $E$. Then for any $u = \sum \alpha_iv_i + \sum\beta_iw_i$, we have $f(u) = \sum\alpha_if(v_i)+\sum\beta_if(w_i) = \sum\alpha_if(v_i)$ since $w_i\in\mathrm{Ker}(f)$, and note that this lies in $\langle f(v_1),\ldots,f(v_r)\rangle$, so $\mathrm{Im}(f)\subseteq\langle f(v_1),\ldots,f(v_r)\rangle$. The reverse inclusion is simple.

For the reverse implication, suppose that $E \neq \mathrm{Ker}(f) + \langle v_1,\ldots,v_r\rangle$. Then there is some $u \in E$ that does not lie in $\mathrm{Ker}(f) + \langle v_1,\ldots,v_r\rangle$. Now, if $f(u)$ lies in $\langle f(v_1),\ldots,f(v_r)\rangle$, then there are some $\alpha_1,\ldots,\alpha_r$ such that $f(u) = \sum\alpha_if(v_i) = f(\sum\alpha_iv_i)$. Thus, $w := u - \sum\alpha_iv_i$ has $f(w) = f(u) - f(\sum\alpha_iv_i) = 0$, so $w\in\mathrm{Ker}(f)$. But then, $u = w + \sum\alpha_iv_i$, so $u\in\mathrm{Ker}(f)+\langle v_1,\ldots,v_r\rangle$, a contradiction. Thus, $f(u)\not\in\langle f(v_1),\ldots,f(v_r)\rangle$, so $\mathrm{Im}(f)\neq \langle f(v_1),\ldots,f(v_r)\rangle$.

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Let $U=\langle v_1,\dots,v_r\rangle$ and $K=\ker f$.

Grassmann’s formula tells you that $\dim U+\dim K=\dim(U+K)+\dim(U\cap K)$, so $$ \dim(U+K)=\dim U+\dim K-\dim(U\cap K)=r+n-k $$ where $n=\dim K$ and $k=\dim(K\cap N)$; let $s$ be the rank of $f$, so $\dim E=s+n$ by the rank-nullity theorem.

We have $E=U+K$ if and only if $r+n-k=s+n$, that is, if and only if $r=k+s$.

We can also consider $f'\colon U\to F$ (the restriction of $f$). The nullity of $f'$ is $k$; if $s'$ is the rank of $f'$, then $r=k+s'$. Clearly the image of $f'$ is a subspace of the image of $f$ and is spanned by $f(v_1),\dots,f(v_r)$.

Therefore $r=k+s$ if and only if $s=s'$ that is, if and only if $\operatorname{im}f=\operatorname{im}f'=\langle f(v_1),\dots,f(v_r)\rangle$.

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