Elementary linear algebra and linear maps exercise.

Question

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.

Answer 1

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.

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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 $\mathrm{Im}(f)\neq \langle f(v_1),\ldots,f(v_r)\rangle$.

Answer 2

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 $\operatorname{im}f=\operatorname{im}f'=\langle f(v_1),\dots,f(v_r)\rangle$.