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Let $V, W$ vector spaces and $f, g:V\rightarrow W$, linear transformations.
$\ker f \subset \ker g$.

Now, let $\{v_1,...,v_n\}$ a basis for $\ker f$ and we'll complete it with $\{u_1,...,u_m\}$ to a basis of $\ker g$. Also, with the addition of $\{y_1,...,y_k\}$ we'll have a basis for $V$.

I want to show that: $\{f(u_1),...f(u_m),f(y_1),...f(y_k)\}$ is a basis for $Im f$

Easy to show that this set spans the image so I'll skip this step.

Now, it's left to show that this set is linearly independent.

$$\sum\limits_{i=1}^{m}{\alpha_i f(u_i)} + \sum\limits_{i=1}^{k}{\beta_i f(y_i)} = 0$$

From linearity of $f$:

$$f(\sum\limits_{i=1}^{m}{\alpha_i u_i + } \sum\limits_{i=1}^{k}{\beta_i y_i}) = 0$$

Hence, $$\sum\limits_{i=1}^{m}{\alpha_i u_i + } \sum\limits_{i=1}^{k}{\beta_i y_i} \in \ker f$$

Now, How to show $\alpha_i, \beta_i$ are all $0$s?

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Express $\sum\limits_{i=1}^{m}{\alpha_i u_i + } \sum\limits_{i=1}^{k}{\beta_i y_i}$ as a linear combination of $\{v_1,...,v_n\}$. This can be done as $\sum\limits_{i=1}^{m}{\alpha_i u_i + } \sum\limits_{i=1}^{k}{\beta_i y_i} \in \ker f$. And use the linear independence of the basis $\lbrace v_1,v_2,\ldots, v_n, u_1,u_2,\ldots,u_m,y_1,y_2,\ldots,y_k\rbrace$ of V.

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  • $\begingroup$ Oh lovely. Thank you! $\endgroup$ – Elimination Jun 26 '14 at 9:57
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As $\sum_i \alpha_i u_i + \sum_j \beta_j y_j \in \ker f$, there are $\gamma_k \in K$ ($K$ denotes the base field) such that $$ \sum_i \alpha_i u_i + \sum_j \beta_j y_j = \sum_l \gamma_l v_l\iff \sum_i \alpha_i u_i + \sum_j \beta_j y_j + \sum_l (-\gamma_l) v_l = 0$$ As $\{u_1, \ldots, u_m, v_1, \ldots, v_n, y_1,\ldots, y_k\}$ is a basis for $V$, this is linear independent set, hence all $\alpha_i$, $\beta_j$ and $\gamma_l$ are zero.

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