Exercise I.B.1

Question

Let $V,W$ be vector spaces, and let $\phi:V	o W$ be a linear transformation. Show that \begin{enumerate} \item $\ker(\phi)$ is a vector subspace of $V$ \item $\phi(V)$ is a vector subspace of $W$ \end{enumerate}

Toğrul Topal · Accepted Answer

\begin{itemize}
    \item By Proposition $I.1$ to show that the subset $\ker(\phi) \subseteq V$ is a vector subspace of $V$ it suffices to show that $\ker(\phi)$ is closed under linear combinations. Let $v_1,\ldots,v_n \in \ker(\phi)$ and $a_1,\ldots,a_n \in \mathbb{K}$ be arbitrary. Then
      $$\phi(a_1v_1 + \ldots + a_nv_n) = a_1\phi(v_1) + \ldots + a_n\phi(v_n) = 0 + \ldots + 0 = 0$$
      Thus, $a_1v_1 + \ldots + a_nv_n \in \ker(\phi)$ by definition.
    \item Let $w_1,\ldots,w_n \in \phi(V)$ and $a_1,\ldots,a_n \in \mathbb{K}$ be arbitrary. Then there exist $v_1,\ldots,v_n \in \ker(\phi)$ such that $\phi(v_1) = w_1, \ldots, \phi(v_n) = w_n$. We then have
      $$a_1w_1 + \ldots + a_nw_n = a_1\phi(v_1) + \ldots + a_n\phi(v_n) = \phi(a_1v_1 + \ldots + a_nv_n) \in \phi(V)$$
      since $V$ is closed under linear combinations itself.
\end{itemize}

Exercise I.B.1

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Exercise I.B.1

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