Remark 1 on Kernel and Image

Beweis

$\text{Im}(f)$ is a subspace:

It holds that $0 = f(0) \in \text{Im}(f)$.

Let $y_1 = f(x_1)$ and $y_2 = f(x_2)$ be in $\text{Im}(f)$, where $x_1, x_2 \in V$. Then

\[

y_1 + y_2 = f(x_1) + f(x_2) = f(x_1 + x_2) \in \text{Im}(f).

\]

Let $\lambda \in K$ and $y = f(x) \in \text{Im}(f)$. Then

\[

\lambda y = \lambda f(x) = f(\lambda x) \in \text{Im}(f).

\]

We have shown that $\text{Im}(f)$ contains $0$ and is closed under addition and scalar multiplication. Therefore, $\text{Im}(f)$ is a subvector space. The assertion for $\text{Ker}(f)$ is shown in the exercise. $\square$