🍈 Zettelkasten

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Apr 19, 20251 min read

Theorem

For any linear map $T : V \to W$ , the $I ma g e (T)$ is a subspace of $W$ .

Apply Subspace Test
Notice $I ma g e (T)$ is non-empty. This is because $T (0 v) = 0 w \in I ma g e (T)$
Then notice that any two vectors $w_{1}, w_{2} \in I ma g e (T)$ and for any scalar $c \in F$ . $w_{1} = T (v_{1})$ , $w_{2} = T (v_{2})$
Thus, we have: $c (w_{1}) + w_{2} = c T (v_{1}) + T (v_{2})$
$= T (c v_{1} + v_{2})$ by Linearity
Note that this is also in the image. $T (c v_{1} + v_{2}) \in I ma g e (T)$
Therefore, $I ma g e (T)$ is a Subspace of $W$