🍈 Zettelkasten

❯

❯

Apr 19, 20252 min read

If $V, W$ are finite dimensional vector spaces
With $d im (V) = n$
If $V = s p an {u_{1}, \dots, u_{k}, v_{k + 1}, \dots, v_{n}}$ s.t:
- $k er (T) = s p an {u_{1}, \dots, u_{k}}$
Then: $I ma g e (T) = s p an {T (v_{k + 1}), \dots, T (v_{n})}$

Proof

Pick a basis for $k er (T) = {u_{1}, \dots, u_{k}}$
We know that $V$ is finitely dimensional, and that $d im (V) = n$
Thus, we can extend to a basis $V = s p an {u_{1}, \dots, u_{k}, v_{k + 1}, \dots, v_{n}}$
We want to verify tat $I ma g e (T) = s p an {T (v_{k + 1}), \dots, T (v_{n})}$
1. Proving $s p an {T (v_{k + 1}), \dots, T (v_{n})} \subset I ma g e (T)$
2. $s p an {T (v_{k + 1}), \dots, T (v_{n})} \in I ma g e (T)$ is straight forward
3. Pick $w \in s p an {T (v_{k + 1}), \dots, T (v_{n})}$ , then we have:
4. $w = a_{k + 1} T (v_{k + 1}) + \dots + a_{n} T (v_{n})$ by definition of $s p an$
5. $w = T (a_{k + 1} v_{k + 1} + \dots + a_{n} v_{n})$ by Linearity
6. Thus, $w \in I ma g e (T)$
7. Proving $I ma g e (T) \subset s p an {v_{k + 1}, \dots, v_{n}}$
8. Pick $w \in I ma g e (T)$
9. Then, we have $w = T (v)$ for some $v \in V$
10. Thus, $w = T (a_{1} u_{1} + \dots + a_{k} u_{k} + v_{k + 1} u_{k + 1} + \dots + a_{n} v_{n})$
11. $= a_{1} T (u_{1}) + \dots + a_{k} T (u_{k}) + a_{k + 1} T (v_{k + 1}) + \dots + a_{n} T (v_{n})$ by Linearity
12. As $u_{1}, \dots u_{k} \in k er (T) ⟹ T (u_{1}), \dots, T (u_{k}) = 0$
13. $= 0 + \dots + 0 + a_{k + 1} T (v_{k + 1}) + \dots + T (v_{n})$
14. $\in s p an {T (v_{k} + 1), \dots, T (v_{n})}$