Proof Explanation: $\rho(\tau)=\rho(\tau\sigma)+dim[Im(\sigma)\cap K(\tau)]$

Question

$\tau$ and $\sigma$ are linear transformations. $K(\tau)$ means kernel of $\tau$. $\rho$ is the rank. $dim$ means dimension

The domain and codomain wasn't stated clearly in the book so i think we'll assume $\sigma: U\rightarrow V$ and $\tau:V\rightarrow W$

Proof:

Let $\tau'$ be a new linear transformation such that $\tau':Im(\sigma)\rightarrow W$, and

$\forall \overline{\alpha}\in Im(\sigma), \tau'(\overline{\alpha})=\tau(\overline{\alpha})$... Does this mean that $Im(\tau')=Im(\tau)$?

.

Then $K(\tau')=Im(\sigma)\cap K(\tau)$... Would this be because $Im(\sigma)\cap K(\tau)=K(\tau)?$ and $Im(\tau')=Im(\tau)$?

and

$\rho(\tau')=dim\{\tau[Im(\sigma)]\}=dim(\tau\sigma(U))=\rho(\tau\sigma)$... Is this also because $Im(\tau')=Im(\tau)$?

Then Theorem 1.6 takes the form, $\rho(\tau')+v(\tau')=dim\{Im(\sigma)\}$ or $\rho(\tau)=\rho(\tau\sigma)+dim[Im(\sigma)\cap K(\tau)]$. $\Bbb{QED}$

Answer 1

So it seems as if $\sigma(U)=Im(\sigma)\subseteq V$. Then

$\forall \overline{\alpha}\in Im(\sigma),\tau'(\overline{\alpha})=\tau(\overline{\alpha})$

would mean that $Im(\tau')\subseteq Im(\tau).$ In fact $Im(\tau')=Im(\tau|_{Im(\sigma)})$.

Also $$K(\tau')=\{x: x\in Im(\sigma)\wedge \tau'(x)=0\}=\{x: x\in Im(\sigma)\wedge \tau(x)=0\}=\{x: x\in Im(\sigma)\wedge x\in K(\tau)\}=Im(\sigma)\cap K(\tau)$$ and $$\rho(\tau')=dim(Im(\tau'))=dim(Im(\tau|_{Im(\sigma)}))=dim(\tau\sigma(U))=\rho(\tau\sigma)$$