How to prove $max\{0, \rho(\sigma)+\rho(\tau)-m\}\leq \rho(\tau\sigma)\leq min\{\rho(\tau), \rho(\sigma)\}$?

Question

Let $\sigma:U\rightarrow V$ and $\tau:V\rightarrow W$ such that $dimU=n$, $dimV=m$. Define $v(\tau)$ to be the nullity of $\tau$, $\sigma$ and $\tau$ are linear transformations, and $\rho$ means rank of the linear transformation.

proof (attempt 1):

By Corollary 1.11, $\rho(\sigma)=\rho(\tau\sigma)+v(\tau)$ $\Leftrightarrow$ $\rho(\sigma)-v(\tau)=\rho(\tau\sigma)$. Note that $v(\tau)\leq m$ \begin{align}\Rightarrow \rho(\sigma)-m &\leq \rho(\tau\sigma)\\ \Rightarrow \rho(\sigma)-(\rho(\tau)+v(\tau)) &\leq \rho(\tau\sigma)\\ \Rightarrow \rho(\sigma)-\rho(\tau)-m &\leq \rho(\tau\sigma)\end{align}

proof (attempt 2):

By Corollary 1.11, $\rho(\sigma)-m\leq \rho(\tau\sigma)$. This implies $\rho(\sigma)-m+\rho(\tau)\leq\rho(\tau\sigma)+\rho(\tau)\Rightarrow \rho(\sigma)+\rho(\tau)-m\leq\rho(\tau\sigma)+\rho(\tau)$.

I am stuck. Is it possible to this relationship intuitively?

Answer 1

Note that $\rho(\sigma)+\rho(\tau)-m=\rho(\sigma)-(m-\rho(\tau))=\rho(\sigma)-v(\tau)$ and $\rho(\tau\sigma)+v(\tau\sigma)=n\Rightarrow \rho(\tau\sigma)=n=v(\tau\sigma)$. By Theorem 1.4., $\rho(\sigma)\leq min\{m,n\}\Rightarrow \rho(\sigma)\leq n\Rightarrow \rho(\sigma)-v(\tau)\leq n-v(\tau)$. Since $(\tau\sigma)\subset K(\tau)\Rightarrow v(\tau\sigma)\leq v(\tau)\Rightarrow -v(\tau\sigma)\geq -v(\tau)$. Then \begin{align}n-v(\tau)\leq n-v(\tau\sigma)&\Rightarrow \rho(\sigma)-v(\tau)\leq n-v(\tau\sigma)\\&\Rightarrow \rho(\sigma)-(m-\rho(\tau))\leq \rho(\tau\sigma)\\&\Rightarrow \rho(\sigma)+\rho(\tau)-m\leq \rho(\tau\sigma)\end{align} and $\rho(\tau\sigma)\leq min\{\rho(\sigma),\rho(\tau)\}$ directly follows from Theorem 1.4.

$\Bbb{Q.E.D.}$