Proving a statement about the rank of transformations

[Terrell]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

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?

 

[fresh_42]

Say we have $U_n \stackrel{\sigma}{\longrightarrow} V_m \stackrel{\tau}{\longrightarrow} W_k$.

Then - I think - we already have the general formulas $n=\rho(\sigma) + \nu(\sigma)\, , \,n=\rho(\tau \sigma) + \nu(\tau \sigma)$ and $m=\rho(\tau) + \nu(\tau)$. Now all you need is some substitution and $\nu(\tau \sigma) \leq \nu(\sigma)+\nu(\tau)$. The second inequality is more or less obvious as you cannot gain linear independent vectors by a transformation. The images can only shrink.

 

[Terrell]

Using the general formulas... please check the proof that follows:
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 $K(\tau\sigma)\subset K(\tau)\Rightarrow v(\tau\sigma)\leq v(\tau)\Rightarrow -v(\tau\sigma)\geq -v(\tau)$. 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.
$\Bbb{Q.E.D.}$

 

[Terrell]

Now all you need is some substitution and ν(τσ)≤ν(σ)+ν(τ)

I don't know how this inequality gets into the picture, but it's part of the next proof i want to write so.. proving the inequality i have:
$K(\tau\sigma)=K(\tau\vert_{\sigma(U_{n})})\Rightarrow K(\tau\sigma)\subset K(\tau)\Rightarrow v(\tau\sigma)\leq v(\tau)\Rightarrow v(\tau\sigma)\leq v(\tau)+v(\sigma)$

 

[fresh_42]

Using the general formulas... please check the proof that follows:
Note that $\rho(\sigma)+\rho(\tau)-m=\rho(\sigma)-(m-\rho(\tau))=\rho(\sigma)-v(\tau)$ and $\rho(\tau\sigma)+v(\tau\sigma)=n\Rightarrow \rho(\tau\sigma)=n=v(\tau\sigma)$.

Typo.

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 $K(\tau\sigma)\subset K(\tau)\Rightarrow v(\tau\sigma)\leq v(\tau)\Rightarrow -v(\tau\sigma)\geq -v(\tau)$.

Assuming $K$ shall denote the kernel, we have $K(\tau \sigma) \subseteq U_n$ and $K(\tau) \subseteq V_m\,,$ so how can $K(\tau\sigma)\subset K(\tau)$ be?
Furthermore, let's assume $n>m>0\, , \,V_m=W_k\, , \,\tau = \operatorname{id}|_{V_m}$ and $\sigma = 0$. Then $\nu(\tau \sigma) = n > m > 0 = \nu(\tau)$.

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.

 

[fresh_42]

I don't know how this inequality gets into the picture, but it's part of the next proof i want to write so.. proving the inequality i have:
$K(\tau\sigma)=K(\tau\vert_{\sigma(U_{n})})\Rightarrow K(\tau\sigma)\subset K(\tau)\Rightarrow v(\tau\sigma)\leq v(\tau)\Rightarrow v(\tau\sigma)\leq v(\tau)+v(\sigma)$

See my previous correction.

 

[Terrell]

how can K(τσ)⊂K(τ)K(τσ)⊂K(τ)K(\tau\sigma)\subset K(\tau) be?

I think I ran into trouble here by assuming that $K(\tau\sigma)=K(\tau\vert_{\sigma(U_n)})$, where $K(\tau\vert_{\sigma(U_n)})$ means Kernel of $\tau$ strictly on $\sigma(U_n)$. It's an incorrect assumption? Because certainly, $K(\tau\vert_{\sigma(U_n)})\subset K(\tau)$.

 

[Terrell]

Then - I think - we already have the general formulas n=ρ(σ)+ν(σ),n=ρ(τσ)+ν(τσ)n=ρ(σ)+ν(σ),n=ρ(τσ)+ν(τσ)n=\rho(\sigma) + \nu(\sigma)\, , \,n=\rho(\tau \sigma) + \nu(\tau \sigma) and m=ρ(τ)+ν(τ)m=ρ(τ)+ν(τ)m=\rho(\tau) + \nu(\tau). Now all you need is some substitution and ν(τσ)≤ν(σ)+ν(τ)ν(τσ)≤ν(σ)+ν(τ)\nu(\tau \sigma) \leq \nu(\sigma)+\nu(\tau).

I think this is the way to go. Now to show $v(\tau\sigma)\leq v(\tau)+v(\sigma)$, I started by defining $K(\tau\sigma)=\{\zeta\in U_n\mid \sigma(\zeta)\in K(\tau)\lor \sigma(\zeta)=0_V\}$.

 

[fresh_42]

I think I ran into trouble here by assuming that $K(\tau\sigma)=K(\tau\vert_{\sigma(U_n)})$, where $K(\tau\vert_{\sigma(U_n)})$ means Kernel of $\tau$ strictly on $\sigma(U_n)$. It's an incorrect assumption? Because certainly, $K(\tau\vert_{\sigma(U_n)})\subset K(\tau)$.

Yes, but not the first assumption on $K(\tau \sigma)$. That's why I wrote $\nu(\tau \sigma) \leq \nu(\tau) + \nu(\sigma)$, if a vector is sent to zero by a composite function, then either by the first or by the second, so the sum is the secure side. But your argumentation is fine for the second inequality in your initial post, and with images instead of kernels. $\operatorname{im}(\tau \sigma) \subseteq \operatorname{im}(\tau)$ gives one inequality, and that the $\tau$ cannot increase the rank of $\sigma$ gives the other inequality which is needed for the minimum.

Do you think my current approach is futile?

No, just a bit more cautious. I think you've been on the right track. Just substitute all three ranks by the formula $\rho(\alpha)+\nu(\alpha)=\dim X$ for a $\alpha:X \rightarrow \,\ldots$ and then use what I said above and in post #2: $\nu(\tau \sigma)\leq \nu(\tau) + \nu(\sigma)$.

 

[Terrell]

Yes, but not the first assumption on K(τσ)K(τσ)K(\tau \sigma). That's why I wrote ν(τσ)≤ν(τ)+ν(σ)

So $K(\tau\vert_{\sigma(U_n)})=K(\tau\sigma)$?

 

[fresh_42]

So $K(\tau\vert_{\sigma(U_n)})=K(\tau\sigma)$?

Is this correct with $\sigma =0$ and $\tau = \operatorname{id}\,$?

 

[Terrell]

I found the perfect theorem from the book I'm working on that fits the proof, perfectly!
By that theorem, $\rho(\sigma)=\rho(\tau\sigma)+dim[\sigma(U)\cap K(\tau)]$ \begin{align}\Rightarrow dim[\sigma(U)\cap K(\tau)]&=\rho(\sigma)-\rho(\tau\sigma)\\&=(n-v(\sigma))-(n-v(\sigma))\\&=v(\tau\sigma)-v(\sigma)\\&=dim[K(\tau\vert_{\sigma(U_n)})]\\&=v(\tau\vert_{\sigma(U_n)})\\ \Rightarrow v(\tau\sigma)=v(\tau\vert_{\sigma(U_n)}+v(\sigma)\leq v(\tau)+v(\sigma).\end{align} $\Bbb{Q.E.D.}$