MHB Rank of composition of linear maps

[mathmari]

Hey! :giggle:

Question 1:
Let $C$ be a $\mathbb{R}$-vector space, $1\leq n\in \mathbb{N}$ and let $\phi_1, \ldots , \phi_n:V\rightarrow V$ be linear maps.
I have shown by induction that $\phi_1\circ \ldots \circ \phi_n$ is then also a linear map.
I want to show now by induction that if $V$ is finite then $\text{Rank}(\phi_1\circ \ldots \circ \phi_n)\leq \min \{\text{Rank}(\phi_i)\mid 1\leq i\leq n\}$.

Base Case : For $n=1$ we have that $\text{Rank}(\phi_1)\leq \min \{\text{Rang}(\phi_1)\}$, so the equality holds.
Inductive Hypothesis : We suppose that it holds for $n=m$, so $\text{Rank}(\phi_1\circ \ldots \circ \phi_m)\leq \min \{\text{Rank}(\phi_i)\mid 1\leq i\leq m\}$. (IV)
Inductive Step : We want to show that it holds for $n=m+1$, i.e. that $\text{Rank}(\phi_1\circ \ldots \circ \phi_m\circ \phi_{m+1})\leq \min \{\text{Rank}(\phi_i)\mid 1\leq i\leq m+1\}$.
Does it hold in general that $\text{Im}(f\circ g)\subseteq \text{Im}(f)$ and $\text{Im}(f\circ g)\subseteq \text{Im}(g)$ and so we get $\text{Rank}(f\circ g)\leq \text{Rank}(f)$ and $\text{Rank}(f\circ g)\leq \text{Rank}(g)$ ?

:unsure:

 

[I like Serena]

Does it hold in general that $\text{Im}(f\circ g)\subseteq \text{Im}(f)$ and $\text{Im}(f\circ g)\subseteq \text{Im}(g)$ and so we get $\text{Rank}(f\circ g)\leq \text{Rank}(f)$ and $\text{Rank}(f\circ g)\leq \text{Rank}(g)$ ?

Hey mathmari!

I'm afraid not. (Shake)

Consider $f:v\mapsto e_1$ and $g:v\mapsto e_2$. Then $\text{Im}(f\circ g)\not\subseteq \text{Im}(g)$.

 

[mathmari]

I'm afraid not. (Shake)

Consider $f:v\mapsto e_1$ and $g:v\mapsto e_2$. Then $\text{Im}(f\circ g)\not\subseteq \text{Im}(g)$.

Ah ok.. But how can we continue then the inductive step? :unsure:

 

[I like Serena]

Ah ok.. But how can we continue then the inductive step?

The rank of a function with a specific domain is the number of independent vectors in the image.
And the number of independent vectors in the image is at most the number of independent vectors in the domain.

 

[mathmari]

The rank of a function with a specific domain is the number of independent vectors in the image.
And the number of independent vectors in the image is at most the number of independent vectors in the domain.

But how do we use that? I got stuck right now. :unsure:

 

[I like Serena]

But how do we use that? I got stuck right now.

Suppose $\operatorname{Rank}(f\circ g)>\operatorname{Rank}(g)$.
Then there must be more independent vectors in $(f\circ g)(V)$ than in $g(V)$.
But there can only be at most as many independent vectors in $f(g(V))$ as there are in $g(V)$.
Contradiction.
Therefore $\operatorname{Rank}(f\circ g)\le\operatorname{Rank}(g)$.

 

[mathmari]

Suppose $\operatorname{Rank}(f\circ g)>\operatorname{Rank}(g)$.
Then there must be more independent vectors in $(f\circ g)(V)$ than in $g(V)$.
But there can only be at most as many independent vectors in $f(g(V))$ as there are in $g(V)$.
Contradiction.
Therefore $\operatorname{Rank}(f\circ g)\le\operatorname{Rank}(g)$.

Ahh ok (Malthe)

And in general it holds that $\operatorname{Rank}(f\circ g)\le\operatorname{Rank}(f)$.

So we have that $\operatorname{Rank}(f\circ g)\le\operatorname{Rank}(f)$ and $\operatorname{Rank}(f\circ g)\le\operatorname{Rank}(g)$ and then we take the minimum? :unsure:

 

[I like Serena]

And in general it holds that $\operatorname{Rank}(f\circ g)\le\operatorname{Rank}(f)$.

So we have that $\operatorname{Rank}(f\circ g)\le\operatorname{Rank}(f)$ and $\operatorname{Rank}(f\circ g)\le\operatorname{Rank}(g)$ and then we take the minimum?

Yep. (Nod)