# Rank of composition of linear maps

• MHB
• mathmari
In summary, we have discussed a proof by induction that shows if $V$ is finite, then the rank of the composition of linear maps $\phi_1, \dots, \phi_n$ is less than or equal to the minimum rank of each individual map. We have also determined that the rank of a function is the number of independent vectors in its image, and that the number of independent vectors in the image is at most the number of independent vectors in the domain. Therefore, the rank of the composition of linear maps is less than or equal to the minimum rank of each individual map.
mathmari
Gold Member
MHB
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:

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mathmari said:
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 $\text{Im}(f\circ g)\not\subseteq \text{Im}(g)$.

Klaas van Aarsen said:
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:

mathmari said:
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.

Klaas van Aarsen said:
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:

mathmari said:
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 $\operatorname{Rank}(f\circ g)\le\operatorname{Rank}(g)$.

Klaas van Aarsen said:
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:

mathmari said:
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)

## 1. What is the rank of composition of linear maps?

The rank of composition of linear maps refers to the number of linearly independent columns or rows in the resulting matrix after composing two or more linear maps. In other words, it is the dimension of the vector space spanned by the columns or rows of the resulting matrix.

## 2. How is the rank of composition of linear maps calculated?

The rank of composition of linear maps can be calculated by multiplying the ranks of each individual linear map. For example, if the rank of linear map A is 3 and the rank of linear map B is 4, then the rank of the composition of A and B would be 3 x 4 = 12.

## 3. What is the significance of the rank of composition of linear maps?

The rank of composition of linear maps is significant because it determines the dimension of the vector space that the linear maps can span. It also provides information about the linear independence of the maps and can be used to solve systems of linear equations.

## 4. Can the rank of composition of linear maps be greater than the rank of the individual maps?

Yes, it is possible for the rank of composition of linear maps to be greater than the rank of the individual maps. This can happen when the composition of the maps results in a larger vector space that can be spanned by the individual maps alone.

## 5. How does the rank of composition of linear maps relate to the nullity of the maps?

The rank of composition of linear maps and the nullity of the maps are related through the rank-nullity theorem, which states that the sum of the rank and nullity of a linear map is equal to the dimension of the domain of the map. In other words, if the rank of the composition of linear maps is known, the nullity of the maps can be calculated by subtracting the rank from the dimension of the domain.

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