How Can We Prove the Inequality of Rank Matrices?

[mathmari]

Hey!

Let $\mathbb{K}$ be a fiels and $A\in \mathbb{K}^{p\times q}$ and $B\in \mathbb{K}^{q\times r}$.
I want to show that $\text{Rank}(AB)\leq \text{Rank}(A)$ and $\text{Rank}(AB)\leq \text{Rank}(B)$.

We have that every column of $AB$ is a linear combination of the columns of $A$, or not? (Wondering)

What can we do to show the inequality? (Wondering)

 

[Euge]

It follows from the fact that the column space of $AB$ is contained in the column space of $A$, and the row space of $AB$ is contained in the row space of $B$. I'll prove the latter. Given $y$ in the row space of $AB$, there is a row vector $x$ such that $y=xAB$. The vector $z := xA$ is a row vector such that $y=zB$. Thus, $y$ belongs to the row space of $B$.

 

[mathmari]

Given $y$ in the row space of $AB$, there is a row vector $x$ such that $y=xAB$.

The row space of $AB$ contains the rows of $AB$ that spans the rows, right? (Wondering)
Then $y\in R(AB)$ means that we can write $y$ a linear combination of the vectors of $R(AB)$, right? (Wondering)
We can write this as $y=xAB$ because:
Let $x^T=(x_1, x_2, \ldots , x_{n-1}, x_n)$, then we have that $$xAB=\left (x_1 \cdot (1. \text{ row of }AB))+(x_2 \cdot (2. \text{ row of }AB))+\ldots +(x_{n-1} \cdot ((n-1). \text{ row of }AB))+(x_n \cdot (n. \text{ row of }AB))\right )$$ where some of the $x_i$'s might be $0$.
Is this correct? (Wondering)

 

[Euge]

It looks good, although $x^T$ should just be written as $x$, since $x$ is already a row vector.

 

[mathmari]

It looks good, although $x^T$ should just be written as $x$, since $x$ is already a row vector.

Ah ok!

Let $y\in C(AB)$ then there is a column vector $x$ such that $y=ABx$. The vector $z:Bx$ is a column vector such that $y=Az$. Therefore, $y$ belongs to the column space of $A$.

Is this correct? (Wondering) Having that $C(AB)\subseteq C(A)$ and $R(AB)\subseteq R(B)$, how exactly does it imply that $\text{Rank}(AB)\leq \text{Rank}(A)$ and $\text{Rank}(AB)\leq \text{Rank}(B)$ ? (Wondering)

 

[Euge]

Your work is correct. To finish the argument, use the fact that the row rank of a matrix equals its column rank.

 

[mathmari]

Your work is correct. To finish the argument, use the fact that the row rank of a matrix equals its column rank.

So, we have that $\text{Rank}(AB)=|C(AB)|=|R(AB)|$ and $\text{Rank}(A)=|C(A)|$ and $\text{Rank}(B)=|R(B)|$, right? (Wondering)

Since $C(AB)\subseteq C(A)$, it follows that $|C(AB)|\leq |C(A)|$, or not? (Wondering)

And since $R(AB)\subseteq R(B)$, it follows that $|R(AB)|\leq |R(B)|$.

Therefore, we have that $\text{Rank}(AB)=|C(AB)|\leq |C(A)|=\text{Rank}(A)$ and $\text{Rank}(AB)=|R(AB)|\leq |R(B)|=\text{Rank}(B)$.

Is this correct? (Wondering)

 

[Euge]

Looks great!

 

[mathmari]

Thank you very much! (Smile)