Is the Rank of AB Related to the Ranks of A and B?

  • POTW
  • Thread starter Euge
  • Start date
  • Tags
    Inequality
In summary, "An Inequality of Sylvester" is a mathematical concept named after James Joseph Sylvester and states that the sum of the squares of two real numbers is always greater than or equal to twice their product. Its significance lies in its various applications in fields such as geometry, optimization, and statistics, and it can be derived using algebraic manipulations and the Pythagorean theorem. This inequality can also be extended to multiple variables, known as the Sylvester's inequality or the Sylvester's criterion. In real-life situations, it can be used to find the maximum area of a rectangle with a fixed perimeter, maximize the volume of a box with a fixed surface area, or determine the shortest distance between two points in a coordinate plane
  • #1
Euge
Gold Member
MHB
POTW Director
2,073
244
Let ##F## be a field. If ##A \in M_{n\times k}(F)## and ##B\in M_{k\times n}(F)##, show that $$\operatorname{rank}(A) + \operatorname{rank}(B) - k \le \operatorname{rank}(AB) \le \min\{\operatorname{rank}(A), \operatorname{rank}(B)\}$$
 
  • Like
Likes topsquark
Physics news on Phys.org
  • #2
First some preliminaries. The column space of a matrix ##M##, denoted ##\mathcal{C} (M)##, is the span of the columns of ##M## and the row space of a matrix ##M##, denoted ##\mathcal{R} (M)##, is the span of the rows of ##M##. The column rank of a matrix ##M## is the dimension of the column space of ##M##, while the row rank of ##M## is the dimension of the row space of ##M##. It can be shown that the column rank is equal to the row rank, which is denoted ##\text{rank} (M)##. Note ##\mathcal{C} (M)## is the ##\text{range} (M)##.

For ##M \in M_{k \times n} (F)## we have by the rank-nullity theorem:

\begin{align*}
\dim \ker (M) + \dim \text{range} (M) = n .
\end{align*}

But ##\dim \text{range} (M) = \dim \mathcal{C} (M) = \text{rank} (M)##. Therefore,

\begin{align*}
\dim \ker (M) + \text{rank} (M) = n \qquad (*)
\end{align*}

We have the inequality

\begin{align*}
\dim \ker (AB) \leq \dim \ker (A) + \dim \ker (B) \qquad (**)
\end{align*}

as

\begin{align*}
\dim \ker (AB) =\dim [\ker (A) \cap \text{range} (B)] + \dim \ker (B) .
\end{align*}

Using ##(*)## in ##(**)##,

\begin{align*}
[n - \text{rank} (AB)] \leq [k - \text{rank} (A)] + [n - \text{rank} (B)]
\end{align*}

or

\begin{align*}
\text{rank} (A) + \text{rank} (B) - k \leq \text{rank} (AB)
\end{align*}

Next, the columns of ##AB## are linear combinations of columns of ##A## and are hence in its span so ##\text{rank} (AB)## is less than or equal to ##\text{rank} (A)##. The rows of ##AB## are linear combinations of rows of ##B## and are hence in its span so ##\text{rank} (AB)## is less than or equal to ##\text{rank} (B)##. Therefore,

\begin{align*}
\text{rank} (AB) \leq \min \{ \text{rank} (A) , \text{rank} (B) \} .
\end{align*}
 
Last edited:
  • Like
Likes Euge and topsquark
  • #3
Supplementary

We elaborate on the proof of:

\begin{align*}
\dim \ker (AB) \leq \dim \ker (A) + \dim \ker (B) .
\end{align*}First, we have by the rank-nullity theorem:

\begin{align*}
\dim \ker (AB) = n - \dim \text{range} (AB) \qquad (*)
\end{align*}

Now, consider the linear transformation ##A_{| \text{range} (B)} : \text{range} (B) \rightarrow F^n## defined by ##A_{| \text{range} (B)} x = A x## for all ##x \in \text{range} (B) \subseteq F^k##. Note ##\text{range} (A_{| \text{range} (B)}) = \text{range} (AB)##. By the rank-nullity theorem applied to ##A_{| \text{range} (B)}##:

\begin{align*}
\dim \text{range} (AB) = \dim \text{range} (A_{| \text{range} (B)}) = \dim \text{range} (B) - \dim \ker (A_{| \text{range} (B)})
\end{align*}

Note, by the rank-nullity theorem, ##\dim \text{range} (B) = n - \dim \ker (B)##, therefore by the previous result

\begin{align*}
\dim \text{range} (AB) = n - \dim \ker (B) - \dim \ker (A_{| \text{range} (B)}) \quad (**)
\end{align*}

Using ##(**)## in ##(*)## we have

\begin{align*}
\dim \ker (AB) = \dim \ker (A_{| \text{range} (B)}) + \dim \ker (B)
\end{align*}

As ##\ker (A_{| \text{range} (B)}) = \ker (A) \cap \text{range} (B)##,

\begin{align*}
\dim \ker (AB) = \dim [\ker (A) \cap \text{range} (B)] + \dim \ker (B) .
\end{align*}

Finally, as ##\ker (A) \cap \text{range} (B) \subseteq \ker (A)##, we have ##\dim [\ker (A) \cap \text{range} (B)] \leq \dim \ker (A)##, and thus

\begin{align*}
\dim \ker (AB) \leq \dim \ker (A) + \dim \ker (B) .
\end{align*}
 
  • Like
Likes Euge

FAQ: Is the Rank of AB Related to the Ranks of A and B?

What is "An Inequality of Sylvester"?

"An Inequality of Sylvester" is a mathematical theorem that was discovered by James Joseph Sylvester in the 19th century. It is a fundamental result in linear algebra and states that the rank of a matrix plus the nullity of its transpose is always greater than or equal to the number of columns of the matrix.

What is the significance of "An Inequality of Sylvester"?

This inequality has many applications in mathematics and engineering, particularly in the field of linear algebra. It is used to prove the existence of solutions to systems of linear equations and is also important in the study of linear transformations and vector spaces.

How is "An Inequality of Sylvester" proven?

The proof of this inequality involves using the rank-nullity theorem, which states that the rank of a matrix plus its nullity is equal to the number of columns of the matrix. By manipulating this equation, we can arrive at the inequality of Sylvester.

Can "An Inequality of Sylvester" be extended to higher dimensions?

Yes, this inequality can be extended to higher dimensions. In fact, it can be generalized to any rectangular matrix, not just square matrices. The only requirement is that the matrix must have more columns than rows.

Are there any real-world applications of "An Inequality of Sylvester"?

Yes, this inequality has many real-world applications. It is used in fields such as engineering, physics, and economics to solve problems involving systems of linear equations. It is also used in computer science for data compression and error correction algorithms.

Similar threads

Replies
1
Views
1K
Replies
1
Views
1K
Replies
1
Views
2K
Replies
10
Views
1K
Replies
7
Views
2K
Replies
5
Views
2K
Replies
15
Views
2K
Replies
8
Views
3K
Replies
11
Views
2K
Back
Top