Rank of a Matrix
By Catalin David
The rank of a matrix with m rows and n columns is a number r with the following properties:
- r is less than or equal to the smallest number out of m and n.
- r is equal to the order of the greatest minor of the matrix which is not 0.
Determining the Rank of a Matrix
- We pick an element of the matrix which is not 0.
- We calculate the order 2 minors which contain that element until we find a minor which is not 0.
- If every order 2 minor is 0, then the rank of the matrix is 1.
- If there is any order 2 minor which is not 0, we calculate the order 3 minors which contain the previous minor until we find one which is not 0.
- If every order 3 minor is 0, then the rank of the matrix is 2.
- If there is any order 3 minor which is not 0, we calculate the order 4 minors until we find one which is not 0.
- We keep doing this until we get minors of an order equal to the smallest number out of the number of rows and the number of columns.
Example 42
$A=\begin{pmatrix}
1 & 2 & 4\\
3 & 6 & 5
\end{pmatrix}$
The matrix has 2 rows and 3 columns, so the greatest possible value of its rank is 2. We pick any element which is not 0.
$\begin{pmatrix} \color{red}{1} & 2 & 4\\ 3 & 6 & 5 \end{pmatrix}$
We form an order 2 minor containing 1.
$\begin{pmatrix}
\color{red}{1} & \color{red}{2} & 4\\
\color{red}{3} & \color{red}{6} & 5
\end{pmatrix}$
We calculate this minor.
$\begin{vmatrix}
\color{red}{1} & \color{red}{2}\\
\color{red}{3} & \color{red}{6}
\end{vmatrix}=6 - 6 = 0$
We form another order 3 minor containing 1. $A=\begin{pmatrix} \color{blue}{1} & 2 & \color{blue}{4}\\ \color{blue}{3} & 6 & \color{blue}{5} \end{pmatrix}$
We calculate this minor.
$\begin{vmatrix}
\color{blue}{1} & \color{blue}{4}\\
\color{blue}{3} & \color{blue}{5}
\end{vmatrix}= 5 - 12 = -7 \neq 0.$
The rank is 2.
Example 43
$B=\begin{pmatrix}
1 & 1 & 1\\
1 & 1 & 1\\
1 & 1 & 1\\
\end{pmatrix}$
We pick an element which is not 0.
$\begin{pmatrix}
1 & 1 & 1\\
1 & 1 & 1\\
1 & \color{red}{1} & 1
\end{pmatrix}$
We calculate order 2 minors containing this element. $\begin{pmatrix} 1 & 1 & 1\\ \color{red}{1} & \color{red}{1} & 1\\ \color{red}{1} & \color{red}{1} & 1 \end{pmatrix}$
$\begin{vmatrix} \color{red}{1} & \color{red}{1} \\ \color{red}{1} & \color{red}{1} \end{vmatrix}=0 $ (because it has two equal rows)
Every other order 2 minor is 0 because it's the same as the others. In this case, the rank of the matrix is 1.
Example 44
$B=\begin{pmatrix}
3 & 8 & 2\\
2 & 1 & 1\\
5 & 3 & 4\\
7 & 4 & 5
\end{pmatrix}$
The matrix has 4 rows and 3 columns, so the greatest possible value of its rank is 3.
We pick an element which is not 0.
$\begin{pmatrix}
3 & 8 & 2\\
2 & 1 & 1\\
5 & 3 & \color{red}{4}\\
7 & 4 & 5
\end{pmatrix}$
We calculate an order 2 minor containing 4.
$ \begin{pmatrix}
3 & 8 & 2\\
2 & \color{red}{1} & \color{red}{1}\\
5 & \color{red}{3} & \color{red}{4}\\
7 & 4 & 5
\end{pmatrix}$
$\begin{vmatrix} \color{red}{1} & \color{red}{1}\\ \color{red}{3} & \color{red}{4}\\ \end{vmatrix} = 4 - 3 = 1$
We form an order 3 minor containing the previous minor. $\begin{pmatrix} 3 & 8 & 2\\ \color{red}{2} & \color{red}{1} & \color{red}{1}\\ \color{red}{5} & \color{red}{3} & \color{red}{4}\\ \color{red}{7} & \color{red}{4} & \color{red}{5} \end{pmatrix}$
We calculate this minor.
$\begin{pmatrix}
\color{red}{2} & \color{red}{1} & \color{red}{1}\\
\color{red}{5} & \color{red}{3} & \color{red}{4}\\
\color{red}{7} & \color{red}{4} & \color{red}{5}
\end{pmatrix}=0 $ because $ R_{1}+R_{2}=R_{3}$
We calculate another order 3 minor containing the previous minor.
$\begin{pmatrix}
\color{blue}{3} & \color{blue}{8} & \color{blue}{2}\\
\color{blue}{2} & \color{blue}{1} & \color{blue}{1}\\
\color{blue}{5} & \color{blue}{3} & \color{blue}{4}\\
7 & 4 & 5
\end{pmatrix}$
$\begin{vmatrix}
\color{blue}{3} & \color{blue}{8} & \color{blue}{2}\\
\color{blue}{2} & \color{blue}{1} & \color{blue}{1}\\
\color{blue}{5} & \color{blue}{3} & \color{blue}{4}
\end{vmatrix} =$ $12 + 12 +40 -10 -9 -64 =-19 \neq 0 $
The rank of the matrix is 3.
Example 45
$D=\begin{pmatrix}
1 & 5 & 1 & 6\\
2 & 3 & 2 & 5\\
6 & 1 & 6 & 7
\end{pmatrix}$
D is a matrix with 3 rows and 4 columns, so the greatest possible value of the rank is 3.
We pick an element which is not 0.
$\begin{pmatrix}
1 & \color{red}{5} & 1 & 6\\
2 & 3 & 2 & 5\\
6 & 1 & 6 & 7
\end{pmatrix}$
We form an order 2 minor containing 5.
$\begin{pmatrix}
\color{red}{1} & \color{red}{5} & 1 & 6\\
\color{red}{2} & \color{red}{3} & 2 & 5\\
6 & 1 & 6 & 7
\end{pmatrix}$
$\begin{vmatrix} 1 & 5\\ 2 & 3 \end{vmatrix}= 3 - 10 = -7 \neq 0$
We form an order 3 minor containing the previous minor.
$\begin{pmatrix}
\color{red}{1} & \color{red}{5} & \color{red}{1} & 6\\
\color{red}{2} & \color{red}{3} & \color{red}{2} & 5\\
\color{red}{6} & \color{red}{1} & \color{red}{6} & 7
\end{pmatrix}$
$\begin{vmatrix} \color{red}{1} & \color{red}{5} & \color{red}{1}\\ \color{red}{2} & \color{red}{3} & \color{red}{2}\\ \color{red}{6} & \color{red}{1} & \color{red}{6} \end{vmatrix} = 0 $ (because it has 2 equal columns)
In this case, we form another order 3 minor containing the order 2 minor which is not 0.
$\begin{pmatrix}
\color{blue}{1} & \color{blue}{5} & 1 & \color{blue}{6}\\
\color{blue}{2} & \color{blue}{3} & 2 & \color{blue}{5}\\
\color{blue}{6} & \color{blue}{1} & 6 & \color{blue}{7}
\end{pmatrix}$
$\begin{pmatrix} \color{blue}{1} & \color{blue}{5} & \color{blue}{6}\\ \color{blue}{2} & \color{blue}{3} & \color{blue}{5}\\ \color{blue}{6} & \color{blue}{1} & \color{blue}{7} \end{pmatrix} = 0 $ because $ C_{1} + C_{2}=C_{3}$
Since all order 3 minors are 0, the rank of matrix D is 2.