What is the rank of matrix A?
\[ A = \left[\begin{array}{cc} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{array}\right] \]
The rank of the matrix is 4 because there are 4 pivots:
\[ R_2 = R_1+R_2 \left[\begin{array}{cc} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{array}\right] \]
\[ R_4 = R_4-5R_1 \left[\begin{array}{cc} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 1 & -2 & 1 \\ 0 & -6 & -17 & -23 \end{array}\right] \]
\[ R_4 = R_4-5R_1 \left[\begin{array}{cc} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 1 & -2 & 1 \\ 0 & -6 & -17 & -23 \end{array}\right] \]
\[ R_4 = R_4+6R_3 \left[\begin{array}{cc} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 1 & -2 & 1 \\ 0 & 0 & -29 & -17 \end{array}\right] \]
\[ R_2 <-> R_3, R_3 = R_3-2R_2 \left[\begin{array}{cc} 1 & 2 & 3 & 4 \\ 0 & 1 & -2 & 1 \\ 0 & 0 & 8 & 5 \\ 0 & 0 & -29 & -17 \end{array}\right] \]
\[ R_3 = 1/8R_3, R_4 = 1/29R_4 \left[\begin{array}{cc} 1 & 2 & 3 & 4 \\ 0 & 1 & -2 & 1 \\ 0 & 0 & 1 & 5/8 \\ 0 & 0 & -1 & -17/29 \end{array}\right] \]
\[ R_4 = R_4+R_3 \left[\begin{array}{cc} 1 & 2 & 3 & 4 \\ 0 & 1 & -2 & 1 \\ 0 & 0 & 1 & 5/8 \\ 0 & 0 & 0 & -24/29 \end{array}\right] \]
Given an mxn matrix where m > n, what can be the maximum rank? The minimum rank, assuming that the matrix is non-zero?
The rank of a matrix is defined as the maxiumum number of linearly independent column/row vectors in a matrix. Therefore, the maximum rank is the minimum of {n,m} (the number of rows or number of columns). The minimum rank is 1 (this would mean that all rows are linearly dependent).
What is the rank of matrix B?
\[ B = \left[\begin{array}{cc} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2 \end{array}\right] \]
The rank of matrix B is 1. This is because all rows are linearly dependent. \[R_2 = 3 * R_1\] \[R_3 = 2 * R_1\]