\[\mathbf{A} = \left[\begin{array} {rrr} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{array}\right] \]
The rank is 4. The work is shown below.
Let’s find number of non-zero rows by creating upper matrix of A. The number of non-zero rows will be the rank of matrix A.
\[\mathbf{A} = \left[\begin{array} {rrr} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{array}\right] \]
\(R_1\) + \(R_2\) and -5x\(R_1\) + \(R_4\) ==> \[\mathbf{A} = \left[\begin{array} {rrr} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 1 & -2 & 1 \\ 0 & 6 & 17 & 23 \end{array}\right] \]
\(R_2\) - 2X\(R_3\) and 3x\(R_2\) - \(R_4\) ==> \[\mathbf{A} = \left[\begin{array} {rrr} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 0 & 6 & 5 \\ 0 & 0 & -5 & -2 \end{array}\right] \]
5*\(R_3\) + 6x\(R_4\) ==> \[\mathbf{A} = \left[\begin{array} {rrr} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 0 & 6 & 5 \\ 0 & 0 & 0 & 13 \end{array}\right] \]
mxn matrix where m>n, what can be maximum rank? The mininum rank, assuming that the matrix is non-zero?The maximum rank is always min(m,n). In this case it will be n. A non-zero matrix contains at least one element that is not zero. Therefore, there will be at least 1 row that will have non-zero value and thus a minimum rank of 1.
\[\mathbf{B} = \left[\begin{array} {rrr} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2 \end{array}\right] \]
The rank is 1. The work is shown below.
\[\mathbf{B} = \left[\begin{array} {rrr} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2 \end{array}\right] \rightarrow 3R_1 - R_2, 2R_1 - R_3 \rightarrow \left[\begin{array} {rrr} 1 & 2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \]
\[\mathbf{A} = \left[\begin{array} {rrr} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{array}\right] \]
Let’s solve,
\[\mathbf {det}(\left[\begin{array} {rrr} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{array}\right] -\left[\begin{array} {rrr} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{array}\right])=0 \]
\[\mathbf {det}(\left[\begin{array} {rrr} 1-\lambda & 2 & 3 \\ 0 & 4-\lambda & 5 \\ 0 & 0 & 6-\lambda \end{array}\right]=0 \]
\[ (1-\lambda) \begin{vmatrix} 4-\lambda & -5 \\ 0 & 6-\lambda \end{vmatrix} + 2\begin{vmatrix} 0 & 5 \\ 0 & 6-\lambda \end{vmatrix} + 3\begin{vmatrix} 0 & 4-\lambda \\ 0 & 0 \end{vmatrix} \]
\[(1-\lambda)((4-\lambda)(6-\lambda))=0\]
\[(1-\lambda)(\lambda^2-10\lambda+24)=0\]
Charactristic polynomial: \(\lambda^3 - 11\lambda^2 + 34\lambda -24 = 0\)
Eignevalues: \(\lambda_1=1, \lambda_2=4\) and \(\lambda_3=6\)
solving for eigenvectors:
\(\lambda_1=1\)
\[\begin{bmatrix} 0 & 2 & 3 \\ 0 & 3 & 5 \\ 0 & 0 & 5 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = 0 \]
\(5v_3 = 0 \; \rightarrow \; v_3=0\)
\(v_2 = \frac{-5}{3}v_3 \; \rightarrow \; v_2=0\)
\(v_1\) can take value
For eigenvalue: \(\lambda=1\), eigenvector is \(\begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}\)
\(\lambda_1=4\)
\[\begin{bmatrix} -3 & 2 & 3 \\ 0 & 0 & 5 \\ 0 & 0 & 2 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = 0 \]
\(2v_3 = 0 \; or \; 5v_3=0 \; \rightarrow \; v_3=0\)
\(v_2 = \frac{3}{2}v_1 \; \rightarrow \; v_2=\frac{3}{2} \;and \; v_1=1\)
For eigenvalue: \(\lambda=4\), eigenvector is \(\begin{bmatrix}1 \\ \frac{3}{2} \\ 0 \end{bmatrix}\)
\(\lambda_1=6\)
\[\begin{bmatrix} -5 & 2 & 3 \\ 0 & -2 & 5 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = 0 \]
\(2v_2 = 5v_3 \; \rightarrow \; v_2=\frac{5}{2}v_3\)
\(2v_1 = 2v_2 + 3v_3 \; \rightarrow \; v_1=\frac{2}{5}*\frac{5}{2}v_3 + \frac{3}{2} \; \rightarrow v_1=\frac{8}{5}\)
For eigenvalue: \(\lambda=6\), eigenvector is \(\begin{bmatrix}\frac{8}{5} \\ \frac{5}{2} \\ 1 \end{bmatrix}\)
Eigenvectors for matrix A are \(\begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix}1 \\ \frac{3}{2} \\ 0 \end{bmatrix} \; and \; \begin{bmatrix}\frac{8}{5} \\ \frac{5}{2} \\ 1 \end{bmatrix}\)