\(A = \begin{bmatrix} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{bmatrix}\).
To find the rank, I first need to find the Reduced Row Echelon Form of the matrix.
\(\begin{bmatrix} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{bmatrix} \xrightarrow{5R_2 + R4} \begin{bmatrix} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1 \\ 0 & 9 & 3 & 12 \end{bmatrix} \xrightarrow{ \frac 13 R4} \begin{bmatrix} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1 \\ 0 & 3 & 1 & 4 \end{bmatrix}\)
\(\xrightarrow{-3R_3 + R_4} \begin{bmatrix} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1 \\ 0 & 0 & 7 & 1 \end{bmatrix} \xrightarrow{R_1 + R_2} \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 1 & -2 & 1 \\ 0 & 0 & 7 & 1 \end{bmatrix} \xrightarrow{\frac 17 R_4} \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 1 & -2 & 1 \\ 0 & 0 & 1 & \frac 17 \end{bmatrix}\)
\(\xrightarrow{2R_4 + R_3} \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 1 & 0 & \frac 97 \\ 0 & 0 & 1 & \frac 17 \end{bmatrix} \xrightarrow{-R_2 + R_q} \begin{bmatrix} 1 & 0 & -1 & -3 \\ 0 & 2 & 4 & 7 \\ 0 & 1 & 0 & \frac 97 \\ 0 & 0 & 1 & \frac 17 \end{bmatrix} \xrightarrow{R_4 + R_1} \begin{bmatrix} 1 & 0 & 0 & - \frac {20}7 \\ 0 & 2 & 4 & 7 \\ 0 & 1 & 0 & \frac 97 \\ 0 & 0 & 1 & \frac 17 \end{bmatrix}\)
\(\xrightarrow{\frac 12 R_2} \begin{bmatrix} 1 & 0 & 0 & - \frac {20}7 \\ 0 & 1 & 2 & \frac 72 \\ 0 & 1 & 0 & \frac 97 \\ 0 & 0 & 1 & \frac 17 \end{bmatrix} \xrightarrow{-2R_4 + R_2} \begin{bmatrix} 1 & 0 & 0 & - \frac {20}7 \\ 0 & 1 & 0 & \frac {53}{14} \\ 0 & 1 & 0 & \frac 97 \\ 0 & 0 & 1 & \frac 17 \end{bmatrix} \xrightarrow{-R_2 + R_3} \begin{bmatrix} 1 & 0 & 0 & - \frac {20}7 \\ 0 & 1 & 0 & \frac {53}{14} \\ 0 & 0 & 0 & - \frac {35}{14} \\ 0 & 0 & 1 & \frac 17 \end{bmatrix}\)
\(\xrightarrow{- \frac {2}{5}R_3 \leftrightarrow R_4} \begin{bmatrix} 1 & 0 & 0 & - \frac {20}7 \\ 0 & 1 & 0 & \frac {53}{14} \\ 0 & 0 & 1 & \frac 17 \\ 0 & 0 & 0 & 1 \end{bmatrix}\)
A few more trivial row operations leads to:
\(\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}\)
The rank of this matrix is 4 because there are four nonzero rows when the matrix is in reduced row echelon form.
The maximum rank of a matrix is equivalent to the maximum number of linearly independent rows. In the case where \(m > n\), there can only be as many linearly independent rows as there are columns. The maximum possible rank of an \(m \times n\) matrix is \(n\).
Assuming a non-zero matrix, the minimum rank of an \(m \times n\) matrix, where \(m > n\), is 1. In the most trivial case, the matrix would have one row that contains any non-zero elements, thus making the rank 1. Because the matrix cannot be empty, the rank cannot be 0. So, the minimum rank of a non-zero matrix is 1.
\(B = \begin{bmatrix} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2 \end{bmatrix}\)
\(\begin{bmatrix} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2 \end{bmatrix} \xrightarrow{-2R_1 + R_3} \begin{bmatrix} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 0 & 0 & 0 \end{bmatrix} \xrightarrow{-3R_1 + R_2} \begin{bmatrix} 1 & 2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\)
So, with only one linearly independent row, the rank of matrix B is 1.
\(A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix}\)
Characteristic polynomial: \(pc(x) = det(C - \lambda I_2)\)
\(\begin{bmatrix} 1 - \lambda & 2 & 3 \\ 0 & 4 - \lambda & 5 \\ 0 & 0 & 6 - \lambda \end{bmatrix}\)
\(\begin{vmatrix} 1 - \lambda & 2 & 3 \\ 0 & 4 - \lambda & 5 \\ 0 & 0 & 6 - \lambda \end{vmatrix} = 0\)
\(det(A - \lambda I_2) = (1 - \lambda)(4 - \lambda)(6 - \lambda) + (2)(5)(0) + (3)(0)(0) - (2)(0)(6 - \lambda) - (5)(0)(1) - (3)(4 - \lambda)(0)\)
\(det(A - \lambda I_2) = (1 - \lambda)(4 - \lambda)(6 - \lambda) = 0\)
\(\therefore \lambda_1 = 1, \lambda_2 = 4, \lambda_3 = 6\)
\(A - \lambda_1I = \begin{bmatrix} 0 & 2 & 3 \\ 0 & 3 & 5 \\ 0 & 0 & 5 \end{bmatrix}\)
Augmented matrix set equal to 0:
\(\begin{bmatrix} 0 & 2 & 3 & 0 \\ 0 & 3 & 5 & 0 \\ 0 & 0 & 5 & 0 \end{bmatrix}\)
\(\begin{bmatrix} 0 & 2 & 3 & 0 \\ 0 & 3 & 5 & 0 \\ 0 & 0 & 5 & 0 \end{bmatrix} \xrightarrow{\frac 12 R_1} \begin{bmatrix} 0 & 1 & \frac 32 & 0 \\ 0 & 3 & 5 & 0 \\ 0 & 0 & 5 & 0 \end{bmatrix} \xrightarrow{-3R_1 + R_2} \begin{bmatrix} 0 & 1 & \frac 32 & 0 \\ 0 & 0 & \frac 12 & 0 \\ 0 & 0 & 5 & 0 \end{bmatrix}\)
\(\begin{bmatrix} 0 & 1 & \frac 32 & 0 \\ 0 & 0 & \frac 12 & 0 \\ 0 & 0 & 5 & 0 \end{bmatrix} \xrightarrow{2R_2} \begin{bmatrix} 0 & 1 & \frac 32 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 5 & 0 \end{bmatrix} \xrightarrow{-5R_2 + R_3} \begin{bmatrix} 0 & 1 & \frac 32 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \xrightarrow{-\frac 32R_2 + R_1} \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}\)
\(\therefore x_2 = 0, x_3 = 0, x_1 = x_1\)
\(x_1\) can be anything, so let \(x_1 = \lambda_1= 1\)
\(v_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\)
\(A - \lambda_2I = \begin{bmatrix} -3 & 2 & 3 \\ 0 & 0 & 5 \\ 0 & 0 & 2 \end{bmatrix}\)
Augmented matrix set equal to 0:
\(\begin{bmatrix} -3 & 2 & 3 & 0 \\ 0 & 0 & 5 & 0 \\ 0 & 0 & 2 & 0 \end{bmatrix}\)
\(\begin{bmatrix} -3 & 2 & 3 & 0 \\ 0 & 0 & 5 & 0 \\ 0 & 0 & 2 & 0 \end{bmatrix} \xrightarrow{ -\frac 13 R_1} \begin{bmatrix} 1 & - \frac 23 & -1 & 0 \\ 0 & 0 & 5 & 0 \\ 0 & 0 & 2 & 0 \end{bmatrix} \xrightarrow{ -\frac 25 R_2 + R_3} \begin{bmatrix} 1 & - \frac 23 & -1 & 0 \\ 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}\)
\(\begin{bmatrix} 1 & - \frac 23 & -1 & 0 \\ 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \xrightarrow{ \frac 15 R_2} \begin{bmatrix} 1 & - \frac 23 & -1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \xrightarrow{ R_2 + R_1} \begin{bmatrix} 1 & - \frac 23 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}\)
\(\therefore x_1 - \frac 23 x_2 = 0, x_3 = 0\)
\(x_1 = \frac 23 x_2\)
Let \(x_2 = \lambda_2 = 4\), then \(x_1 = \frac 83\).
\(v_2 = \begin{pmatrix} \frac 83 \\ 4 \\ 0 \end{pmatrix}\)
\(A - \lambda_3I = \begin{bmatrix} -5 & 2 & 3 \\ 0 & -2 & 5 \\ 0 & 0 & 0 \end{bmatrix}\)
Augmented matrix set equal to 0:
\(\begin{bmatrix} -5 & 2 & 3 & 0 \\ 0 & -2 & 5 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}\)
\(\begin{bmatrix} -5 & 2 & 3 & 0 \\ 0 & -2 & 5 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \xrightarrow{R_2 + R_1} \begin{bmatrix} -5 & 0 & 8 & 0 \\ 0 & -2 & 5 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \xrightarrow{- \frac 15 R_1} \begin{bmatrix} 1 & 0 & - \frac 85 & 0 \\ 0 & -2 & 5 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \xrightarrow{- \frac 12 R_2} \begin{bmatrix} 1 & 0 & - \frac 85 & 0 \\ 0 & 1 & - \frac 52 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}\)
\(\therefore x_1 = \frac 85 x_3, x2 = \frac 52 x_3\)
Let \(x_3 = \lambda_3 = 6\), then \(x_1 = \frac {48}5, x_2 = 15\)
\(v_3 = \begin{pmatrix} \frac {48}5 \\ 15 \\ 6 \end{pmatrix}\)
\(v_1 = \begin{pmatrix} x_1 \\ 0 \\ 0 \end{pmatrix}\), \(v_2 = \begin{pmatrix} \frac 23x_2 \\ x_2 \\ 0 \end{pmatrix}\), \(v_3 = \begin{pmatrix} \frac 85x_3 \\ \frac 52x_3 \\ x_3 \end{pmatrix}\)