A = \(\begin{bmatrix} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{bmatrix}\)
Solution:
Reduce matrix A to row echelon form
row 1 + row 2 to row 2 = \(\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{bmatrix}\)
-5 * row 1 + row 4 to row 4 = \(\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 1 & -2 & 1 \\ 0 & -6 & -17 & -23 \end{bmatrix}\)
(1/2) * row 2 to row 2 = \(\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 7/2 \\ 0 & 1 & -2 & 1 \\ 0 & -6 & -17 & -23 \end{bmatrix}\)
-1 * row 2 + row 3 to row 3 = \(\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 7/2 \\ 0 & 0 & -4 & -5/2 \\ 0 & -6 & -17 & -23 \end{bmatrix}\)
6 * row 2 + row 4 to row 4 = \(\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 7/2 \\ 0 & 0 & -4 & -5/2 \\ 0 & 0 & -5 & -2 \end{bmatrix}\)
-(1/4) * row 3 to row 3 = \(\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 7/2 \\ 0 & 0 & 1 & 5/8 \\ 0 & 0 & -5 & -2 \end{bmatrix}\)
5 * row 3 + row 4 to row 4 = \(\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 7/2 \\ 0 & 0 & 1 & 5/8 \\ 0 & 0 & 0 & 9/8 \end{bmatrix}\)
(8/9) * row 4 to row 4 = \(\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 7/2 \\ 0 & 0 & 1 & 5/8 \\ 0 & 0 & 0 & 1 \end{bmatrix}\)
-(5/8) * row 4 + row 3 to row 3 = \(\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 7/2 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}\)
-(7/2) * row 4 + row 2 to row 2 = \(\begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}\)
-4 * row 4 + row 1 to row 1 = \(\begin{bmatrix} 1 & 2 & 3 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}\)
-2 * row 3 + row 2 to row 2 = \(\begin{bmatrix} 1 & 2 & 3 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}\)
-3 * row 3 + row 1 to row 1 = \(\begin{bmatrix} 1 & 2 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}\)
-2 * row 2 + row 1 to row 1 = \(\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}\)
Since matrix A is now in reduced row echelon form to find the rank of A count the number of non-zero rows. The matrix is now the Identity matrix \(I_{ 4 }\), so there are 4 non-zero rows. So the rank of the matrix A is 4.
Solution:
The maximum rank for a mxn matrix where m > n is n since the maximum rank is the minimum between the number of rows and the numbers of columns. Since we are told m > n the maximum rank for the matrix would be the value of n. The minimum rank for a mxn matrix where m > n is 1 since we are told to assume that the matrix is non-zero. There is at least one non-zero value in the matrix, so the minimum rank of the matrix would be 1.
B = \(\begin{bmatrix} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2 \end{bmatrix}\)
Solution:
Reduce matrix B to row echelon form
3 * row 1 - row 2 to row 2 = \(\begin{bmatrix} 1 & 2 & 1 \\ 0 & 0 & 0 \\ 2 & 4 & 2 \end{bmatrix}\)
2 * row 1 - row 3 to row 3 = \(\begin{bmatrix} 1 & 2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\)
Since matrix B is now in reduced row echelon form to find the rank of B count the number of non-zero rows. Since only 1 non-zero row remains after reducing to row echelon form, the rank of matrix A is 1.
matrix A = \(\begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix}\)
Find the characteristic polynomial
\(det(A - \lambda I_{3})\) = 0
Solve for \(A - \lambda I_{3}\)
= \(\begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\)
= \(\begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix} - \begin{bmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{bmatrix}\)
= \(\begin{bmatrix} (1 - \lambda) & 2 & 3 \\ 0 & (4 - \lambda) & 5 \\ 0 & 0 & (6 - \lambda) \end{bmatrix}\)
\(A - \lambda I_{3} = \begin{bmatrix} (1 - \lambda) & 2 & 3 \\ 0 & (4 - \lambda) & 5 \\ 0 & 0 & (6 - \lambda) \end{bmatrix}\)
\(det(\begin{bmatrix} (1 - \lambda) & 2 & 3 \\ 0 & (4 - \lambda) & 5 \\ 0 & 0 & (6 - \lambda) \end{bmatrix}) = 0\)
= \((1 - \lambda)((4 - \lambda)(6 - \lambda) - (-5 * 0)) - 2(0(6 - \lambda) - (5 * 0)) + 3((0 * 0) - (4 - \lambda) * 0)) = 0\)
= \((1 - \lambda)((4 - \lambda)(6 - \lambda)) = 0\)
= \((1 - \lambda)(24 - 10 \lambda + \lambda ^2) = 0\)
= \((24 - 10 \lambda + \lambda ^2 - 24 \lambda + 10 \lambda^2 - \lambda ^3)\)
= \((-\lambda ^3 + 11 \lambda^2 - 34 \lambda + 24) = 0\)
multiply by -1 to make leading coefficient positive
\(-1(- \lambda ^3 + 11 \lambda^2 - 34 \lambda + 24)= 0\)
= \((\lambda ^3 - 11 \lambda^2 + 34 \lambda - 24)= 0\)
The characteristic polynomial is \(\lambda ^3 - 11 \lambda^2 + 34 \lambda - 24 = 0\)
Compute Eigenvalues
Since we found the characteristic polynomial, we can use it to find the eigenvalues as the roots of the characteristic polynomial are equal to the eigenvalues
\(\lambda ^3 - 11 \lambda^2 + 34 \lambda - 24 = 0\)
= \((1 - \lambda)(4 - \lambda)(6 - \lambda) = 0\)
= \(1 - \lambda = 0 | 4 - \lambda = 0 | 6 - \lambda = 0\)
= \(\lambda = 1 | \lambda = 4 | \lambda = 6\)
The eigenvalues for the matrix A are \(\lambda = 1\), \(\lambda = 4\), and \(\lambda = 6\)
Compute Eigenvectors
To find the eigenvectors, we use the eigenvalues and plug them into \((A - \lambda I_{3})v = 0\) where v is a vector
\(\lambda = 1\)
\((A - 1I_{3})v = 0\)
= \((\begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix})v = 0\)
= \(\begin{bmatrix} 0 & 2 & 3 \\ 0 & 3 & 5 \\ 0 & 0 & 5 \end{bmatrix}v = 0\)
reduce to row echelon form (I won’t show the steps due to not making my assignment too long and repetitive)
= \(\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}v = 0\)
= \(\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} v1 \\ v2 \\ v3 \end{bmatrix} = 0\)
\(0 * v1 + 1 * v2 + 0 * v3 = 0\)
\(v2 = 0\)
\(0 * v1 + 0 * v2 * 1 * v3 = 0\)
\(v3 = 0\)
Let v1 = 1
\(v1 = 1\)
Eigenvector for \(\lambda = 1\) is \(\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}\)
\(\lambda = 4\)
\((A - 4I_{3})v = 0\)
= \((\begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix} - \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix})v = 0\)
= \(\begin{bmatrix} -3 & 2 & 3 \\ 0 & 0 & 5 \\ 0 & 0 & 2 \end{bmatrix}v = 0\)
reduce to row echelon form (I won’t show the steps due to not making my assignment too long and repetitive)
= \(\begin{bmatrix} 1 & -2/3 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}v = 0\)
= \(\begin{bmatrix} 1 & -2/3 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} v1 \\ v2 \\ v3 \end{bmatrix} = 0\)
= \(1 * v1 - (2/3)v2 + 0 * v3 = 0\)
= \(v1 - (2/3)v2 = 0\)
\(v1 = (2/3)v2\)
\(0 * v1 + 0 * v2 + 1 * v3 = 0\)
\(v3 = 0\)
Let v2 = 1
\(v2 = 1\)
Then solve for v1
\(v1 = (2/3)(1) = 2/3\)
\(v1 = 2/3\)
Eigenvector for \(\lambda = 4\) is \(\begin{bmatrix} 2/3 \\ 1 \\ 0 \end{bmatrix}\)
\(\lambda = 6\)
\((A - 6I_{3})v = 0\)
= \((\begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix} - \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix})v = 0\)
= \(\begin{bmatrix} -5 & 2 & 3 \\ 0 & -2 & 5 \\ 0 & 0 & 0 \end{bmatrix}v = 0\)
reduce to row echelon form (I won’t show the steps due to not making my assignment too long and repetitive)
= \(\begin{bmatrix} 1 & 0 & -8/5 \\ 0 & 1 & -5/2 \\ 0 & 0 & 0 \end{bmatrix}v = 0\)
=\(\begin{bmatrix} 1 & 0 & -8/5 \\ 0 & 1 & -5/2 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} v1 \\ v2 \\ v3 \end{bmatrix} = 0\)
\(1 * v1 + 0 * v2 - (8/5)v3 = 0\)
= \(v1 - (8/5)v3 = 0\)
\(v1 = (8/5)v3\)
\(0 * v1 + 1 * v2 - (5/2)v3 = 0\)
= \(v2 - (5/2)v3 = 0\)
= \(v2 = (5/2)v3\)
Let v3 = 1
\(v3 = 1\)
Then solve for v1 and v2
\(v1 = (8/5)(1) = 8/5\)
\(v1 = 8/5\)
v2 = (5/2)(1) = 5/2
\(v2 = 5/2\)
Eigenvector for \(\lambda = 6\) is \(\begin{bmatrix} 8/5 \\ 5/2 \\ 1 \end{bmatrix}\)
The eigenvectors for the matrix A are \(\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}\), \(\begin{bmatrix} 2/3 \\ 1 \\ 0 \end{bmatrix}\), and \(\begin{bmatrix} 8/5 \\ 5/2 \\ 1 \end{bmatrix}\)