Let \(A\) be a \(4 \times 4\) matrix: \[ A = \begin{bmatrix} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{bmatrix} \]
Now convert to reduced row echelon form
Add row 1 row 2 as \[R1 + R2 -> R2 \]
\[ A = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{bmatrix} \]
Multiply row 1 by -5 and add row 4 as \[-5R1 + R4 -> R4 \]
\[ A = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 2 & 4 & 7 \\ 0 & 1 & -2 & 1 \\ 0 & -6 & -17 & -23 \end{bmatrix} \]
Divide row 2 by 2 and place in row 2 as \[R2/2 -> R2\]
\[ A = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 7/2 \\ 0 & 1 & -2 & 1 \\ 0 & -6 & -17 & -23 \end{bmatrix} \]
Multiply row 2 by -2 and add row 1 as \[-2R2 + R1 -> R1\]
\[ A = \begin{bmatrix} 1 & 0 & -1 & -3 \\ 0 & 1 & 2 & 7/2 \\ 0 & 1 & -2 & 1 \\ 0 & -6 & -17 & -23 \end{bmatrix} \]
Subtract row 2 from row 3 and place row 3 \[R3 - R2 -> R3\]
\[ A = \begin{bmatrix} 1 & 0 & -1 & -3 \\ 0 & 1 & 2 & 7/2 \\ 0 & 0 & -4 & -5/2 \\ 0 & -6 & -17 & -23 \end{bmatrix} \]
\[6R2 + R4 -> R4\]
\[ A = \begin{bmatrix} 1 & 0 & -1 & -3 \\ 0 & 1 & 2 & 7/2 \\ 0 & 0 & -4 & -5/2 \\ 0 & 0 & -5 & -2 \end{bmatrix} \]
\[R3/-4 -> R3\]
\[ A = \begin{bmatrix} 1 & 0 & -1 & -3 \\ 0 & 1 & 2 & 7/2 \\ 0 & 0 & 1 & 5/8 \\ 0 & 0 & -5 & -2 \end{bmatrix} \]
\[ R1 + R3 -> R1\]
\[ A = \begin{bmatrix} 1 & 0 & 0 & -19/8 \\ 0 & 1 & 2 & 7/2 \\ 0 & 0 & 1 & 5/8 \\ 0 & 0 & -5 & -2 \end{bmatrix} \]
Multiply row 3 by -2 and add row 2 \[-2R3 + R2 -> R2\]
\[ A = \begin{bmatrix} 1 & 0 & 0 & -19/8 \\ 0 & 1 & 0 & 9/4 \\ 0 & 0 & 1 & 5/8 \\ 0 & 0 & -5 & -2 \end{bmatrix} \]
Multiply row 3 by 5 and add row 4 \[5R3 + R4 -> R4\]
\[ A = \begin{bmatrix} 1 & 0 & 0 & -19/8 \\ 0 & 1 & 0 & 9/4 \\ 0 & 0 & 1 & 5/8 \\ 0 & 0 & 0 & 9/8 \end{bmatrix} \]
Multiply row 4 by 8/9 \[R4*8/9\]
\[ A = \begin{bmatrix} 1 & 0 & 0 & -19/8 \\ 0 & 1 & 0 & 9/4 \\ 0 & 0 & 1 & 5/8 \\ 0 & 0 & 0 & 1 \end{bmatrix} \]
Multiply row 4 by 19/8 and add row 1 \[R1*19/8 + R1 -> R1\]
\[ A = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 9/4 \\ 0 & 0 & 1 & 5/8 \\ 0 & 0 & 0 & 1 \end{bmatrix} \]
Multiply row 4 by -9/4 and add row 2 \[R4*-9/4 + R2 -> R2\]
\[ A = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 5/8 \\ 0 & 0 & 0 & 1 \end{bmatrix} \]
Multiply row 4 by -5/8 and add row 3 \[R4*-5/8 + R3 -> R3\]
\[ A = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \]
There are four non zero row and it proves that the rank of matrix A = 4
rank(A) = dim(col(A)) = number of pivot columns = 4
Given m X n matrix and m > n
since rank(A) = dim(col(A)) = number of pivot columns
max of rank(A) = number of n pivot columns
Also, it’s not a 0 matrix, the rank is between 1 and n
Let \(A\) be a \(3 \times 3\) matrix:
\[ B = \begin{bmatrix} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2 \end{bmatrix} \]
Multiply row 2 by 1/3 \[ R2 * 1/3 -> R2 \]
\[ B = \begin{bmatrix} 1 & 2 & 1 \\ 1 & 1 & 1 \\ 2 & 4 & 2 \end{bmatrix} \]
\[ R2 - R1 -> R2\]
\[ B = \begin{bmatrix} 1 & 2 & 1 \\ 0 & 0 & 0 \\ 2 & 4 & 2 \end{bmatrix} \]
Switch R2 to R3
\[ B = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 4 & 2 \\ 0 & 0 & 0 \end{bmatrix} \]
Multiply R1 by -2 and add to R2 \[-2R1 +R2 -> R2\]
\[ B = \begin{bmatrix} 1 & 2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \]
There is one pivot non zero row and it proves that the rank of matrix B = 1
rank(B) = dim(col(B)) = number of pivot columns = 1
Compute the eigenvalues and eigenvectors of the matrix A. You’ll need to show your work. You’ll need to write out the characteristic polynomial and show your solution.
To find eigenvalues and eigenvectors, we take the determinant of a special matrix and set it to zero: \[det(A−λI) = 0.\]
A polynomial equation in terms of λ is called the characteristic polynomial.
Let \(A\) be a \(3 \times 3\) matrix:
\[ A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix} \]
The characteristic polynomial \[p(\lambda)\] is given by: \[ p(\lambda) = \det(A - \lambda I) \]
Now calculate \[(λI)\] where I is the identity matrix
\[ \lambda I = \begin{bmatrix} 1-\lambda & 0 & 0 \\ 0 & 1-\lambda & 0 \\ 0 & 0 & 1-\lambda \end{bmatrix} \]
Substitute the values into the determinant:
\[ p(\lambda) = \det\left(\begin{bmatrix} 1 - \lambda & 2 & 3\\ 0 & 4 - \lambda & 5\\ 0 & 0 & 6 - \lambda \end{bmatrix}\right) \]
Expand the determinant:
\[ p(\lambda) = (1 - \lambda)(4 - \lambda)(6 - \lambda)\]
which gives the roots of this polynomial equation which are the eigenvalues
\[ \lambda = 1 \] \[ \lambda = 4 \] \[ \lambda = 6 \]
Now we can solve for \[(A − λI)v = \] and find the eigenvectors v.
\[ Av − λv = (A − λI)v = 0 \]
\[ = \begin{bmatrix} 1-\lambda & 2 & 3 \\ 0 & 4-\lambda & 5 \\ 0 & 0 & 6-\lambda \end{bmatrix} \]
Now solve for \[ λ = 1 \]
\[ = \left( \begin{array}{ccc|c} 1-\lambda & 2 & 3 & 0\\ 0 & 4-\lambda & 5 & 0\\ 0 & 0 & 6-\lambda & 0 \end{array} \right)\]
\[ = \left( \begin{array}{ccc|c} 0 & 2 & 3 & 0\\ 0 & 3 & 5 & 0\\ 0 & 0 & 5 & 0 \end{array} \right)\]
Apply \[1/2R1 -> R1\]
\[ = \left( \begin{array}{ccc|c} 0 & 1 & 3/2 & 0\\ 0 & 3 & 5 & 0\\ 0 & 0 & 5 & 0 \end{array} \right)\]
Apply \[-3R1 + R2 -> R2\]
\[ = \left( \begin{array}{ccc|c} 0 & 1 & 3/2 & 0\\ 0 & 0 & 1/2 & 0\\ 0 & 0 & 5 & 0 \end{array} \right)\]
Apply \[-3R2 + R1 -> R1\]
\[ = \left( \begin{array}{ccc|c} 0 & 1 & 0 & 0\\ 0 & 0 & 1/2 & 0\\ 0 & 0 & 5 & 0 \end{array} \right)\]
Apply \[-1/2R3 -> R3\]
\[ = \left( \begin{array}{ccc|c} 0 & 1 & 0 & 0\\ 0 & 0 & 1/2 & 0\\ 0 & 0 & -5/2 & 0 \end{array} \right)\]
Apply \[5R2 + R3 -> R3\]
\[ = \left( \begin{array}{ccc|c} 0 & 1 & 0 & 0\\ 0 & 0 & 1/2 & 0\\ 0 & 0 & 0 & 0 \end{array} \right)\]
Apply \[2R2 -> R2\]
\[ = \left( \begin{array}{ccc|c} 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end{array} \right)\]
\[ V2 = V3 = 0 \]
and since V1 is free, we can assign \[ V1 = 1 \]
\[ V1 = \begin{bmatrix} 1 & 0 & 0 \\ \end{bmatrix} \]
Next solve for \[ λ = 4 \]
\[ = \left( \begin{array}{ccc|c} 1-\lambda & 2 & 3 & 0\\ 0 & 4-\lambda & 5 & 0\\ 0 & 0 & 6-\lambda & 0 \end{array} \right)\]
\[ = \left( \begin{array}{ccc|c} -3 & 2 & 3 & 0\\ 0 & 0 & 5 & 0\\ 0 & 0 & 2 & 0 \end{array} \right)\]
Apply \[-1/3R1 -> R1\]
\[ = \left( \begin{array}{ccc|c} 1 & -2/3 & -1 & 0\\ 0 & 0 & 5 & 0\\ 0 & 0 & 2 & 0 \end{array} \right)\]
Apply \[1/5R2 -> R2\]
\[ = \left( \begin{array}{ccc|c} 1 & -2/3 & -1 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 2 & 0 \end{array} \right)\]
Apply \[R1 + R2 -> R1\]
\[ = \left( \begin{array}{ccc|c} 1 & -2/3 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 2 & 0 \end{array} \right)\]
Apply \[-2R2 + R3 -> R3\]
\[ = \left( \begin{array}{ccc|c} 1 & -2/3 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end{array} \right)\]
\[ V1 = V3 = 0 \]
\[ V2 = -2/3 \]
Next solve for \[ λ = 6 \]
\[ = \left( \begin{array}{ccc|c} 1-\lambda & 2 & 3 & 0\\ 0 & 4-\lambda & 5 & 0\\ 0 & 0 & 6-\lambda & 0 \end{array} \right)\]
\[ = \left( \begin{array}{ccc|c} -5 & 2 & 3 & 0\\ 0 & -2 & 5 & 0\\ 0 & 0 & 0 & 0 \end{array} \right)\]
Apply \[-1/5R1 -> R1\]
\[ = \left( \begin{array}{ccc|c} 1 & -2/5 & -3/5 & 0\\ 0 & -2 & 5 & 0\\ 0 & 0 & 0 & 0 \end{array} \right)\]
Apply \[-1/2R2 -> R2\]
\[ = \left( \begin{array}{ccc|c} 1 & -2/5 & -3/5 & 0\\ 0 & 1 & -5/2 & 0\\ 0 & 0 & 0 & 0 \end{array} \right)\]
Apply \[2/5R2 + R1 -> R1\]
\[ = \left( \begin{array}{ccc|c} 1 & 0 & -8/5 & 0\\ 0 & 1 & -5/2 & 0\\ 0 & 0 & 0 & 0 \end{array} \right)\]
\[ V1 = V2 = 0 \]
\[ V3 = -8/5 \]
\[ V3 = -5/2 \]