Juan Falck - Assignment 3
Problem Set 1
Question 1
What is the rank of the matrix A?
\(A = \begin{bmatrix}1&2&3&4 \\ -1 & 0 & 1 & 3\\ 0 & 1 & -2 & 1\\ 5 & 4 & -2 & -3 \end{bmatrix}\)
Solution To get the rank we will do row reductions to get the matrix in Reduced Row Echelon Form RREF. In this form the rank would be equal to the number of non-redundant columns, that is columns which have a pivot role in it.
Swap R2 with R3
\(A = \begin{bmatrix}1&2&3&4 \\ 0 & 1 & -2 & 1\\ -1 & 0 & 1 & 3\\ 5 & 4 & -2 & -3 \end{bmatrix}\)
R3=R3+R1 R4=R4 - 5R1
\(A = \begin{bmatrix}1&2&3&4 \\ 0 & 1 & -2 & 1\\ 0 & 2 & 4 & 7\\ 0 & -6 & -17 & -23 \end{bmatrix}\)
R1=R1-2R2
R3=R3-2R2
R4=R4+6R2
\(A = \begin{bmatrix}1&0&7&2 \\ 0 & 1 & -2 & 1\\ 0 & 0 & 8 & 5\\ 0 & 0 & -29 & -17 \end{bmatrix}\)
R3 = R3 / 8
\(A = \begin{bmatrix}1&0&7&2 \\ 0 & 1 & -2 & 1\\ 0 & 0 & 1 & 5/8\\ 0 & 0 & -29 & -17 \end{bmatrix}\)
R1=R1-7R3
R2=R2+2R3
\(A = \begin{bmatrix}1 & 0 & 0 & -17/8 \\ 0 & 1 & 0 & 18/8\\ 0 & 0 & 1 & 5/8\\ 0 & 0 & -29 & -17 \end{bmatrix}\)
R4=R4 + 29R3
\(A = \begin{bmatrix}1 & 0 & 0 & -17/8 \\ 0 & 1 & 0 & 18/8\\ 0 & 0 & 1 & 5/8\\ 0 & 0 & 0 & 9/8 \end{bmatrix}\)
R4 = R4 / (9/8)
R1=R1+17/8R4
R2=R2-18/8R4
R3=R3-5/8R4
\(A = \begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}\)
We ended with a FULL RANK matrix, that means Rank = # Columns. So in this case RANK = 4
Question 2
Given an mxn matrix where m>n,what can be the maximum rank? The minimum rank, assuming that the matrix is non-zero?
Solution The maximum rank of a m x n where m>n would be n, the number of colummns. That is because some rows (n+1, n+2…) would have to be linearly dependant thus if we were to perform RREF the additional rows would be elimintate (all zeros).
The minimum (assuming non-zero) is 1. We can always reduce any matrix to a single row which is not linearly dependant to any other, while eliminating all other rows (which must be linearly dependant to the one)
Question 3
What is the Rank of Matrix B?
Solution \(B = \begin{bmatrix}1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2 & \end{bmatrix}\)
Just by simple observation we can see that R2 is 3R1 and R3 is 2R1, which means R2 and R3 and are linearly dependent so in a RREF they both would be eliminated (all zeros). Therefore RANK =1
Problem Set 2
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.
\(A = \begin{bmatrix}1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 & \end{bmatrix}\)
Solution To finds eigenvalues and eigen vectors we need to solve for:
\(DET ( A - \lambda I_3)\)
or
\(DET (\begin{bmatrix}1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 & \end{bmatrix}-\begin{bmatrix}x & 0 & 0 \\ 0 & x & 0 \\ 0 & 0 & x & \end{bmatrix})\)
\(DET (\begin{bmatrix}1-x & 2 & 3 \\ 0 & 4-x & 5 \\ 0 & 0 & 6-x & \end{bmatrix}\)
\(= (1-x)(4-x)(6-x)\)
This is our characteristic polynomial. We can also express it as:
\(-x^3 + 11x^2 -34x + 24\)
Getting our eigenvalues is relatively easy since we the characteristic polinomial in its factorized form: \(= (1-x)(4-x)(6-x)\) So it just a matter to find its roots. In this case we have three eigenvalues:
\(\lambda_1 = 1\)
\(\lambda_2 = 4\)
\(\lambda_3 = 6\)
To find now our eigenvectors we need to apply our formula: \(DET ( A - \lambda I_3)\) for each of the 3 lambdas (eigenvalues) we found above.
a) For \(\lambda = 1\)
\(\begin{bmatrix}1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 & \end{bmatrix}-\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 & \end{bmatrix} = \begin{bmatrix}0 & 2 & 3 \\ 0 & 3 & 5 \\ 0 & 0 & 5 & \end{bmatrix}\)
This resulting matrix will put in Reduced Row Echelon Form RREF To simply a little presentation of this answe here are the row operations performed and the final matrix in RREF form.
R1=R1/2
R2=R2-3R1
R2=2R2
R1=R1-3/2R2
R3=R3-5R2
RREF Matrix is:
\(\begin{bmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 & \end{bmatrix}\)
From RREF we can get the values of:
x=t
y=0
z=0
First vector for our eigenbasis is: \(\begin{bmatrix}1\\ 0 \\ 0 \end{bmatrix}\)
b) For \(\lambda = 4\)
\(\begin{bmatrix}1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 & \end{bmatrix}-\begin{bmatrix}4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 & \end{bmatrix} = \begin{bmatrix}-3 & 2 & 3 \\ 0 & 0 & 5 \\ 0 & 0 & 2 & \end{bmatrix}\)
This resulting matrix will put in Reduced Row Echelon Form RREF To simply a little presentation of this answe here are the row operations performed and the final matrix in RREF form.
R1/-3
R2/5
R1=R1+R2
R3=R3-2R2
RREF Matrix is:
\(\begin{bmatrix}1 & -2/3 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 & \end{bmatrix}\)
From RREF we can get the values of:
x = 2/3t
y = t
z = 0
Second vector for our eigenbasis is: \(\begin{bmatrix}2/3\\ 1 \\ 0 \end{bmatrix}\)
c) For \(\lambda = 6\)
\(\begin{bmatrix}1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 & \end{bmatrix}-\begin{bmatrix}6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 & \end{bmatrix} = \begin{bmatrix}-5 & 2 & 3 \\ 0 & -2 & 5 \\ 0 & 0 & 0 & \end{bmatrix}\)
This resulting matrix will put in Reduced Row Echelon Form RREF To simply a little presentation of this answe here are the row operations performed and the final matrix in RREF form.
R2/2
R1=R1 + 2/5R2
RREF Matrix is:
\(\begin{bmatrix}1 & 0 & -8/5 \\ 0 & 1 & -5/2 \\ 0 & 0 & 0 & \end{bmatrix}\)
From RREF we can get the values of:
x = 8/5t
y = 5/2t
z = t
Third and final vector for our eigenbasis is: \(\begin{bmatrix}8/5\\ 5/2 \\ 1 \end{bmatrix}\)
** Our complete answer, the eigenbasis for our eigenvectors is:
\(SPAN(\begin{bmatrix}1\\ 0 \\ 0 \end{bmatrix},\begin{bmatrix}2/3\\ 1 \\ 0 \end{bmatrix},\begin{bmatrix}8/5\\ 5/2 \\ 1 \end{bmatrix})\)