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1. Problem set 1
- What is the rank of the matrix A?
\(A\quad =\quad \begin{vmatrix} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{vmatrix}\)
The set of pivot columns of any reduced row echelon form matrix is known as Rank.
Transform matrix A to reduced echelon form, which is:
\(A\quad =\quad \begin{vmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{vmatrix}\)
Looking at the resultant matrix, the rank is 4.
Confirming rank using R
## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4
## [2,] -1 0 1 3
## [3,] 0 1 -2 1
## [4,] 5 4 -2 -3## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 0
## [2,] 0 1 0 0
## [3,] 0 0 1 0
## [4,] 0 0 0 1There are 4 pivot rows, so the rank of A is 4. Double-check with R.
## [1] 4- Given an mxn matrix where m > n, what can be the maximum rank? The mini-mum rank, assuming that the matrix is non-zero?
Since rank of a matrix is defined as the number of linearly independent column vectors and is equal to the number of linearly independent row vectors, given an m×n matrix with m>n, the maximum rank is the lower value \({ rank }_{ max }\quad =\quad n\). Assuming a non-zero matrix, the rank should be at least 1, so \({ rank }_{ min }\quad =\quad 1\).
- What is the rank of matrix B?
\(B\quad =\quad \begin{vmatrix} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2 \end{vmatrix}\)
We can see that row 2 and row 3 are multiples of row 1, so rows 1 and 2 and 1 and 3 are linearly dependent. There is only 1 linearly independent row, so \(rank(B)\quad =\quad 1\)
Confirming rank using R
## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 3 6 3
## [3,] 2 4 2## [1] 12. 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.
\(A\quad =\quad \begin{vmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{vmatrix}\)
Since, A is a triangular matrix, its eigenvalues are values on the diagonal, so λ1=1, λ2=4 and λ3=6. But i am going to show the work as following
Step 1 - Calculate the eigenvalues of A
\(det(A-λI)\quad =\quad 0\)
\(det(\begin{vmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{vmatrix}-\begin{vmatrix} λ & 0 & 0 \\ 0 & λ & 0 \\ 0 & 0 & 0 \end{vmatrix})\quad =\quad 0\)
\(det(\begin{vmatrix} 1-λ & 2 & 3 \\ 0 & 4-λ & 5 \\ 0 & 0 & 6-λ \end{vmatrix})\quad =\quad 0\)
\((1-λ)\quad \begin{vmatrix} 4-λ & 5 \\ 0 & 6-λ \end{vmatrix}\quad =\quad 0\)
\((1−λ)((4−λ)(6−λ)\quad =\quad 0\)
So,The characteristic polynomial is
\(P(λ)=(1−λ)(4−λ)(6−λ)\quad or\quad P(λ)={ λ }^{ 3 }-11{ λ }^{ 2 }+34λ-24\)
We determine that the eigenvalues of A, given by λ, are 1, 4, 6.
We can use R to confirm the result
#define matrix
A = matrix(c(1,0,0, 2,4,0, 3,5,6),ncol = 3)
cp = charpoly(A)
#Characteristic polynomial
cp## [1] 1 -11 34 -24## [1] 6 4 1Step 2 - Calculate the eigenvectors of A using λ
Using λ1 = 1:
\(A−λI\quad =\quad \begin{vmatrix} 1−1 & 2 & 3 \\ 0 & 4-1 & 5 \\ 0 & 0 & 6-1 \end{vmatrix}\quad =\quad \begin{vmatrix} 0 & 2 & 3 \\ 0 & 3 & 5 \\ 0 & 0 & 5 \end{vmatrix}\)
Using Gaussian Elimination on the resulting augmented matrix:
\(\begin{vmatrix} 0 & 2 & 3 & 0 \\ 0 & 3 & 5 & 0 \\ 0 & 0 & 5 & 0 \end{vmatrix}\)
\(R1=1/2(R1)\quad =\quad \begin{vmatrix} 0 & 1 & 3/2 & 0 \\ 0 & 3 & 5 & 0 \\ 0 & 0 & 5 & 0 \end{vmatrix}\)
\(R2\quad =\quad −3(R1)\quad =\quad \begin{vmatrix} 0 & 1 & 3/2 & 0 \\ 0 & 0 & 1/2 & 0 \\ 0 & 0 & 5 & 0 \end{vmatrix}\)
\(R2\quad =\quad 2(R2)\quad =\quad \begin{vmatrix} 0 & 1 & 3/2 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 5 & 0 \end{vmatrix}\)
\(R3\quad =\quad −5(R2)=\quad \begin{vmatrix} 0 & 1 & 3/2 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{vmatrix}\)
\(R1\quad =\quad R1−3/2(R2)\quad =\quad \begin{vmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{vmatrix}\)
For the eigenvalue λ1 = 1, the eigenvector is:
\({ v }_{ 1 }=\quad \begin{pmatrix} { x }_{ 1 } \\ 0 \\ 0 \end{pmatrix}\quad =\quad \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\)
Using λ2 = 4:
\(A−λI\quad =\quad \begin{vmatrix} 1−4 & 2 & 3 \\ 0 & 4-4 & 5 \\ 0 & 0 & 6-4 \end{vmatrix}\quad =\quad \begin{vmatrix} -3 & 2 & 3 \\ 0 & 0 & 5 \\ 0 & 0 & 2 \end{vmatrix}\)
Using Gaussian Elimination on the resulting augmented matrix:
\(\begin{vmatrix} -3 & 2 & 3 & 0 \\ 0 & 0 & 5 & 0 \\ 0 & 0 & 2 & 0 \end{vmatrix}\)
\(R1\quad =\quad −1/3(R1)\quad =\quad \begin{vmatrix} 1 & -2/3 & -1 & 0 \\ 0 & 0 & 5 & 0 \\ 0 & 0 & 2 & 0 \end{vmatrix}\)
\(R2\quad =\quad (R1)/5)\quad =\quad \begin{vmatrix} 1 & -2/3 & -1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 2 & 0 \end{vmatrix}\)
\(R3\quad =\quad R3−3(R2)\quad =\quad \begin{vmatrix} 1 & -2/3 & -1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{vmatrix}\)
\(R1\quad =\quad R1+R2\quad =\quad \begin{vmatrix} 1 & -2/3 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{vmatrix}\)
Solving the system: \({ x }_{ 1 }−2/3{ x }_{ 2 }\quad =\quad 0;\quad { x }_{ 1 }\quad =\quad 2/3{ x }_{ 2 }\\ { x }_{ 3 }\quad =\quad 0\)
For the eigenvalue λ2 = 4, the eigenvector is:
\({ v }_{ 2 }\quad =\quad \begin{pmatrix} 2/3{ x }_{ 2 } \\ { x }_{ 2 } \\ 0 \end{pmatrix}\quad =\quad \begin{pmatrix} 2 \\ 3 \\ 0 \end{pmatrix}\)
Using λ3 = 6:
\(A−λI\quad =\quad \begin{vmatrix} 1−6 & 2 & 3 \\ 0 & 4-6 & 5 \\ 0 & 0 & 6-6 \end{vmatrix}\quad =\quad \begin{vmatrix} -5 & 2 & 3 \\ 0 & -2 & 5 \\ 0 & 0 & 0 \end{vmatrix}\)
Using Gaussian Elimination on the resulting augmented matrix:
\(\begin{vmatrix} -5 & 2 & 3 & 0 \\ 0 & -2 & 5 & 0 \\ 0 & 0 & 0 & 0 \end{vmatrix}\)
\(R1\quad =\quad −1/5(R1)\quad =\quad \begin{vmatrix} 1 & -2/5 & -3/5 & 0 \\ 0 & -2 & 5 & 0 \\ 0 & 0 & 0 & 0 \end{vmatrix}\)
\(R2\quad =\quad −1/2(R2)\quad =\quad \begin{vmatrix} 1 & -2/5 & -3/5 & 0 \\ 0 & 1 & -5/2 & 0 \\ 0 & 0 & 0 & 0 \end{vmatrix}\)
\(R1=R1−2/5(R2)\quad =\quad \begin{vmatrix} 1 & 0 & -8/5 & 0 \\ 0 & 1 & -5/2 & 0 \\ 0 & 0 & 0 & 0 \end{vmatrix}\)
Solving the system:
\({ x }_{ 1 }+0{ x }_{ 2 }-8/5{ x }_{ 3 }\quad =\quad 0\\ { 0x }_{ 1 }+{ x }_{ 2 }-5/2{ x }_{ 3 }\quad =\quad 0\)
\({ x }_{ 1 }=8/5{ x }_{ 3 }\\ { x }_{ 2 }=5/2{ x }_{ 3 }\)
For the eigenvalue λ3 = 6, the eigenvector is: