Problem set 1
- What is the rank of the matrix A? \[ A = \begin{bmatrix} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{bmatrix} \]
## [,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] 4- Given an mxn matrix where m > n, what can be the maximum rank? The minimum rank, assuming that the matrix is non-zero?
In a mxn matrix where m > n the maximum rank will be the minimum between m and n. Hence the maximum rank is n. Assuming the matrix is non-zero, the minimum rank is 1.
- What is the rank of matrix B?
\[ B = \begin{bmatrix} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2 \end{bmatrix} \]
## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 3 6 3
## [3,] 2 4 2## [1] 1Problem 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. \[ A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix} \]
Please show your work using an R-markdown document. Please name your assignment submission with your first initial and last name.
## [,1] [,2] [,3]
## [1,] 1 2 3
## [2,] 0 4 5
## [3,] 0 0 6Eigen values
To find the eigenvalues of A, we must solve det(λI-A) =0 for λ.Where I is Identitity Matrix.
\[ det(λ\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\0 & 0 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix}) = 0 \] \[ det(\begin{bmatrix} λ & 0 & 0 \\ 0 & λ & 0 \\0 & 0 & λ \end{bmatrix} - \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix}) = 0 \]
\[ det(\begin{bmatrix} λ-1 & -2 & -3 \\ 0 & λ-4 & -5 \\0 & 0 & λ-6 \end{bmatrix}) = 0 \] \[ =(λ-1) det(\begin{bmatrix} λ-4 & -5 \\ 0 & λ-6 \end{bmatrix}) - (-2) det(\begin{bmatrix}0 & -5 \\ 0 & λ-6 \end{bmatrix}+(-3)det(\begin{bmatrix}0 & λ -4 \\ 0 & 0 \end{bmatrix}) \]
\[ =(λ-1) det(\begin{bmatrix} λ-4 & -5 \\ 0 & λ-6 \end{bmatrix}) + 2 det(\begin{bmatrix}0 & -5 \\ 0 & λ-6 \end{bmatrix}-3 det(\begin{bmatrix}0 & λ -4 \\ 0 & 0 \end{bmatrix}) \]
\[ = -λ^3 + 11λ^2 -34λ +24 \]
\[ = λ^2 - 10λ +24 \] \[ = λ^2 -4λ -6λ +24 \] \[ = (λ -1)(λ -4) (λ -6) \]
Hence, The EigenValues for λ are:- λ = 1 λ = 4 λ = 6
Let us calculate the Eigenvectors for the 3 values of λ :-
For λ = 1
det(λI-A) = \[det(1.\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\0 & 0 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix}) \]
\[ = det(1.\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\0 & 0 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix}) \]
\[= \begin{bmatrix} 0 & -2 & -3 \\ 0 & -3 & -5 \\ 0 & 0 & -5 \end{bmatrix}\]
Reduce Matrix to reduced row echelon form gives us:-
## [1] 0.0 1.0 1.5## [1] 0.0 0.0 -0.5## [1] 0 0 1## [1] 0 1 0## [1] 0 0 0## [,1] [,2] [,3]
## [1,] 0 1 0
## [2,] 0 0 1
## [3,] 0 0 0\[ (λ I - A)v = \begin{bmatrix}0 & 1 & 0 \\0 & 0 & 1 \\0 & 0 & 0 \end{bmatrix}\begin{bmatrix}v_{1} \\v_{2} \\v_{3}\end{bmatrix}\]
\(v_{2} = 0\)
\(v_{3} = 0\)
Let, \(v_{1} =a\)
a be any whole number
\[E_{λ =1}=a\begin{bmatrix}1 \\0 \\0\end{bmatrix}\]
For λ = 4
\[ = det(4.\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\0 & 0 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix}) \]
\[=\begin{bmatrix}3 & -2 & -3 \\0 & 0 & -5 \\0 & 0 & -2\end{bmatrix}\begin{bmatrix}v_{1} \\v_{2} \\v_{3}\end{bmatrix}=\begin{bmatrix}0 \\0 \\0\end{bmatrix}\] Reduce Matrix to reduced row echelon form gives us:-
## [1] 1.0000000 -0.6666667 -1.0000000## [1] 0 0 1## [1] 1.0000000 -0.6666667 0.0000000## [1] 0 0 0## [,1] [,2] [,3]
## [1,] 1 -0.6666667 0
## [2,] 0 0.0000000 1
## [3,] 0 0.0000000 0\[(λ I - A)v = \begin{bmatrix}1 & -0.6666667 & 0 \\0 & 0 & 1 \\0 & 0 & 0 \end{bmatrix}\begin{bmatrix}v_{1} \\v_{2} \\v_{3}\end{bmatrix}\]
\(v_{1} - 0.6666667v_{2} = 0\)
\(v_{3} = 0\)
Let, \(v_{2} =a\)
\(ie. 0.6666667a= v_{1}\)
\[E_{λ =4}=a\begin{bmatrix}0.6666667 \\1 \\0\end{bmatrix}\]
For λ = 6
\[ = det(6.\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\0 & 0 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix}) \]
\[\begin{bmatrix}5 & -2 & -3 \\0 & 2 & -5 \\0 & 0 & 0\end{bmatrix}\begin{bmatrix}v_{1} \\v_{2} \\v_{3}\end{bmatrix}=\begin{bmatrix}0 \\0 \\0\end{bmatrix}\] Reduce Matrix to reduced row echelon form gives us:-
## [1] 1.0 -0.4 -0.6## [1] 0.0 1.0 -2.5## [1] 1.0 0.0 -1.6## [,1] [,2] [,3]
## [1,] 1 0 -1.6
## [2,] 0 1 -2.5
## [3,] 0 0 0.0\[(λ I - A)v = \begin{bmatrix}1 & -0.4 & -1.6 \\0 & 1 & -2.5 \\0 & 0 & 0 \end{bmatrix}\begin{bmatrix}v_{1} \\v_{2} \\v_{3}\end{bmatrix}\]
\(v_{1} - 1.6v_{3} = 0\)
\(v_{2} - 2.5v_{3}= 0\)
Let, \(v_{3} =a\)
\(ie. 1.6a= v_{1}\)
\(2.5a= v_{2}\)
\[E_{λ =6}=a\begin{bmatrix}1.6 \\2.5 \\0\end{bmatrix}\]
Verify using R
A <- matrix(c(1, 2, 3, 0, 4, 5, 0, 0, 6), nrow = 3, byrow = T)
t <- eigen(A)
values <- t$values
vectors <- t$vectors
values## [1] 6 4 1## [,1] [,2] [,3]
## [1,] 0.5108407 0.5547002 1
## [2,] 0.7981886 0.8320503 0
## [3,] 0.3192754 0.0000000 0