HW3:
1.1:
Find the rank of Matrix A:
\(A= \begin{bmatrix} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \\ \end{bmatrix}\)
JB:
Using the library pracma, I’ll employ the function rref() to get the row-reduced echelon form of the Matrix A - this will indicate which columns of A are linearly independent and form the column space of A or \(C(A)=\{\vec{a_1}, \vec{a_2}, \vec{a_3}, \vec{a_4}\}\) where \(\vec{a_i}\) are linearly independent:
A <- matrix(c(1, 2, 3, 4, -1, 0, 1, 3, 0, 1, -2, 1, 5, 2, -2, -3),
nrow=4, ncol=4, byrow=T)
my_R <- rref(A)
my_R## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 0
## [2,] 0 1 0 0
## [3,] 0 0 1 0
## [4,] 0 0 0 1As shown above, given that each column has been reduced to a pivot column, I know that each column of the original matrix is linearly independent. Given this, I know that the column space of matrix A, \(C(A)\) is:
\(C(A)=\{\begin{bmatrix} 1 \\ -1 \\ 0 \\ 5 \end{bmatrix}, \begin{bmatrix} 2 \\ 0 \\ 1 \\ 4 \end{bmatrix}, \begin{bmatrix} 3 \\ 1 \\ -2 \\ -2 \end{bmatrix}, \begin{bmatrix} 4 \\ 3 \\ 1 \\ -3 \end{bmatrix} \}\)
Then, since I know that the rank is the dimension of the space spanned by its column (or row) vectors: \(dim(C(A))=rank(A)\) and since \(dim(C(A))=4\), \(\therefore\) the \(rank(A)=4\). In this case, the matrix \(A\) is full rank where full rank is when all of the vectors in a matrix are linearly independent.
1.2:
Given an \(m \times n\) matrix where \(m > n\), what can be the maximum rank? The minimum rank, assuming that the matrix is non-zero?
JB:
ROM (Rank Of a Matrix), as defined on page 325, states that: Suppose that \(A\) is an \(m \times n\) matrix. Then the rank of \(A\), \(r(A)\) is the dimension of the column space of \(A\), \(C(A)\). That is: \(r(A)=dim(C(A))\).
Further, using the Theorem CRN, page 326, Computing Rank and Nullity, given \(A\) is an \(m \times n\) matrix and \(B\) is a row-equivalent matrix in reduced row-echelon form. Let \(r\) denote the number of pivot columns (or the number of nonzero rows). Then \(r(A)=r\) and \(n(A)=n-r\) which leads to theorem RPNC (Rank Plus Nullity is Columns) that states that for an \(m \times n\) matrix \(A\), \(r(A)+n(A)=n\). That is, the rank of \(A\) plus the null space of \(A\) is equal to \(n\), the number of columns in the matrix.
\(\therefore\) the maximum rank for an \(m \times n\) matrix \(A\) is less than or equal to \(n\) and assuming non-zero, the minimum rank would be 1.
1.3:
What is the rank of Matrix B?:
\(B= \begin{bmatrix} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2 \\ \end{bmatrix}\)
JB:
Again, employing the rref() function within the pracma library:
B <- matrix(c(1, 2, 1, 3, 6, 3, 2, 4, 2),
nrow=3, ncol=3, byrow=T)
my_R1 <- rref(B)
my_R1## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 0 0 0
## [3,] 0 0 0As shown above, given that only the 1st column has been reduced to a pivot column, and that \(r(B)=dim(C(B))\), then \(r(B)=1\). This makes intuitive sense because, by inspection, each row of \(B\) is just a multiple of the 1st row.
2:
Compute the eigenvalues and eigenvectors of the matrix A. You’ll need to show your work:
\(A= \begin{bmatrix} 1 & 2 & 1 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \\ \end{bmatrix}\)
JB:
Below, please find the work used to find the eigenvalues of \(A\):
Part I
Note that I used rref to find the row-reduced version of each matrix of: \(\lambda_{1,2,3}=1,4,6\):
B1 <- matrix(c(0, -2, -1, 0, -3, -5, 0, 0, -5),
nrow=3, ncol=3, byrow=T)
B4 <- matrix(c(3, -2, -1, 0, 0, -5, 0, 0, -2),
nrow=3, ncol=3, byrow=T)
B6 <- matrix(c(5, -2, -1, 0, 2, -5, 0, 0, 0),
nrow=3, ncol=3, byrow=T)
rref(B1) ## [,1] [,2] [,3]
## [1,] 0 1 0
## [2,] 0 0 1
## [3,] 0 0 0rref(B4)## [,1] [,2] [,3]
## [1,] 1 -0.6666667 0
## [2,] 0 0.0000000 1
## [3,] 0 0.0000000 0rref(B6)## [,1] [,2] [,3]
## [1,] 1 0 -1.2
## [2,] 0 1 -2.5
## [3,] 0 0 0.0With the rref versions above, I was then able to find the eigenvectors of \(A\):
Part II