Solution 1

A <- matrix(c(1,-2,0,5,2,0,1,4,3,1,-2,-2,4,3,1,-3), ncol=4)
A

##      [,1] [,2] [,3] [,4]
## [1,]    1    2    3    4
## [2,]   -2    0    1    3
## [3,]    0    1   -2    1
## [4,]    5    4   -2   -3

det(A)

## [1] 52

Since \(Det(A)\neq 0\), we say there is no linear dependence of any role or column.

Rank of a matrix A is the number of role or column that are not a linear combination of another role of column.

qr(A)$rank

## [1] 4

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 2

For a Non-Zero Matrix, the Maximum rank can’t be greater than the the smaller dimension of the matrix which is n and the mininum rank can’t be less than 1.
Therefore, the rank is between 1 and n (inclusive). Maximum Rank = n Mininum Rank = 1

3. What is the rank of matrix B?

\[ B = \begin{bmatrix} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2 \\ \end{bmatrix}\]

Solution 3

B <- matrix(c(1,3,2,2,6,4,1,3,2), nrow = 3)
B

##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    3    6    3
## [3,]    2    4    2

det(B)

## [1] 0

The rank of B is 1, because:

1. The determinant of this square matrix is 0, which hints at linear dependence.
2. Each row and column is a linear combination of other rows and column. We manipulate the matrix to get the number of NOn-zero role to be 1. Hence, the rank is 1

Solution

\(Det (A - \lambda I)= 0\)

\(Det\left ( \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \\ \end{bmatrix}- \lambda \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} \right ) = 0\)

\(= Det\begin{bmatrix} 1-\lambda & 2 & 3 \\ 0 & 4-\lambda & 5 \\ 0 & 0 & 6-\lambda \\ \end{bmatrix}\)

\(= (1-\lambda )((4-\lambda)(6-\lambda)-0\times 5) - 2((6-\lambda)\times 0-0\times 5) + 3((0\times 0)-(0\times (4-\lambda)))\)

simplifying the above equation, we get the characteristic polynomial as:

\(P(\lambda )= -\lambda ^3 + 11\lambda^2 - 34\lambda + 24\)

Equating the characteristiv polynomial to zero(0) and Solving to get the Eigenvalues, we get:

\((-\lambda + 6)(\lambda - 4)(\lambda - 1) = 0\)

Eigenvalues of matrix A are: \(\lambda = 1, \lambda = 4, \lambda = 6\)

To calculate Eigenvectors, we use the formular \((AX-\lambda I) = 0\)

using R to reduce to matrix

A <- matrix(c(1,2,3,0,4,5,0,0,6), nrow=3, byrow=TRUE)

If \(\lambda=1\), then \(A - 1I_3\) is row-reduced to

rref(A - 1 * diag(3))

##      [,1] [,2] [,3]
## [1,]    0    1    0
## [2,]    0    0    1
## [3,]    0    0    0

\[ \begin{bmatrix} 0 &1 &0\\ 0 &0 &1\\ 0&0&0 \end{bmatrix} \begin{bmatrix} v_1\\ v_2\\ v_3 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} \]

Then \(v_1=v_1\) and \(v_2=0\) and \(v_3=0\). The eigenvector for \(\lambda=1\) is

\[ X_{1}= \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix} \]

If \(\lambda=4\), then \(A - 4I_3\) is row-reduced to

rref(A - 4 * diag(3))

##      [,1]       [,2] [,3]
## [1,]    1 -0.6666667    0
## [2,]    0  0.0000000    1
## [3,]    0  0.0000000    0

\[ \begin{bmatrix} 1 &-\frac{2}{3} &0\\ 0 &0 &1\\ 0&0&0 \end{bmatrix} \begin{bmatrix} v_1\\ v_2\\ v_3 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} \]

Then \(v_1 - \frac{2}{3}v_2=0\) and \(v_3=0\).

Or \(v_1=v_1\) and \(v_2=\frac{3}{2}v_1=1.5v_1\) and \(v_3=0\).

The eigenvector for \(\lambda=4\) is

\[ X_{2}= \begin{bmatrix} 1\\ 1.5\\ 0 \end{bmatrix} \]

Finally, if \(\lambda=6\), then \(A - 6I_3\) is row-reduced to

rref(A - 6 * diag(3))

##      [,1] [,2] [,3]
## [1,]    1    0 -1.6
## [2,]    0    1 -2.5
## [3,]    0    0  0.0

\[ \begin{bmatrix} 1 &0 &-1.6\\ 0 &1 &-2.5\\ 0&0&0 \end{bmatrix} \begin{bmatrix} v_1\\ v_2\\ v_3 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} \]

Then \(v_1-1.6v_3=0\) and \(v_2-2.5v_3=0\).

Or \(v_1=1.6v_3\) and \(v_2=2.5v_3\) and \(v_3=v_3\).

The eigenvector for \(\lambda=6\) is:

\[ X_{3}= \begin{bmatrix} 1.6\\ 2.5\\ 1 \end{bmatrix} \]

Eigenvalues of matrix A are: \(\lambda_{1} = 1, \lambda_{2} = 4, \lambda_{3} = 6\)

Eigenvectors of matrix A are: \(X_{1}=\begin{bmatrix}1\\0\\0\end{bmatrix}\) , \(X_{2}=\begin{bmatrix}1\\1.5\\0\end{bmatrix}\) and \(X_{3}=\begin{bmatrix}1.6\\2.5\\1\end{bmatrix}\)