1. What is the rank of the matrix A?
A <- matrix(c(1,2,3,4,-1,0,1,3,0,1,-2,1,5,4,-2,-3), nrow=4, byrow=TRUE)
A
##      [,1] [,2] [,3] [,4]
## [1,]    1    2    3    4
## [2,]   -1    0    1    3
## [3,]    0    1   -2    1
## [4,]    5    4   -2   -3

The maximum number of linearly independent vectors in a matrix is equal to the number of non-zero rows in its row echelon matrix. Therefore, to find the rank of a matrix, we simply transform the matrix to its row echelon form and count the number of non-zero rows.

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

Answer:

There are four (4) non-zero rows, therefore Rank(A) = 4. This is confirmed by the Rank(A), above.

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  1. Given an mxn matrix where m > n, what can be the maximum rank? The minimum rank, assuming that the matrix is non-zero?

Answer:

The maximum number of linearly independent rows in a matrix A is called the row rank of A, and the maximum number of linearly independent columns in A is called the column rank of A. If A is an m by n matrix, that is, if A has m rows and n columns, then it is obvious that

row rank of A <= m

column rank of A <=n

Additionally, for any matrix A, the row rank of A = the column rank of A, the common value is simply called the rank of the matrix. Therefore, if A is m x n, it follows that:

rank(A) <= min(m,n)

For example, the rank of a 3 x 5 matrix can be no more than 3, and the rank of a 4 x 2 matrix can be no more than 2. Assuming a non-zero matrix, the rank should be at least 1.

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  1. What is the rank of matrix B?
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    3    6    3
## [3,]    2    4    2
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    0    0    0
## [3,]    0    0    0
## [1] 1

Answer:

There is one (1) non-zero row, therefore Rank(B) = 1. This is confirmed by the Rank(B) above.

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Problem Set 2

  1. 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.
##      [,1] [,2] [,3]
## [1,]    1    2    3
## [2,]    0    4    5
## [3,]    0    0    6

A is a triangular matrix. Therefore its eigenvalues are values on the diagonal, so \(\lambda_1=1\), \(\lambda_2=4\) and \(\lambda_3=6\). This is confirmed below with eigen(A).

## [1] 6 4 1

The characteristic polynomial is:

\(p_A(\lambda) = (1-\lambda)(4-\lambda)(6-\lambda)\) or

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

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

##      [,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 eigenspace is

\[ E_{\lambda=1}= \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix} \]

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

\[ E_{\lambda=4}= \begin{bmatrix} 1\\ 1.5\\ 0 \end{bmatrix} \]

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

##      [,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 eigenspace is

\[ E_{\lambda=6}= \begin{bmatrix} 1.6\\ 2.5\\ 1 \end{bmatrix} \]