(1) What is the rank of matrix A?

\[ A = \left(\begin{array}{cccc} 1 & 2 & 3 & 4\\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1\\ 5 & 4 & -2 & -3\\ \end{array}\right) \] \[ \left(\begin{array}{cccc} 1 & 2 & 3 & 4\\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1\\ 5 & 4 & -2 & -3\\ \end{array}\right) = \left(\begin{array}{cccc} 1 & 2 & 3 & 4\\ 0 & 2 & 4 & 7 \\ 0 & 1 & -2 & 1\\ 5 & 4 & -2 & -3\\ \end{array}\right) = \] \[ \left(\begin{array}{cccc} 1 & 2 & 3 & 4\\ 0 & 2 & 4 & 7 \\ 0 & 1 & -2 & 1\\ 0 & -6 & -17 & -23\\ \end{array}\right) = \left(\begin{array}{cccc} 1 & 2 & 3 & 4\\ 0 & 1 & 2 & 7/2 \\ 0 & 0 & -4 & -5/2\\ 0 & -6 & -17 & -23\\ \end{array}\right) =\] \[ \left(\begin{array}{cccc} 1 & 2 & 3 & 4\\ 0 & 1 & 2 & 7/2 \\ 0 & 0 & -4 & -5/2\\ 0 & 0 & -5 & -2\\ \end{array}\right) = \left(\begin{array}{cccc} 1 & 2 & 3 & 4\\ 0 & 2 & 4 & 7 \\ 0 & 0 & 1 & 5/8\\ 0 & 0 & 0 & 9/8\\ \end{array}\right) = \left(\begin{array}{cccc} 1 & 2 & 3 & 4\\ 0 & 2 & 4 & 7 \\ 0 & 0 & -4 & -5/2\\ 0 & 0 & 0 & 9/8\\ \end{array}\right) = R4 \]

A = matrix(c(1,-1,0,5,2,0,1,4,3,1,-2,-2,4,3,1,-3),nrow=4,ncol = 4)
Matrix::rankMatrix(A)[1]
## [1] 4

(2) Given an m x n matrix where m > n, what can be the maximum rank? The minimum rank, assuming that the matrix is non-zero?

Maximum rank = m
Minmum rank = 1

For an m x n Matrix :

If m is less than n…the maximum rank of the matrix is m

If m is greater than n…the maximum rank of the matrix is n

Matrix rank can only be 0 if there are 0 non-zero elements, otherwise a minimum rank of 1

(3) What is the rank of matrix B?

\[ B = \left(\begin{array}{ccc} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2 \end{array}\right) \]

Eyeballing this one; 1 is a multiple of 2 & 3, and 2 is a multiple of 6 & 4…This matrix has a rank of 1.

(4) Compute the eigenvalues and eigenvectors of the matrix C. You’ll need to show your work. You’ll need to write out the characterstic polynomial and show your solution.

\[ C = \left(\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{array}\right) \]

Step 1 : Find eigenvalues of A by identifying roots of the characteristic polynomial equation.

\[ det(C-\lambda\cdot I) = 0 \]

\[ det ( \left(\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{array}\right) - \lambda \left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right) ) = -(\lambda-6)(\lambda-4)(\lambda-1) \]

Eigenvalues identified : [6,4,1]

Step 2 : Find eigenvectors by solving the system of equations that arise by plugging in our newly found eigenvalues…One by one.

\[\lambda = 1\]

\[ det ( \left(\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{array}\right) - 1 \left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right) ) \longrightarrow \left(\begin{array}{ccc} 0 & 2 & 3 \\ 0 & 3 & 5 \\ 0 & 0 & 5 \end{array}\right) \cdot \left(\begin{array}{c} x\\ y \\ z \end{array}\right) = \left(\begin{array}{c} 0\\ 0 \\ 0 \end{array}\right) \]

\[ \left(\begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right) \cdot \left(\begin{array}{c} x\\ y \\ z \end{array}\right) = \left(\begin{array}{c} 0\\ 0 \\ 0 \end{array}\right) \rightarrow x \neq 0 \] \[ \lambda = 1 \longrightarrow \left(\begin{array}{c} 1\\ 0 \\ 0 \end{array}\right) \]

\[\lambda = 6\]

\[ det ( \left(\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{array}\right) - 6 \left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right) ) \longrightarrow \left(\begin{array}{ccc} -5 & 2 & 3 \\ 0 & -2 & 5 \\ 0 & 0 & 0 \end{array}\right) \cdot \left(\begin{array}{c} x\\ y \\ z \end{array}\right) = \left(\begin{array}{c} 0\\ 0 \\ 0 \end{array}\right) \]

\[ \left(\begin{array}{ccc} 1 & 0 & -8/5 \\ 0 & 1 & -5/2 \\ 0 & 0 & 0 \end{array}\right) \cdot \left(\begin{array}{c} x\\ y \\ z \end{array}\right) = \left(\begin{array}{c} 0\\ 0 \\ 0 \end{array}\right) \rightarrow z \neq 0 \] \[ \lambda = 6 \longrightarrow \left(\begin{array}{c} 8/5\\ 5/2 \\ 1 \end{array}\right) \]

\[\lambda = 4\]

\[ det ( \left(\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{array}\right) - 4 \left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right) ) \longrightarrow \left(\begin{array}{ccc} -3 & 2 & 3 \\ 0 & -2 & 5 \\ 0 & 0 & 2 \end{array}\right) \cdot \left(\begin{array}{c} x\\ y \\ z \end{array}\right) = \left(\begin{array}{c} 0\\ 0 \\ 0 \end{array}\right) \]

\[ \left(\begin{array}{ccc} 1 & -2/3 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right) \cdot \left(\begin{array}{c} x\\ y \\ z \end{array}\right) = \left(\begin{array}{c} 0\\ 0 \\ 0 \end{array}\right) \rightarrow y \neq 0 \] \[ \lambda = 4 \longrightarrow \left(\begin{array}{c} 2/3\\ 1 \\ 0 \end{array}\right) \]

So all our eigenvectors would be…

\[ v_1,v_2,v_3 = \left(\begin{array}{c} 1\\ 0 \\ 0 \end{array}\right) \left(\begin{array}{c} 8/5\\ 5/2 \\ 1 \end{array}\right) \left(\begin{array}{c} 2/3\\ 1 \\ 0 \end{array}\right) \]

or

\[ v_1,v_2,v_3 = \left(\begin{array}{c} 1\\ 0 \\ 0 \end{array}\right) \left(\begin{array}{c} 16\\ 25 \\ 10 \end{array}\right) \left(\begin{array}{c} 2\\ 3 \\ 0 \end{array}\right) \]