matA <- matrix(c(1,-1,0,5,2,0,1,4,3,1,-2,-2,4,3,1,3), nrow = 4)
matA## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4
## [2,] -1 0 1 3
## [3,] 0 1 -2 1
## [4,] 5 4 -2 3matALower <- lu(matA)$U
matALower## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4.000
## [2,] 0 2 4 7.000
## [3,] 0 0 -4 -2.500
## [4,] 0 0 0 7.125By reducing the matrix to upper triangular form, we see that there are 4 pivot columns. Therefore the Rank=4
For a \(mxn\) matrix where \(m>n\), the maximum rank(\(r\)) is \(r=n\) and mimimun rank is \(r=1\) assuming the matrix is non-zero.
What is the rank of Matrix B?
matB <- matrix(c(1,2,1,3,6,3,2,4,2), nrow = 3, byrow = T)
matB## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 3 6 3
## [3,] 2 4 2Row-reduced echelon form
rref(matB)## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 0 0 0
## [3,] 0 0 0There is only 1 pivot column so the rank=1
Computer the eigenvalues and eigenvectors of the Matrix A. Eigenvalues of a matrix are roots of characteristic polynomials. \(det(A-\lambda I)=0\)
Matrix A \[ \left| \begin{array}{ccc} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{array}\right| - \left| \begin{array}{cccc} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{array} \right| = \left| \begin{array}{ccc} 1-\lambda & 2 & 3 \\ 0 & 4-\lambda & 5 \\ 0 & 0 & 6-\lambda \end{array}\right| \]
Finding the determinent: \((1 -\lambda)(4-\lambda)(6-\lambda)=0\)
Since the matrix is already in upper triangular form, the determinent is the multiplication of the diagonal. Finding the roots are easy. The eigenvalues are:
\(\lambda=1\)
\(\lambda=4\)
\(\lambda=6\)
Characteristic Polynomial is: \(-\lambda^{3}+11\lambda^{2}-34\lambda+24=0\)
To find the eigenvectors, we substitute in the eigenvalues:
The eigenspace for eigenvalue \(\lambda = 1\): \[ E_1 = \left| \begin{array}{ccc} 0 & 2 & 3 \\ 0 & 3 & 5 \\ 0 & 0 & 5 \end{array} \right| \to RREF \to \left| \begin{array}{ccc} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0\end{array} \right| \left| \begin{array}{c} v_{1}\\ v_{2}\\ v_{3} \end{array} \right|= \left| \begin{array}{c} 0\\ 0\\ 0 \end{array} \right| \] \(v_{2}= 0\)
\(v_{3}= 0\)
Let \(v_{1}=1\). This gives us an eigenvector of \[ \left| \begin{array}{c} v_{1}\\ v_{2}\\ v_{3} \end{array} \right| \to \left| \begin{array}{c} 1\\ 0\\ 0 \end{array} \right| \]
The eigenspace for eigenvalue \(\lambda = 4\): \[ E_4 = \left| \begin{array}{ccc} -3 & 2 & 3 \\ 0 & 0 & 5 \\ 0 & 0 & 2 \end{array} \right| \to RREF \to \left| \begin{array}{ccc} -3 & 2 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0\end{array} \right| \left| \begin{array}{c} v_{1}\\ v_{2}\\ v_{3} \end{array} \right|= \left| \begin{array}{c} 0\\ 0\\ 0 \end{array} \right| \] \(-3v_{1} + 2v_{2}=0 \to v_{1}=(2/3)v_{2}\)
\(v_{3}=0\)
Let \(v_{2}=3\). This gives us an eigenvector of \[ \left| \begin{array}{c} v_{1}\\ v_{2}\\ v_{3} \end{array} \right| \to \left| \begin{array}{c} 2\\ 3\\ 0 \end{array} \right| \]
The eigenspace for eigenvalue \(\lambda = 6\):
\[ E_6 = \left| \begin{array}{ccc} -5 & 0 & 8 \\ 0 & -2 & 5 \\ 0 & 0 & 0 \end{array} \right| \to RREF \to \left| \begin{array}{ccc} -3 & 2 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0\end{array} \right| \left| \begin{array}{c} v_{1}\\ v_{2}\\ v_{3} \end{array} \right|= \left| \begin{array}{c} 0\\ 0\\ 0 \end{array} \right| \]
\(-5v_{1} + 8v_{3}=0 \to v_{1}=(8/5)v_{3}\)
\(-2v_{2} + 5v_{3}=0 \to v_{2}=(5/2)v_{3}\)
Since the common denominator of 2 and 5 is 10. I set \(v_{3}=10\) to make the math cleaner.
This gives us an eigenvector of \[ \left| \begin{array}{c} v_{1}\\ v_{2}\\ v_{3} \end{array} \right| \to \left| \begin{array}{c} 16\\ 25\\ 10 \end{array} \right| \]