Rank is the number of independent columns. that is a column cant be a factor of another or a combination of 2 rows. In order to find the rank of the Matrix, the matrix needs to be tranformed into row echelon form and count the no of non zero diagonal elements.
Lets tranform A into its row echelon form.
## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4
## [2,] -1 0 1 3
## [3,] 0 1 -2 1
## [4,] 5 4 -2 -3## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4
## [2,] -1 0 1 3
## [3,] 0 1 -2 1
## [4,] 5 4 -2 -3## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4
## [2,] 0 2 4 7
## [3,] 0 1 -2 1
## [4,] 0 -6 -17 -23## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4
## [2,] 0 2 4 7
## [3,] 0 0 -8 -5
## [4,] 0 0 -5 -2## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4
## [2,] 0 2 4 7
## [3,] 0 0 -8 -5
## [4,] 0 0 0 -9You can see that the number of non-zero diagonal element is 4. Thus rank of Matrix A is 4.
Maximun Rank = n Minimum Rank = 1
## [,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 0The number of non zero diagonal elements in the row echelon form is 1. So the rank is 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.
\[\mathbf{A} = \left[\begin{array} {rrr} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{array}\right] \]
First, To find the eigen value, we need to construct the characteristic polynomial using below equation.
\[det(A−λI)=0\]
\[det(\left[\begin{array} {rrr} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{array}\right] - \left[\begin{array} {rrr} λ & 0 & 0 \\ 0 & λ & 0 \\ 0 & 0 & λ \end{array}\right])=0 \] ### Equation(1) After subtracting the Iλ matrix, you will get below equation, we
\[det(\left[\begin{array} {rrr} 1-λ & 2 & 3 \\ 0 & 4-λ & 5 \\ 0 & 0 & 6-λ \end{array} \right] )=0 \]
After determining the determinent of above matrix,
\[(1-λ)((4-λ)(6-λ)-0) - 2(0) - 3 (0) = 0\]
This becomes below equation.
\[(1-λ)(4-λ)(6-λ) = 0 \] Now, we got the 3 Eigen Values as given below.
\[ λ = 1, λ =4 , λ = 6 \]
Now inorder to find the eigen matrix, we need to substitute λ in above Equation(1)
\[\left[\begin{array} {rrr} 0 & 2 & 3 \\ 0 & 3 & 5 \\ 0 & 0 & 5 \end{array}\right] * \left[\begin{array} {rrr} v1 \\ v2 \\ v3 \end{array} \right] = \left[\begin{array} {rrr} 0 \\ 0 \\ 0 \end{array} \right] \]
We get 3 equations
\[2v2 + 3v3 = 0 \]
\[3v2 + 5v3 = 0, \] \[5v3 =0, v3 = 0\]
Substituting v3 = 0 in above equations, we get v2 = 0 and v1 can take any value. Lets assume v1 be 1
\[\left[\begin{array} {rrr} v1 \\ v2 \\ v3 \end{array} \right] =\left[\begin{array} {rrr} 1 \\ 0 \\ 0 \end{array} \right] \]
\[\left[\begin{array} {rrr} -3 & 2 & 3 \\ 0 & 0 & 5 \\ 0 & 0 & 2 \end{array}\right] * \left[\begin{array} {rrr} v1 \\ v2 \\ v3 \end{array} \right] = \left[\begin{array} {rrr} 0 \\ 0 \\ 0 \end{array} \right] \]
We get 3 equations
\[-3v1 + 2v2 + 3v3 = 0 \]
\[5v3 = 0,\] \[2v3 =0, v3 = 0\]
Substituting v3 = 0 in above equations, we dont have v2. Lets assume v2 be 1, then v1 will ve -2/3
\[\left[\begin{array} {rrr} v1 \\ v2 \\ v3 \end{array} \right] =\left[\begin{array} {rrr} -2/3 \\ 1 \\ 0 \end{array} \right] \]
\[\left[\begin{array} {rrr} -5 & 2 & 3 \\ 0 & -2 & 5 \\ 0 & 0 & 0 \end{array}\right] * \left[\begin{array} {rrr} v1 \\ v2 \\ v3 \end{array} \right] = \left[\begin{array} {rrr} 0 \\ 0 \\ 0 \end{array} \right] \]
We get 3 equations
\[-5v1 + 2v2 + 3v3 = 0 \]
\[-2v2 + 5v3 = 0,\] \[2v3 =0, v3 = 0\]
Lets assume v2 be 1, v3 = 2/5 and v1 = - 16/25
\[\left[\begin{array} {rrr} v1 \\ v2 \\ v3 \end{array} \right] =\left[\begin{array} {rrr} -16/25 \\ 1 \\ 2/5 \end{array} \right] \]