\(A=\begin{bmatrix} 1 & 2 & 3 & 4\\ -1 & 0 & 1 & 3\\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3\end{bmatrix}\)
First, let’s find the reduced row echelon form of the matrix:
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 -3rref(A)## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 0
## [2,] 0 1 0 0
## [3,] 0 0 1 0
## [4,] 0 0 0 1All four columns of the matrix A are linearly independent; all four columns are pivot columns. So,
\(rank(A)=4\)
\(m\times n\) matrix has \(m\) rows and \(n\) columns, so the number of rows is more than the number of columns.
Since the \(rank(A) = dim(C(A))\) which is the dimension of the column space of (A), then the maximum rank is the number of columns of the matrix, namely \(n\)
For the same reason; there are \(n\) columns in the matrix, then the rank cannot be greater than \(n\), and also the minimum rank will be the number of linearly independent columns, so the minimum rank is \(n\).
Let’s use the square matrix A from question (1) as an example to find the minimum and the maximum \(rank(A)\).
max_rank <- min(dim(A))
max_rank## [1] 4rref(A)## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 0
## [2,] 0 1 0 0
## [3,] 0 0 1 0
## [4,] 0 0 0 1Based on the reduced row echelon form, there are 4 linearly independent columns, so \(min(rank(A))=4\)
To demonstrate the \(max\) and the \(min\) rank, let’s use \(4\times 5\) matrix Z:
Z <- matrix(c(1, 0, -1, 0, 4, 2, 1, 0, 0, 9, -1, 2, 5, 1, -5, 1, -1, -3, -2, 9), nrow =4, byrow=TRUE)
Z## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 0 -1 0 4
## [2,] 2 1 0 0 9
## [3,] -1 2 5 1 -5
## [4,] 1 -1 -3 -2 9max_rank <- min(dim(Z))
max_rank## [1] 4rref(Z)## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 0 -1 0 4
## [2,] 0 1 2 0 1
## [3,] 0 0 0 1 -3
## [4,] 0 0 0 0 0The number of the pivot columns (linearly independent) is 3 which less than 4. So the minimum rank is less than 4, 3 in this case.
\(B=\begin{bmatrix} 1 & 2 & 1\\ 3 & 6 & 3\\ 2 & 4 & 2 \end{bmatrix}\)
We find the \(rref(B)\) first:
B <- matrix(c(1, 2, 1 , 3, 6, 3, 2, 4, 2), nrow = 3, byrow = TRUE)
B## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 3 6 3
## [3,] 2 4 2rref(B)## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 0 0 0
## [3,] 0 0 0There is only one pivot column, the first column, so:
\(rank(B)=1\)
Compute the eigenvalues and eigenvectors of the matrix \(A=\begin{bmatrix}1 &2 & 3\\ 1 & 4 & 5\\ 0 & 0 & 6 \end{bmatrix}\)
I will name it \(A_1\) because I used matrix A for problem set 1:
A_1 <- matrix(c(1, 2, 3, 0, 4, 5, 0, 0, 6), nrow=3, byrow=TRUE)
A_1## [,1] [,2] [,3]
## [1,] 1 2 3
## [2,] 0 4 5
## [3,] 0 0 6To compute the eigenvalues and the eigenvectors, we need to find the roots/solution of \(det(\lambda I_3 - A)=0\)
First, let’s find the matrix \(\lambda I_3 -A\):
\(\Rightarrow \lambda I_3 - A = \left(\begin{bmatrix} \lambda & 0 & 0\\ 0 & \lambda & 0\\ 0 & 0 & \lambda\end{bmatrix}-\begin{bmatrix} 1 & 2 & 3\\ 0 & 4 & 5\\ 0 & 0 & 6\end{bmatrix}\right)\)
\(\Rightarrow det(\lambda I_3 - A)= det\begin{bmatrix} \lambda-1 & -2 & -3\\ 0 & \lambda-4 & -5\\ 0 & 0 & \lambda-6\end{bmatrix}\)
\(\Rightarrow det(\lambda I_3 - A)= \left(\lambda-1\right) [\left(\lambda -4\right) \left(\lambda - 6\right)]\)
\(\Rightarrow det(\lambda I_3 - A)= \left(\lambda - 1\right) \left(\lambda^2 - 6 \lambda - 4 \lambda + 24\right)\)
\(\Rightarrow det(\lambda I_3 - A)= \left(\lambda - 1\right) \left(\lambda^2 - 10 \lambda + 24\right)\)
\(\Rightarrow det(\lambda I_3 -A) = \lambda^3 - 10 \lambda^2 + 24 \lambda - \lambda^2 + 10 \lambda - 24\)
\(\Rightarrow det(\lambda I_3 - A) = \lambda^3 - 11 \lambda^2 + 34 \lambda -24\)
Let’s double check if I get the characteristic polynomial correct, with the following function:
char_poly_1 <- charpoly(A_1)
char_poly_1## [1] 1 -11 34 -24The characteristic polynomial \(p(x) = \lambda^3 - 11 \lambda^2 + 34 \lambda - 24\) \(\surd\) >
Second, let’s solve \(p(x) = 0\) to find the eigenvalues of matrix \(A_1\),
\(\lambda^3 -11 \lambda^2 + 34 \lambda - 24 = 0\)
In factors form, the characteristic polynomial was; \((\lambda -1)(\lambda -4)(\lambda - 6) = 0\)
\(\Rightarrow \lambda -1 = 0\) OR \(\lambda - 4 = 0\) OR \(\lambda - 6 =0\)
\(\Rightarrow\) The roots of \(p(x)\) are:
\(\lambda = 1\)
\(\lambda = 4\)
\(\lambda = 6\)
To make sure that those values are the roots of \(p(x)\), we can use this function to compute each of the values of \(\lambda\) into the equation \(p(x)=0\):
Compute_value <- function(lambda) {
value <- lambda^3 - 11*lambda^2 + 34*lambda -24
return(value)
}
value1 <- Compute_value(1)
print(value1)## [1] 0value2 <- Compute_value (4)
print(value2)## [1] 0value3 <- Compute_value(6)
print(value3)## [1] 0So, \(\lambda_1 = 1\) , \(\lambda_2 = 4\) and \(\lambda_3 = 6\) are the eigenvalues for matrix \(A_1 = \begin{bmatrix} 1 & 2 & 3\\ 0 & 4 & 5\\ 0 & 0 & 6\end{bmatrix}\)
Now, let’s find the eigenspace (\(E_\lambda\)), the set of vectors that satisfy the equation:
\(A \vec{v} = \lambda \vec{v}\) \(\Rightarrow \vec{0} = \lambda \vec{v} - A \vec{v}\) \(\Rightarrow \vec{0} = (\lambda I_n - A) \vec{v}\)
So, for each \(\lambda\), we are going to find the null space(N) of \((\lambda I_n - A)\)
\(\bigstar\) The eigenpace for \(\lambda=1\)
\(\underline{\lambda = 1}\): \(E_1 = N(\lambda I_3 - A)\)
\(\Rightarrow E_1 = N\left( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix} \right)\)
\(\Rightarrow E_1 = N \left(\begin{bmatrix} 0 & -2 & -3 \\ 0 & -3 & -5 \\ 0 & 0 & -5 \end{bmatrix} \right)\)
\(\Rightarrow \begin{bmatrix} 0 & -2 & -3 \\ 0 & -3 & -5 \\ 0 & 0 & -5 \end{bmatrix} \vec{v} = \vec{0}\)
Using elementary rows operations on the above matrix, let’s find the \(rref\);
\(\begin{bmatrix} 0 & -2 & -3 \\ 0 & -3 & -5 \\ 0 & 0 & -5 \end{bmatrix}\) \(\Rightarrow R_1 \rightarrow -R_2 + R_1\) \(\begin{bmatrix} 0 & 1 & 2 \\ 0 & -3 & -5 \\ 0 & 0 & -5 \end{bmatrix}\)
\(R_2 \rightarrow R_2 +3 R_1 \begin{bmatrix} 0 & 1 & 2 \\ 0 & 0 & 1 \\ 0 & 0 & -5 \end{bmatrix}\)
\[ \begin{array}{cc} R_1 \rightarrow R_1 - 2R_2 \\ R_3 \rightarrow R_3 + 5R_2 \\ \end{array} \Rightarrow \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \]
Using the R_function \(rref()\), let’s check our answers:
E_1Matrix <- matrix(c(0, -2, -3, 0, -3, -5, 0, 0, -5), nrow = 3, byrow = TRUE)
E_1Matrix## [,1] [,2] [,3]
## [1,] 0 -2 -3
## [2,] 0 -3 -5
## [3,] 0 0 -5rref(E_1Matrix)## [,1] [,2] [,3]
## [1,] 0 1 0
## [2,] 0 0 1
## [3,] 0 0 0This means that \(v_2 = 0\) and \(v_3=0\)
So, the eigenvectors corresponding to \(\lambda = 1\) are any multiple of \(\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\)
\(\Rightarrow E_{\lambda=1} = \left\{\begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = t \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\right\}\)
\(\bigstar\) The eigenpace for \(\lambda=4\)
\(\underline{\lambda = 4}\): \(E_4 = N(\lambda I_3 - A)\)
\(\Rightarrow E_4 = N\left( \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix} - \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix} \right)\)
\(\Rightarrow E_4 = N \left(\begin{bmatrix} 3 & -2 & -3 \\ 0 & 0 & -5 \\ 0 & 0 & -2 \end{bmatrix} \right)\)
\(\Rightarrow \begin{bmatrix} 3 & -2 & -3 \\ 0 & 0 & -5 \\ 0 & 0 & -2 \end{bmatrix} \vec{v} = \vec{0}\)
Using elementary rows operations on the above matrix, let’s find the \(rref\);
\[ \begin{array}{ccc} R_1 \rightarrow \frac{R_1}{3} \\ R_2 \rightarrow \frac{R_2}{-5} \\ R_3 \rightarrow \frac{R_3}{-2} \\ \end{array} \Rightarrow \begin{bmatrix} 1 & \frac{-2}{3} & -1 \\ 0 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix} \] \[ \begin{array}{cc} R_1 \rightarrow R_1 + R_2 \\ R_3 \rightarrow R_3 - R_2 \\ \end{array} \Rightarrow \begin{bmatrix} 1 & \frac{-2}{3} & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \]
Using the R_function \(rref()\), let’s check our answers:
#rref in decimals
E_4Matrix <- matrix(c(3, -2, -3, 0, 0, -5, 0, 0, -2), nrow = 3, byrow = TRUE)
E_4Matrix## [,1] [,2] [,3]
## [1,] 3 -2 -3
## [2,] 0 0 -5
## [3,] 0 0 -2rref(E_4Matrix)## [,1] [,2] [,3]
## [1,] 1 -0.6666667 0
## [2,] 0 0.0000000 1
## [3,] 0 0.0000000 0#rref in fractions
E_4Matrix <- matrix(c(3, -2, -3, 0, 0, -5, 0, 0, -2), nrow = 3, byrow = TRUE)
fractions(rref(E_4Matrix))## [,1] [,2] [,3]
## [1,] 1 -2/3 0
## [2,] 0 0 1
## [3,] 0 0 0This means that \(v_1 - \frac{2}{3} v_2 =0\) and \(V_3 =0\)
So, the eigenvectors corresponding to \(\lambda = 4\) are any multiple of \(\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\)
\(\Rightarrow E_{\lambda=4} = \left\{\begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = t \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\right\}\)
\(\bigstar\) The eigenpace for \(\lambda=6\)
\(\underline{\lambda = 6}\): \(E_6 = N(\lambda I_3 - A)\)
\(\Rightarrow E_6 = N\left( \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} - \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix} \right)\)
\(\Rightarrow E_6 = N \left(\begin{bmatrix} 5 & -2 & -3 \\ 0 & 2 & -5 \\ 0 & 0 & 0 \end{bmatrix} \right)\)
\(\Rightarrow \begin{bmatrix} 5 & -2 & -3 \\ 0 & 2 & -5 \\ 0 & 0 & 0 \end{bmatrix} \vec{v} = \vec{0}\)
Using elementary rows operations on the above matrix, let’s find the \(rref\);
\[ \begin{array}{cc} R_1 \rightarrow \frac{R_1}{5} \\ R_2 \rightarrow \frac{R_2}{2} \\ \end{array} \Rightarrow \begin{bmatrix} 1 & \frac{-2}{5} & \frac{-3}{5} \\ 0 & 1 & \frac{-5}{2} \\ 0 & 0 & 0 \end{bmatrix} \] \(\Rightarrow R_1 \rightarrow R_1 + \frac{2}{5} R_2\) \(\Rightarrow \begin{bmatrix} 1 & 0 & \frac{-8}{5} \\ 0 & 1 & \frac{-5}{2} \\ 0 & 0 & 0 \end{bmatrix}\)
Using the R_function \(rref()\), let’s check our answers:
#rref in decimals
E_6Matrix <- matrix(c(5, -2, -3, 0, 2, -5, 0, 0, 0), nrow = 3, byrow = TRUE)
E_6Matrix## [,1] [,2] [,3]
## [1,] 5 -2 -3
## [2,] 0 2 -5
## [3,] 0 0 0rref(E_6Matrix)## [,1] [,2] [,3]
## [1,] 1 0 -1.6
## [2,] 0 1 -2.5
## [3,] 0 0 0.0#rref in fractions
E_6Matrix <- matrix(c(5, -2, -3, 0, 2, -5, 0, 0, 0), nrow = 3, byrow = TRUE)
fractions(rref(E_6Matrix))## [,1] [,2] [,3]
## [1,] 1 0 -8/5
## [2,] 0 1 -5/2
## [3,] 0 0 0This means that \(v_1 - \frac{8}{5} v_3 =0\) and \(v_2 - \frac{5}{2} v_3 =0\)
$v_1 = v_3 $ and \(v_2 = \frac{5}{2} v_3\)
So, the eigenvectors corresponding to \(\lambda = 6\) are any multiple of \(\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\)
let \(v_3 =t\): \(\Rightarrow E_{\lambda=6} = \left\{\begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = t \begin{bmatrix} \frac{8}{5} \\ \frac{5}{2} \\ 1 \end{bmatrix}\right\}\)