\[
A = \begin{bmatrix}
1 & 2 & 3 & 4 \\
-2 & 0 & 1 & 3 \\
0 & 1 & -2 & 1 \\
5 & 4 & -2 & -3 \\
\end{bmatrix}
\]
Solution
A <- matrix(c(1,-2,0,5,2,0,1,4,3,1,-2,-2,4,3,1,-3), ncol=4)
A## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4
## [2,] -2 0 1 3
## [3,] 0 1 -2 1
## [4,] 5 4 -2 -3# rank of the matrix, A
rankA <- qr(A)$rank
rankA## [1] 4Therefore the rank of the matrix is 4.
The minimum rank, assuming that the matrix is non-zero?
Solution
For a non-zero matrix, the minimum and maximum rank is given by:
Minimum rank = 1
Maximum rank = n
\[ B = \begin{bmatrix} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2 \\ \end{bmatrix} \]
Solution Looking at the matrix, the rank is 1 given than there is only one linearly independent row for the matrix, B. The rank of a matrix is simply the number of linearly independent rows of the matrix. Given that matrix B has rows 2 and 3 as multiples of row 1, it means that there is only one linearly independent row for the matrix and thus the rank is 1.
# checking the matrix using R code:
B <- matrix(c(1,3,2,2,6,4,1,3,2), nrow = 3)
B## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 3 6 3
## [3,] 2 4 2# rank of matrix B
rankB <- qr(B)$rank
rankB## [1] 1Compute 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.
\[ A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \\ \end{bmatrix} \]
Solution
The eigenvalue(s) is \(\lambda\) is
defined as \(Ax = \lambda x\) where
\(x\) is the eigen vector associated
with the eigenvalue \(\lambda\).
\(|A-\lambda I| = 0\) where \(I\) is a \(3
\times 3\) identity square matrix.
Therefore: \[ \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \\ \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} = \begin{bmatrix} 1-\lambda & 2 & 3 \\ 0 & 4-\lambda & 5 \\ 0 & 0 & 6-\lambda \\ \end{bmatrix} \]
Find the determinant of the resulting matrix and set it equal to zero to find the values of \(\lambda\)
\[ det \begin{bmatrix} 1-\lambda & 2 & 3 \\ 0 & 4-\lambda & 5 \\ 0 & 0 & 6-\lambda \\ \end{bmatrix} = 0 \]
Multiplying along the first column to take advantage of the zeros in
the first columns:
\((1-\lambda)[(4-\lambda)(6-\lambda) - 0] =
0\)
\(=> (1-\lambda)(4-\lambda)(6-\lambda) =
0\) This is the characteristic polynomial and it’s better to keep
it unexpanded since it makes it a lot easier to see the values of \(\lambda\).
Therefore \(\lambda = 1, 4,
6\).
Thus, the eigenvalues are \(1, 4,\) and
\(6\).
# verify using R
A <- matrix(c(1, 0, 0, 2, 4, 0, 3, 5, 6), nrow=3, ncol=3)
eigen_values <- eigen(A)
eigen_values$values## [1] 6 4 1 Eigen-vector for \(\lambda =
1\);
\((A - \lambda I)x = 0\); \(x\) is the eigen-vector
For \(\lambda = 1\);
\[
\begin{bmatrix}
1-1 & 2 & 3 \\
0 & 4-1 & 5 \\
0 & 0 & 6-1 \\
\end{bmatrix}
\times
\begin{bmatrix}
u \\
v \\
w \\
\end{bmatrix} = 0
\]
\[
\begin{bmatrix}
0 & 2 & 3 \\
0 & 3 & 5 \\
0 & 0 & 5 \\
\end{bmatrix}
\times
\begin{bmatrix}
u \\
v \\
w \\
\end{bmatrix} = 0;
\] \(5w = 0; w = 0\); \(3v + 5w = 0; 3v + 5(0) = 0; v = 0\); \(0u + 2v + 3w = 0; 0 + 0 + 0 = 0;
0=0\)
This gives a trivial solution for x;
Thus the eigen-vector associated with \(\lambda = 1\) is given by:
\[
x = u
\begin{bmatrix}
1 \\
0 \\
0 \\
\end{bmatrix}
\]
\(\forall\) \(u\) \(\epsilon\) \(\mathbb{C}\)
Eigen-vector For \(\lambda =
4\)
\[
\begin{bmatrix}
1-4 & 2 & 3 \\
0 & 4-4 & 5 \\
0 & 0 & 6-4 \\
\end{bmatrix}
\times
\begin{bmatrix}
u \\
v \\
w \\
\end{bmatrix} = 0
\] \[
\begin{bmatrix}
-3 & 2 & 3 \\
0 & 0 & 5 \\
0 & 0 & 2 \\
\end{bmatrix}
\times
\begin{bmatrix}
u \\
v \\
w \\
\end{bmatrix} = 0;
\]
\(2w = 0; w=0\); \(0v + 2w = 0; 0v + 0 + 0 = 0; 0 = 0\) =>
\(v\) appears to result in a trivial
solution.
\(-3u + 2v + 3w = 0; -3u + 2v + 0 =
0\); \(u = \frac{2}{3}v\).
Thus the eigen-vector associated with \(\lambda = 4\) is given by:
\[ x = v \begin{bmatrix} \frac{2}{3} \\ 1 \\ 0 \\ \end{bmatrix} \]
\(\forall\) \(v\) \(\epsilon\) \(\mathbb{C}\)
Eigen-vector For \(\lambda =
6\)
\[
\begin{bmatrix}
1-6 & 2 & 3 \\
0 & 4-6 & 5 \\
0 & 0 & 6-6 \\
\end{bmatrix}
\times
\begin{bmatrix}
u \\
v \\
w \\
\end{bmatrix} = 0
\]
\[
\begin{bmatrix}
-5 & 2 & 3 \\
0 & -2 & 5 \\
0 & 0 & 0 \\
\end{bmatrix}
\times
\begin{bmatrix}
u \\
v \\
w \\
\end{bmatrix} = 0;
\]
Clearly, the last row provides a trivial solution.
\(-2v + 5w = 0; v = \frac{5}{2}w; -5u + 2v +
3w = 0\); \(-5u + 2\times\frac{5}{2}w +
3w = 0\); \(-5u + 5w + 3w = 0\);
\(u = \frac{8}{5}w\)
Thus the eigen-vector associated with \(\lambda = 6\) is given by:
\[ x = w \begin{bmatrix} \frac{8}{5} \\ \frac{5}{2} \\ 1 \\ \end{bmatrix} \]
\(\forall\) \(w\) \(\epsilon\) \(\mathbb{C}\)
Hence, the eigen vectors corresponding to \(\lambda = 1, 4, 6\) are:
\[ x = u \begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix}; v \begin{bmatrix} \frac{2}{3} \\ 1 \\ 0 \\ \end{bmatrix}; w \begin{bmatrix} \frac{8}{5} \\ \frac{5}{2} \\ 1 \\ \end{bmatrix} \] \(\forall\) \(u, v, w\) \(\epsilon\) \(\mathbb{C}\) respectively.