###Problem Set 1
\[ A = \begin{vmatrix} 1 & 2 & 3 & 4 \\ -1 & 0 & 1 & 3 \\ 0 & 1 & -2 & 1 \\ 5 & 4 & -2 & -3 \end{vmatrix} \]
Using the function Matrix::rankMatrix, we find that the rank of this matrix is 4. This indicates that all four rows of the Matrix \(A\) are linearly independent of one another.
A <- matrix(c(1, -1, 0, 5, 2, 0, 1, 4, 3, 1, -2, -2, 4, 3, 1, 3), nrow=4, ncol=4, byrow=FALSE)
rankA <- rankMatrix(A)
rankA[1]## [1] 4Since the maximum rank of an \(m \times n\) matrix cannot be greater than \(min(m, n)\), in the case where \(m > n\) , the rank can be no greater than \(n\).
The minimum rank of a nonzero matrix is 1.
\[ B = \begin{vmatrix} 1 & 2 & 1 \\ 3 & 6 & 3 \\ 2 & 4 & 2 \end{vmatrix} \]
Using the function Matrix::rankMatrix, we find that the rank of this matrix is 1. We can verify this by noticing that rows 1 and 3 are both multiples of row 2. Row 1 goes into row 2 three times, while row 3 goes into row 2 one and one half times.
B <- matrix(c(1, 2, 1, 3, 6, 3, 2, 4, 2), nrow=3, ncol=3, byrow=TRUE)
rankB <- rankMatrix(B)
rankB[1]## [1] 1###Problem Set 2
Compute the eigenvalues and eigenvecotrs of the matrix \(A\):
\[A = \begin{vmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{vmatrix}\]
We need to find the characteristic polynomial \(p_A(\lambda) = det(A - \lambda I)\).
\[A - \lambda I = \begin{vmatrix} 1 - \lambda & 2 & 3 \\ 0 & 4 - \lambda& 5 \\ 0 & 0 & 6 - \lambda \end{vmatrix}\]
\[p_A(\lambda) = det(A - \lambda I) = det\left(\begin{vmatrix} 1 - \lambda & 2 & 3 \\ 0 & 4 - \lambda& 5 \\ 0 & 0 & 6 - \lambda \end{vmatrix}\right) \]
Since column 1 has only 1 nonzero value equal to \(1 - \lambda\), we’ll expand along it to reduce the calculations necessary. This give us:
\[p_A(\lambda) = (1 - \lambda) \times det\left(\begin{vmatrix} 4 - \lambda& 5 \\ 0 & 6 - \lambda \end{vmatrix} \right) \]
Since we now have a \(2 \times 2\) matrix, its determinant is easy to find following the formula \(ad - bc\). What’s more, \(bc = 5(0) = 0\) in this case, so:
\[\begin{align}p_A(\lambda) &= (1 - \lambda)((4 - \lambda)(6 - \lambda)) \\ \\ &= (1 - \lambda)(\lambda^2 - 10\lambda + 24) \\ \\ &= -\lambda^3 + 10\lambda^2 - 24\lambda + \lambda^2 - 10\lambda +24 \\ \\ &= -\lambda^3 + 11\lambda^2 - 34\lambda + 24\end{align}\]
We will find the eigenvalues by setting this polynomial equal to zero and finding its roots. \[\begin{align} 0 &= -\lambda^3 + 11\lambda^2 - 34\lambda + 24 \\ \\ &= \lambda^3 - 11\lambda^2 + 34\lambda - 24 \end{align}\]
We know that 1 is a root because:
\[\begin{align} 1^3 - 11(1)^2 + 34(1) - 24 = 0 \end{align}\]
So, let’s factor out \(\lambda - 1\):
\[\begin{align} &= \lambda^3 - 11\lambda^2 + 34\lambda - 24 \\ \\ &= (\lambda - 1) (\lambda^2 + 10\lambda + 24) \\ \\ &= (\lambda - 1) (\lambda + 4) (\lambda + 6) \end{align}\]
So, -4 and -6 are also roots and therefore are also eigenvalues.
Now, to find the eigenvectors, we need to plug these values back into the equation for eigen, \((A - \lambda I) v = 0\).
Doing so for 1 and 4 yields only zero vectors - not very interesting. But when we plug in 6…
\[\left(\begin{vmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{vmatrix} - \begin{vmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{vmatrix} \right)\begin{vmatrix} v_1 \\ v_2 \\ v_3 \end{vmatrix}= 0\]
\[\left(\begin{vmatrix} -5 & 2 & 3 \\ 0 & -2 & 5 \\ 0 & 0 & 0 \end{vmatrix}\right)\begin{vmatrix} v_1 \\ v_2 \\ v_3 \end{vmatrix}= 0\]
\[\begin{align} -5 v_1 + 2 v_2 + 3 v_3 &= 0\\ -2 v_2 + 5 v_3 &= 0\\ \end{align}\]
Subtracting the second from the first row yields:
\[\begin{align} -5v_1 + 3 v_3 &= 0\\ \end{align}\]
or:
\[\begin{align} v_1 = \frac{3}{5} v_3 \\ \end{align}\]
So therefore, the only eigenvector beside \(\begin{vmatrix}0 \\ 0\\ 0\\\end{vmatrix}\) for this matrix is:
\(\begin{vmatrix} 1 \\ 0\\ \frac{5}{3}\\\end{vmatrix}\)