\[A = \left[ \begin{array}{cccc}1 & 2 & 3 & 4 \\-1 & 0 & 1 & 3 \\0 & 1 & -2 & 1 \\5 & 4 & -2 & -3 \end{array} \right]\]
\(A\) is a square (\(4 \times 4\)) matrix. Visible inspection shows that the rows of \(A\) are not linear combinations of one another. As such, it is invertible, and its rank is 4.
For a rectangular m \(\times\) n matrix, the largest possible rank is the smaller of the two dimensions. In the case where m \(\gt\) n, the largest possible rank is n.
A non-zero matrix has a minimum of one entry, meaning it has a minimum of one pivot. As the rank is equivalent to the number of pivots, the minimum possible rank is 1.
\[B = \left[ \begin{array}{ccc}1 & 2 & 1 \\3 & 6 & 3 \\2 & 4 & 2 \end{array} \right]\]
The second and third rows of \(B\) are linear combinations of the first row. As such, the rank of \(B\) is 1.
\[A = \left[ \begin{array}{cc}1 & 2 & 3 \\0 & 4 & 5 \\0 & 0 & 6 \end{array} \right]\]
\[A - \lambda I = \left[ \begin{array}{cc}1 & 2 & 3 \\0 & 4 & 5 \\0 & 0 & 6 \end{array} \right] - \lambda \left[ \begin{array}{ccc}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1 \end{array} \right] = \left[ \begin{array}{cc}1 - \lambda & 2 & 3 \\0 & 4 - \lambda & 5 \\0 & 0 & 6 - \lambda \end{array} \right]\]
Moving down the first column for simplicity,
\[\begin{split} det \left( \left[ \begin{array}{ccc}1 - \lambda & 2 & 3 \\0 & 4 - \lambda & 5 \\0 & 0 & 6 - \lambda \end{array} \right] \right) ={}& (1 - \lambda) \times det \left( \left[ \begin{array}{cc}4 - \lambda & 5 \\0 & 6 - \lambda \end{array} \right] \right) \\ & - \ 0 \times det \left( \left[ \begin{array}{cc}2 & 3 \\0 & 6 - \lambda \end{array} \right] \right) \\ & + \ 0 \times det \left( \left[ \begin{array}{cc}2 & 3 \\4 - \lambda & 5 \end{array} \right] \right) \\ = & \left( 1 - \lambda \right) \times \Big( \left( 4 - \lambda \right) \times \left( 6 - \lambda \right) - 5 \times 0 \Big) \\ = & \left( 1 - \lambda \right) \times \left( 4 - \lambda \right) \times \left( 6 - \lambda \right) \end{split}\]
Setting this determinant equal to zero, the roots of the equation, and therefore the eigenvalues, are apparent: \(\mathbf{\lambda_1 = 1; \ \lambda_2 = 4; \ \lambda_3 = 6}\)
Substituting in the first eigenvalue, \[A - \lambda_1 I = \left[ \begin{array}{ccc}0 & 2 & 3 \\0 & 3 & 5 \\0 & 0 & 5 \end{array} \right] = \left[ \begin{array}{c}0 \\0 \\0 \end{array} \right]\] Solving the bottom row gives \(v_{1,3} = 0\). Inserting this into the second row gives \(v_{1,2} = 0\). Substituting both of these into the first row gives the identity, \(0 = 0\) – this means that \(v_{1,1}\) can take any value. A value of 1 is used to create a unit vector. \[v_1 = \left[ \begin{array}{c}1 \\0 \\0 \end{array} \right]\]
Substituting in the second eigenvalue, \[A - \lambda_2 I = \left[ \begin{array}{ccc}-3 & 2 & 3 \\0 & 0 & 5 \\0 & 0 & 2 \end{array} \right] = \left[ \begin{array}{c}0 \\0 \\0 \end{array} \right]\] Again, solving the bottom row gives \(v_{2,3} = 0\). Inserting this into the second row gives \(0 = 0\) – as before, any value can be used; 1 is used so that \(v_{2,2} = 1\). Substituting both of these into the first row gives \(-3 v_{2,1} + 2 \times 1 + 3 \times 0 = 0\) — solving this gives \(v_{2,1} = \frac{2}{3}\). Converting this to a unit vactor gives \[v_2 = \left[ \begin{array}{c}0.5547 \\0.8321 \\0 \end{array} \right]\]
Substituting in the second eigenvalue, \[A - \lambda_2 I = \left[ \begin{array}{ccc}-5 & 2 & 3 \\0 & -2 & 5 \\0 & 0 & 0 \end{array} \right] = \left[ \begin{array}{c}0 \\0 \\0 \end{array} \right]\] The bottom row gives, the identity – as before, a value of 1 is used; \(v_{3,3} = 1\). Substituting this into the second equation gives \(v_{3,2} = \frac{5}{2}\). Substituting these two values into the first row gives \(-5 v_{3,1} + 2 \times \frac{5}{2} + 3 \times 1\) — solving this gives \(v_{3,1} = \frac{8}{5}\). Converting this to a unit vector, \[v_3 = \left[ \begin{array}{c}0.5108 \\0.7982 \\0.3193 \end{array} \right]\]
Using the built-in eigen function, the calculated results can be verified:
A <- matrix(c(1, 2, 3, 0, 4, 5, 0, 0, 6), nrow = 3, byrow = TRUE)
eigen(A)$values
[1] 6 4 1
$vectors
[,1] [,2] [,3]
[1,] 0.5108407 0.5547002 1
[2,] 0.7981886 0.8320503 0
[3,] 0.3192754 0.0000000 0