Find the eigenvalues, eigenspaces, algebraic multiplicities and geometric multiplicities for the matrix below. It is possible to do all these computations by hand, and it would be instructive to do so.
(C <- matrix(c(-1, 2, -6, 6), nrow = 2, byrow = TRUE))## [,1] [,2]
## [1,] -1 2
## [2,] -6 6First, let’s get the eigenvalues:
\[det\left( \begin{bmatrix} -1 & 2 \\ -6 & 6 \end{bmatrix}-\begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix} \right) =0\]
\[det\left( \begin{bmatrix} -1-\lambda & 2 \\ -6 & 6-\lambda \end{bmatrix} \right) =0\]
\[\left( -1-\lambda \right)\left( 6-\lambda \right) - \left( 2*-6 \right) =0\]
\[-6+\lambda-6\lambda+{ \lambda }^{ 2 }+12 = { \lambda }^{ 2 }-5\lambda+6\]
\[\left( x-3 \right) \left( x-2 \right) =0\]
The eigenvalues for this matrix are 2, and 3. Since these values only used once, the algebraic multiplicity for both eigen values is 1. Now, let’s compute the eigenspaces:
For eigenvalue = 2:
library(pracma)
(A_R <- rref(C - 2 * diag(2)))## [,1] [,2]
## [1,] 1 -0.6666667
## [2,] 0 0.0000000\[S=\left\{ x-\frac { 2 }{ 3 } y=0 \right\} =\left\{ x=\frac { 2 }{ 3 } y \right\} =\begin{Bmatrix} x=\frac { 2 }{ 3 } \\ y=1 \end{Bmatrix}\]
v = c(2/3, 1)
(v_unit = v/sqrt(sum(v^2)))## [1] 0.5547002 0.8320503For eigenvalue = 3:
(A_R <- rref(C - 3 * diag(2)))## [,1] [,2]
## [1,] 1 -0.5
## [2,] 0 0.0\[S=\left\{ x-\frac { 1 }{ 2 } y=0 \right\} =\left\{ x=\frac { 1 }{ 2 } y \right\} =\begin{Bmatrix} x=\frac { 1 }{ 2 } \\ y=1 \end{Bmatrix}\]
v = c(1/2, 1)
(v_unit = v/sqrt(sum(v^2)))## [1] 0.4472136 0.8944272For each eigenvalue, there was only one row that had a solution; therefore, both eigenvalues have a geometric multiplicity of 1.