Eigenvalues Chapter p. 388 C23t Find the eigenvalues, eigenspaces, algebraic and geometric multiplicities for

\[A = \begin{pmatrix}1\:&\:1\\1\:&1\end{pmatrix}\]

To find eigenvalues, solve the following equation: det(\(\lambda\)\(I_n\)-A) = 0

\[det(\begin{pmatrix}\lambda\:&\:0\:\\0\:&\lambda\end{pmatrix} - \begin{pmatrix}1\:&\:1\\1\:&1\end{pmatrix})=0\]

\[det\begin{pmatrix}\lambda-1&-1\\-1\:&\:\lambda-1\end{pmatrix}=0\]

(\(\lambda\)-1)(\(\lambda\)-1)-1 = 0

\(\lambda\)\(^2\) - 2\(\lambda\) = 0

\(\lambda\) = 0 or \(\lambda\) = 2 <- eigenvalues

Substitute in the eigenvalue = 0 to find the eigenvectors: \[ \begin{pmatrix}-1\:&\:-1\\-1\:&-1\end{pmatrix}\] Put the matrix in reduced row eschelon form \[ \begin{pmatrix}1\:&\:1\\0\:&0\end{pmatrix}\] \(v_1\) + \(v_2\) = 0
\(v_1\) = -\(v_2\)

eigenspace = span\[\begin{pmatrix}1\\-1\end{pmatrix}\]

algebraic and geometric multiplicities = 1

Substitute in the eigenvalue = 2 to find the eigenvectors: \[ \begin{pmatrix}1\:&\:-1\\-1\:&1\end{pmatrix}\] Put the matrix in reduced row eschelon form \[ \begin{pmatrix}1\:&\:-1\\0\:&0\end{pmatrix}\] \(v_1\) - \(v_2\) = 0
\(v_1\) = \(v_2\)

eigenspace = span\[\begin{pmatrix}1\\1\end{pmatrix}\]

algebraic and geometric multiplicities = 1

A <- matrix(c(1,1,1,1),2,2)
A
##      [,1] [,2]
## [1,]    1    1
## [2,]    1    1
eigen(A)
## eigen() decomposition
## $values
## [1] 2 0
## 
## $vectors
##           [,1]       [,2]
## [1,] 0.7071068 -0.7071068
## [2,] 0.7071068  0.7071068

When R calculated the eigenvectors, it used the convention that the length should equal 1. The values I calculated are equivalent to this.