• Given the characteristic polynomial, the eigenvalues are simply the values at which the polynmoial expression equates to zero. Therefore

\[ P_{A}= (4-x)(1-x)^{2}=0\]

\[ -x+4=0=(\lambda=4)\] \[(1-x=0)=(\lambda=1)\]

  • The eigenspaces are the eigenvectors that satisfy our equation. We know we that there should be 2 eigenvectors when eigenvalue=1, and 1 eigenvector when eigenvlaue= 4.
  • To solve we need to plug the eigenvalue back into the matrix that results from

\[ \lambda I_{n}-A\]

Solve at \(\lambda=4\)

\[\lambda I_{n} -A | \lambda=4 \]

\[\begin{equation*} \mathbf{}\left[\begin{matrix} \lambda -2 & -1 & -1\\ -1 & \lambda -2 & -1\\ -1 & -1 & \lambda -2 \end{matrix}\right]==\left[\begin{matrix} \ 2&-1&-1\\ -1 &2& -1\\-1&-1&2 \end{matrix}\right] \end{equation*}\]

\[ Get RREF\]

library(pracma)
AA <- matrix(c(2,-1,-1,-1,2,-1,-1,-1,2),ncol=3)
rref(AA)
##      [,1] [,2] [,3]
## [1,]    1    0   -1
## [2,]    0    1   -1
## [3,]    0    0    0

\[ x-z=0\]

\[y-z=0\] \[x=y=z \]

  • Therefore given eigen value of 4; x=y=z and therefore our only non trivial eigenvector can be expressed as

\[ x \begin{bmatrix} 1 \\ 1 \\ 1\\ \end{bmatrix} \]

Solve at \(\lambda=1\)

\[\lambda I_{n} -A | \lambda=1 \]

\[\begin{equation*} \mathbf{}\left[\begin{matrix} \lambda -2 & -1 & -1\\ -1 & \lambda -2 & -1\\ -1 & -1 & \lambda -2 \end{matrix}\right]==\left[\begin{matrix} \ -1&-1&-1\\ -1 &-1& -1\\-1&-1&-1 \end{matrix}\right] \end{equation*}\]

\[ Get RREF\]

library(pracma)
A <- matrix(c(-1,-1,-1,-1,-1,-1,-1,-1,-1),ncol=3)
rref(A)
##      [,1] [,2] [,3]
## [1,]    1    1    1
## [2,]    0    0    0
## [3,]    0    0    0

\[ x+y+z=0\]

\[ Given x=1\] \[ y=-z-1 |z=-y-1 \]

substiution

\[y=(-y-1)-1 = 2y=-2= (y=-1,z=0,x=1) \] \[z=(-z-1)-1 = 2z=-2= (z=-1,y=0,x=1) \] \[ \begin{bmatrix} 1 \\ 0 \\ -1\\ \end{bmatrix}|\begin{bmatrix} 1 \\ -1 \\ 0\\ \end{bmatrix}\]