C26 p306

For matrix A = \(\begin{bmatrix}2&1&1\\1&2&1\\1&1&2\end{bmatrix}\), the characteristic polynomial of A is pA(x) = (4 −x)(1 − x)^2. Find the eigenvalues and corresponding eigenspaces of A.

Since they gave the characteristic polynomial, I can solve that to get the eigenvalues:
(4 −x)(1 − x)^2 = pA(x)
(4 −λ)(1 − λ)^2 = 0
4 − λ = 0, 1 − λ = 0
4 = λ, 1 = λ
This gives us the eigenvalues of 1 and 4. (1 twice because of the exponent)

We can then verify the eigenvalues by using the eigen function below:

A <- matrix(c(2,1,1,1,2,1,1,1,2), nrow = 3, ncol = 3)
A
##      [,1] [,2] [,3]
## [1,]    2    1    1
## [2,]    1    2    1
## [3,]    1    1    2
eigen(A)$values
## [1] 4 1 1

To find the eigenspaces, we solve for (A-λI)v = 0 where I is the identify matrix, v is a vector and 0 is the zero vector.
For λ = 1:
A-1I = \(\begin{bmatrix}2&1&1\\1&2&1\\1&1&2\end{bmatrix}\) - \(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\) = \(\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}\)
The next step is to put the matrix into reduced row echelon form:
\(\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}\) – R2 - R1 & R3 - R1 –> \(\begin{bmatrix}1&1&1\\0&0&0\\0&0&0\end{bmatrix}\)
We can put it back into (A-1I)v = 0 to get:
\(\begin{bmatrix}1&1&1\\0&0&0\\0&0&0\end{bmatrix}\)\(\begin{bmatrix}v1\\v2\\v3\end{bmatrix}\) = \(\begin{bmatrix}0\\0\\0\end{bmatrix}\)
The equation we can make from this is:
1v1 + 1v2 + 1v3 = 0 –> v1 + v2 + v3 = 0 –> v1 = -v2 - v3
Let’s say v2 = x and v3 = y so we have:
\(\begin{bmatrix}-v2-v3\\x\\y\end{bmatrix}\) –> \(\begin{bmatrix}-x-y\\x\\y\end{bmatrix}\)
If x = 1 & y = 0:
\(\begin{bmatrix}-1\\1\\0\end{bmatrix}\)x
If x = - & y = 1:
\(\begin{bmatrix}-1\\0\\1\end{bmatrix}\)y
With this, the eigenspaces of A when λ = 1 are \(\begin{bmatrix}-1\\1\\0\end{bmatrix}\)A & \(\begin{bmatrix}-1\\0\\1\end{bmatrix}\)A

For λ = 4:
A-1I = \(\begin{bmatrix}2&1&1\\1&2&1\\1&1&2\end{bmatrix}\) - \(\begin{bmatrix}4&0&0\\0&4&0\\0&0&4\end{bmatrix}\) = \(\begin{bmatrix}-2&1&1\\1&-2&1\\1&1&-2\end{bmatrix}\)
The next step is to put the matrix into reduced row echelon form:
\(\begin{bmatrix}-2&1&1\\1&-2&1\\1&1&-2\end{bmatrix}\) – R1 <–> R2 –> \(\begin{bmatrix}1&-2&1\\-2&1&1\\1&1&-2\end{bmatrix}\) – R2 + 2R1 & R3 - R1 –> \(\begin{bmatrix}1&-2&1\\0&-3&3\\0&3&-3\end{bmatrix}\) – R3/3 & R3 + R2 –> \(\begin{bmatrix}1&-2&1\\0&-1&1\\0&0&0\end{bmatrix}\) – R1 - R2 –> \(\begin{bmatrix}1&-1&0\\0&-1&1\\0&0&0\end{bmatrix}\)
We can put it back into (A-4I)*v = 0 to get:
\(\begin{bmatrix}1&-1&0\\0&-1&1\\0&0&0\end{bmatrix}\)\(\begin{bmatrix}v1\\v2\\v3\end{bmatrix}\) = \(\begin{bmatrix}0\\0\\0\end{bmatrix}\)
The equations we can make from this are:
1v1 - 1v2 + 0v3 = 0 –> v1 - v2 = 0 –> v1 = v2
0v1 - 1v2 + 1v3 = 0 –> -v2 + v3 = 0 –> v3 = v2
Since v1 = v2 & v3 = v2, v1 = v2 = v3 must be true.
With this is the eigenspace of A when λ = 4 is \(\begin{bmatrix}1\\1\\1\end{bmatrix}\)A

We can then verify the eigenvectors by using the eigen function below:

eigen(A)$vectors
##            [,1]       [,2]       [,3]
## [1,] -0.5773503  0.0000000  0.8164966
## [2,] -0.5773503 -0.7071068 -0.4082483
## [3,] -0.5773503  0.7071068 -0.4082483