Find the eigen values and eigen vectors of the following [4, 0, 1; -2, 1, 0; -2, 0, 1]
To find the eigenvalues and eigenvectors of the matrix
\[
A = \begin{bmatrix}
4 & 0 & 1 \\
-2 & 1 & 0 \\
-2 & 0 & 1
\end{bmatrix},
\]
we begin by calculating the eigenvalues. The eigenvalues can be found by solving the characteristic polynomial, which is given by
\[
\det(A - \lambda I = 0),
\]
where \( \lambda \) is the eigenvalue and \( I \) is the identity matrix. Thus, we first compute \( A - \lambda I \):
\[
A - \lambda I = \begin{bmatrix}
4 - \lambda & 0 & 1 \\
-2 & 1 - \lambda & 0 \\
-2 & 0 & 1 - \lambda
\end{bmatrix}.
\]
Next, we calculate the determinant of this matrix:
\[
\det(A - \lambda I) = (4 - \lambda)\det\begin{bmatrix}
1 - \lambda & 0 \\
0 & 1 - \lambda
\end{bmatrix} - 1\det\begin{bmatrix}
-2 & 0 \\
-2 & 1 - \lambda
\end{bmatrix}.
\]
The determinant of the \(2 \times 2\) matrix is computed as follows:
\[
\det\begin{bmatrix}
1 - \lambda & 0 \\
0 & 1 - \lambda
\end{bmatrix} = (1 - \lambda)(1 - \lambda) = (1 - \lambda)^2.
\]
Now, for the second \(2 \times 2\) matrix:
\[
\det\begin{bmatrix}
-2 & 0 \\
-2 & 1 - \lambda
\end{bmatrix} = (-2)(1 - \lambda) - 0(-2) = -2(1 - \lambda) = -2 + 2\lambda.
\]
Putting these together, we calculate the determinant:
\[
\det(A - \lambda I) = (4 - \lambda)(1 - \lambda)^2 - (-2 + 2\lambda).
\]
We simplify this expression:
\[
\det(A - \lambda I) = (4 - \lambda)(1 - 2\lambda + \lambda^2) + 2 - 2\lambda.
\]
Expanding \( (4 - \lambda)(1 - 2\lambda + \lambda^2) \):
\[
= (4 - \lambda)(\lambda^2 - 2\lambda + 1) = 4\lambda^2 - 8\lambda + 4 - \lambda^3 + 2\lambda^2 - \lambda.
\]
Combining like terms gives:
\[
-\lambda^3 + 6\lambda^2 - 9\lambda + 4 + 2 - 2\lambda = -\lambda^3 + 6\lambda^2 - 11\lambda + 6.
\]
Thus, the characteristic polynomial is:
\[
-\lambda^3 + 6\lambda^2 - 11\lambda + 6 = 0.
\]
To solve \( -\lambda^3 + 6\lambda^2 - 11\lambda + 6 = 0 \), we can use the Rational Root Theorem to test possible rational roots. Trying \( \lambda = 1 \):
\[
-1^3 + 6(1)^2 - 11(1) + 6 = -1 + 6 - 11 + 6 = 0.
\]
So \( \lambda = 1 \) is a root. We can factor \( \lambda - 1 \) out of the polynomial:
Using synthetic division for \( -\lambda^3 + 6\lambda^2 - 11\lambda + 6 \) with root \( \lambda = 1 \):
\[
\begin{array}{r|rrrr}
1 & -1 & 6 & -11 & 6 \\
& & -1 & 5 & -6 \\
\hline
& -1 & 5 & -6 & 0 \\
\end{array}
\]
We get \( -\lambda^2 + 5\lambda - 6 = -(\lambda^2 - 5\lambda + 6) = -(\lambda - 2)(\lambda - 3) \).
This means the factored polynomial is:
\[
-(\lambda - 1)(\lambda - 2)(\lambda - 3) = 0.
\]
Thus, the eigenvalues are:
\[
\lambda_1 = 1, \quad \lambda_2 = 2, \quad \lambda_3 = 3.
\]
Next, we find the eigenvectors corresponding to each eigenvalue:
1. **Eigenvector for \( \lambda_1 = 1 \)**:
Solve \( (A - I)\mathbf{v} = 0 \):
\[
A - I = \begin{bmatrix}
3 & 0 & 1 \\
-2 & 0 & 0 \\
-2 & 0 & 0
\end{bmatrix}.
\]
This simplifies to:
\[
\begin{bmatrix}
3 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix}.
\]
From the first row, \( 3x + z = 0 \) which gives \( z = -3x \). We can take \( x = t \), leading to:
\[
\begin{bmatrix}
t \\
y \\
-3t
\end{bmatrix} = t\begin{bmatrix}
1 \\
0 \\
-3
\end{bmatrix} + y\begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix}.
\]
So eigenvectors can be expressed as any scalar multiple of
\[
\begin{bmatrix}
1 \\
0 \\
-3
\end{bmatrix}
\text{ and }
\begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix}.
\]
2. **Eigenvector for \( \lambda_2 = 2 \)**:
Solve \( (A - 2I)\mathbf{v} = 0 \):
\[
A - 2I = \begin{bmatrix}
2 & 0 & 1 \\
-2 & -1 & 0 \\
-2 & 0 & -1
\end{bmatrix}.
\]
This reduces to:
\[
\begin{bmatrix}
2 & 0 & 1 \\
0 & 1 & 0 \\
0 & 0 & 0
\end{bmatrix}.
\]
From the first row, \( 2x + z = 0 \) which gives \( z = -2x \).
The second row gives \( y = 0 \).
Setting \( x = t\), we get:
\[
\begin{bmatrix}
t \\
0 \\
-2t
\end{bmatrix} = t\begin{bmatrix}
1 \\
0 \\
-2
\end{bmatrix}.
\]
4. **Eigenvector for \( \lambda_3 = 3 \)**:
Solve \( (A - 3I)\mathbf{v} = 0 \):
\[
A - 3I = \begin{bmatrix}
1 & 0 & 1 \\
-2 & -2 & 0 \\
-2 & 0 & -2
\end{bmatrix}.
\]
This reduces to:
\[
\begin{bmatrix}
1 & 0 & 1 \\
0 & 2 & 0 \\
0 & 0 & 0
\end{bmatrix}.
\]
From the first row, \( x + z = 0 \) leading to \( z = -x \), and the second gives \( y = 0 \)).
We can express this as:
\[
\begin{bmatrix}
t \\
0 \\
-t
\end{bmatrix}.
\]
Thus, we have eigenvalues and corresponding eigenvectors:
- Eigenvalue \( \lambda = 1 \), with eigenvector \( \begin{bmatrix} 1 \\ 0 \\ -3 \end{bmatrix} \) and \( \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \)
- Eigenvalue \( \lambda = 2 \), with eigenvector \( \begin{bmatrix} 1 \\ 0 \\ -2 \end{bmatrix} \)
- Eigenvalue \( \lambda = 3 \), with eigenvector \( \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} \)
The specific eigenvectors can be scaled, so there are infinitely many multiples for each eigenvector.