For the standard matrix
{ 1 -2 0}
{-2 -1 1}
{0 0 1}
I found the eigenvalues to be -1, and +/- 5^1/2
I am having problems finding the eigenvectors of the root 5 values. Can someone set me straight??
Thanks
To find the eigenvectors corresponding to the eigenvalues of the square root of 5, we can start by setting up the equation: (A - λI)X = 0, where A is the given matrix, λ is the eigenvalue, I is the identity matrix, and X is the eigenvector.
Let's work with one eigenvalue at a time. For the eigenvalue -√5, we have:
(A - (-√5)I)X = 0
Substituting the values, we get:
⎡1 -2 0⎤ ⎡-√5 0 0⎤ ⎡x₁⎤ ⎡0⎤
⎢-2 -1 1⎥ - ⎢0 -√5 0⎥ ⋅ ⎢x₂⎥ = ⎢0⎥
⎣0 0 1⎦ ⎣0 0 -√5⎦ ⎣x₃⎦ ⎣0⎦
Simplifying, we have:
⎡1 + √5 -2 0 ⎤ ⎡x₁⎤ ⎡0⎤
⎢-2 1 - √5 1 ⎥ ⎢x₂⎥ = ⎢0⎥
⎣ 0 0 1 ⎦ ⎣x₃⎦ ⎣0⎦
This augmented matrix represents a system of linear equations. We can solve this system to find the corresponding eigenvector(s).
The goal is to row-reduce the augmented matrix to its reduced row-echelon form. To do this, we can perform elementary row operations. Let's start by subtracting twice the first row from the second row:
R₂ ← R₂ - 2R₁
This gives:
⎡1 + √5 -2 0 ⎤ ⎡x₁⎤ ⎡0⎤
⎢ 0 5 - √5 1 ⎥ ⎢x₂⎥ = ⎢0⎥
⎣ 0 0 1 ⎦ ⎣x₃⎦ ⎣0⎦
Next, we can multiply the second row by (5 - √5)^(-1):
R₂ ← (5 - √5)^(-1)R₂
This gives:
⎡1 + √5 -2 0 ⎤ ⎡x₁⎤ ⎡0⎤
⎢ 0 1 1/(5 - √5) ⎥ ⎢x₂⎥ = ⎢0⎥
⎣ 0 0 1 ⎦ ⎣x₃⎦ ⎣0⎦
Finally, we can subtract the first row multiplied by (√5 + 1) from the second row:
R₂ ← R₂ - (√5 + 1)R₁
After simplifying, we have:
⎡1 + √5 -2 0 ⎤ ⎡x₁⎤ ⎡0⎤
⎢ 0 1 0 ⎥ ⎢x₂⎥ = ⎢0⎥
⎣ 0 0 1 ⎦ ⎣x₃⎦ ⎣0⎦
This reduced row-echelon form tells us that x₂ = 0, x₃ = 0, and x₁ can take any value.
Therefore, for the eigenvalue -√5, the corresponding eigenvector can be represented as:
X = [x₁, 0, 0]
We can repeat these steps to find the eigenvector(s) for the eigenvalue √5. The resulting eigenvector(s) will be independent of the ones we have already obtained.