for the matrix:
{ 1 -2 0}
{-2 -1 1}
{ 0 0 -1}
I found the eigenvalue to be -1 and +/- 5^1/2
Is this correct and can someone show me how to put the root 5 values as eigenvectors?
Please and Thanks
To verify if your eigenvalues are correct, we can compute the characteristic equation of the matrix and solve for the eigenvalues.
The characteristic equation is given by:
det(A - λI) = 0
where A is the given matrix, λ is the eigenvalue, and I is the identity matrix.
For the given matrix:
A = {{1, -2, 0},
{-2, -1, 1},
{0, 0, -1}}
Substituting into the characteristic equation:
det(A - λI) = 0
We get:
det({{1-λ, -2, 0},
{-2, -1-λ, 1},
{0, 0, -1-λ}}) = 0
Expanding the determinant:
(1-λ)[(-1-λ)(-1-λ) - 0] - 2[-2(-1-λ) - 0] + 0 = 0
Simplifying:
(1-λ)[(λ+1)(λ+1)] - 2(2+2λ) = 0
Expanding and collecting terms:
(λ - 1)(λ^2 + 2λ + 5) = 0
Solving for eigenvalues by setting each factor to zero:
λ - 1 = 0 --> λ = 1
λ^2 + 2λ + 5 = 0
To solve the quadratic equation, we can either use the quadratic formula or factorize it. In this case, the quadratic equation does not factorize into simple linear factors, so let's use the quadratic formula:
λ = (-b ± √(b^2 - 4ac))/(2a)
For the equation λ^2 + 2λ + 5 = 0:
a = 1
b = 2
c = 5
Calculating:
λ = (-2 ± √(2^2 - 4(1)(5)))/(2(1))
= (-2 ± √(4 - 20))/(2)
= (-2 ± √(-16))/(2)
= (-2 ± 4i)/(2)
Simplifying:
λ = -1 ± 2i
So the eigenvalues for the given matrix are -1, -1 + 2i, and -1 - 2i.
To find the eigenvectors for these eigenvalues, we need to solve the system of equations (A - λI)v = 0, where v is the eigenvector.
Let's find the eigenvector for the eigenvalue -1.
For the matrix A and eigenvalue -1 (λ = -1), we have:
(A - λI) = {{2, -2, 0},
{-2, 0, 1},
{0, 0, 0}}
Setting up the system of equations:
2x - 2y = 0
-2x + 0y + z = 0
0x + 0y + 0z = 0
We can see that x = y, and z is arbitrary. Let's assign a value to z, say z = 1.
Using x = y, we get:
2x - 2x = 0
x = 0
Therefore, the eigenvector corresponding to the eigenvalue -1 is (0, 0, 1).
To find the eigenvectors for the complex eigenvalues -1 + 2i and -1 - 2i, we can follow a similar process. However, in this case, we will have non-zero values for x, y, and z.
I hope this helps! Let me know if you have any further questions.