A linear operator A is a zero of its characteristic polynomial. In practical calculations, we set the characteristic polynomial equal to zero, giving the characteristic equation
det|A - λI| = 0
The zeros of the characteristic polynomial, which are the solutions to this equation, are the eigenvalues of the matrix A.
Example 1 :
Find the eigenvavlues of
To find the eigen values, solve the following equation.
det|A - λI| = 0
where I is the 2 × 2 identity matrix.
The characteristic polynomial in this case is
det|A - λI| = (5 - λ)(2 - λ) - 18
det|A - λI| = 10 - 5λ - 2λ + λ^{2} - 18
det|A - λI| = λ^{2 }- 7λ - 8
Setting this equal to zero gives the characteristic equation.
λ^{2 }- 7λ - 8 = 0
Solve the above quadratic equation by factoring.
(λ - 8)(λ + 1) = 0
λ = 8 or λ = -1
Therefore, the two eigenvalues of the given matrix are
λ_{1} = 8
λ_{2} = -1
Note :
For a 2 × 2 matrix, we found two eigenvalues. This is because a 2 × 2 matrix leads to a second-order characteristic polynomial.
This is true in general ; an n × n matrix will lead to an n^{th}^{ }order characteristic polynomial with n (not necessarily distinct) solutions.
Example 2 :
Find the eigenvavlues of
To find the eigen values, solve the following equation.
det|B - λI| = 0
where I is the 3 × 3 identity matrix.
The characteristic polynomial in this case is
det|A - λI| = (2 - λ)[(4 - λ)(2 - λ) - 0] - 1[1(2 - λ) - 0] + 0
det|A - λI| = (2 - λ)[8 - 4λ - 2λ + λ^{2}] - 1[2 - λ]
det|A - λI| = (2 - λ)(λ^{2 }- 6λ + 8) - (2 - λ)
Factor (2 - λ) out.
det|A - λI| = (2 - λ)(λ^{2 }- 6λ + 8 - 1)
det|A - λI| = (2 - λ)(λ^{2 }- 6λ + 7)
Setting this equal to zero gives the characteristic equation.
(2 - λ)(λ^{2 }- 6λ + 7) = 0
2 - λ = 0 or λ^{2 }- 6λ + 7 = 0
Solving 2 - λ = 0, we get
λ = 2
Solve λ^{2 }- 6λ + 7 = 0.
The quadratic equation λ^{2 }- 6λ + 7 = 0 can not be solved by factoring. So, we have to use quadratic formula.
Comparing λ^{2 }- 6λ + 7 = 0 and ax^{2 }+ bx + c = 0, we get
a = 1, b = -6, c = 7
Quadratic Formula :
λ = 3 ± √2
λ = 3 + √2 or 3 - √2
Therefore, the three eigenvalues of the given matrix are
λ_{1} = 2
λ_{2} = 3 + √2
λ_{3} = 3 - √2
Notice that since B is a 3 × 3 matrix, it has three eigenvalues.
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