PC←B=(PB←C)−1
Recall that λ is an eigenvalue of a n×n matrix A if Ax=λx for some x in Rn.
Note that this means (A−λI)x=0 has non-trivial solutions.
The null space of A−λI is precisely the collection of x for which (A−λI)x=0.
Definition. The eigenspace of A corresponding to the eigenvalue of λ of A is the null space of A−λI.
Ex. Find a basis for the eigenspace corresponding to λ=2 for A=⎝⎛422−11−1668⎠⎞.
A−2I=⎝⎛422−11−1668⎠⎞−⎝⎛200020002⎠⎞=⎝⎛222−1−1−1666⎠⎞
→→⎝⎛x1x2x3⎠⎞⎝⎛222−1−1−1666000⎠⎞⎝⎛200−100600000⎠⎞⎝⎛100−2100300000⎠⎞=x2⎝⎛2110⎠⎞+x3⎝⎛−301⎠⎞
Thus {⎝⎛2110⎠⎞,⎝⎛−301⎠⎞} is a basis for the required eigenspace.
Observation:
(A−λI)x=0 has non-trivial solutions if det(A−λI)=0. This comes from the Invertible Matrix Theorem. (Recall that eigenvectors are nonzero by definition, so if there are only trivial solutions then A has no eigenvalues.)
Definition: The equations det(A−λI)=0 is called the characteristic equation (for A).
Theorem: A scalar λ is an eigenvalue if and only if λ satisfies the characteristic equation det(A−λI)=0.
Ex. Find the eigenvalues of ⎝⎛300520791⎠⎞ by finding the zeroes of the characteristic equation.
A−λI=⎝⎛300520791⎠⎞−⎝⎛λ000λ000λ⎠⎞=⎝⎛3−λ0052−λ0791−λ⎠⎞
A−λI in this case is a triangular matrix, so its determinant is just the product of the entries on its diagonal.
det(A−λI)λ=(3−λ)(2−λ)(1−λ)=0=3, 2, 1
The eigenvalues of A are 3, 2, and 1.
From this example we can infer that the eigenvalues of any triangular matrix are the entries on its diagonal.
Observation:
The characteristic equation is always a polynomial. This is why it is also called the characteristic polynomial.
If some characteristic polynomial has degree 3, does it have an eigenvalue?
Yes, because eigenvalues occur at the real roots of their characteristic polynomial, and all polynomials of odd degrees have at least one real root by the Intermediate Value Theorem. (Note that polynomials of even degrees may not have any real roots.)
Definition. The (algebraic) multiplicity of an eigenvalue λ is its multiplicity as a root of the characteristic polynomial.
For instance, the characteristic polynomial (λ−2)2(λ−3) has two eigenvalues – 2, which has a multiplicity of 2, and 3, which has a multiplicity of 1.
Ex. Find the eigenvalues and their multiplicities from the characteristic polynomial λ6−4λ5−12λ4.
λ6−4λ5−12λ4=λ4(λ2−4λ−12)=λ4(λ−6)(λ+2)=0
The eigenvalues are 0 (with multiplicity 4), 6 (with multiplicity 1), and -2 (with multiplicity 1).