4. Eigenvalues and -vectors of a matrix

In this section K = C, that is, matrices, vectors and scalars are all complex. Assuming K = R would make the theory more complicated. All the matrices are square matrices (n x n matrices).

4.1. Definitions and terminology

A

,   A

A =

for some.
The number is an eigenvalue of A, if there exists a vector , such that A = . The vector is an eigenvector corresponding to the eigenvalue .

Example 1: An eigenvalue and -vector of a matrix

The set

{ Cⁿ | A = } = { Cⁿ | (A - I) = 0} = N(A - I ) = Ker(A - I)

is called the eigenspace corresponding to the eigenvalue . It is a subspace of the vector space Cⁿ which contains all the eigenvalues corresponding to the eigenvalue and . (The geometric multiplicity of = dim N (A - I) 1). If is an eigenvalue then so is c, c C, c 0. (The same definitions apply to Linear mappings A : V -> Z).

Proposition 1. The eigenvalues of a matrix A are exactly the roots of the equation det (A - I) = 0.

Proof. Let be an eigenvalue of A.
=> : (A - I ) = => det(A - I ) = 0.
(If det (A - I) 0 => (A - I ) ^-1 => "(A - I ) = <=> = (A - I ) ^-1 = " )
Let det(A - I) = 0 => (the rank of (A - I)) < n
=> n - the rank of (A - I) = dim N (A - I) 1
=> : (A - I) = .

Example 2: Computation of eigenvalues and -vectors

If A is an n x n matrix, then det (A - I) is a polynomial of degree n,

det (A - I) = (-1)ⁿⁿ + a_n-1^n-1 + ··· + a₁ + a₀ = p_A (),

called the characteristic polynomial of A. The equation det (A - I) = 0 is the characteristic equation of A.
By the Fundamental Theorem of Algebra, a polynomial of degree n has exactly n roots(counted with multiplicity), some or all of which may be real. Therefore, a (complex) matrix always eigenvalues. The identity matrix I has only one eigenvalue = 1, which has multiplicity n. (det(I - I) = (1 - )ⁿ = 0) By Proposition 1, the eigenvalues of A are the zeros of the characteristic polynomial.

Example 3: Computation of eigenvalues and -vectors

Note. If a real matrix A has a complex eigenvalue and is a corresponding eigenvector, then the complex conjugate is also an eigenvalue with , the conjugate vector of , as a corresponding eigenvector. (Prove!).

Example 4: A complex eigenvalue

Proposition 2. A matrix A has an inverse matrix A ^{- 1} if and only if it does not have zero as an eigenvalue. If ₁ , ..., _n 0 are the eigenvalues of A, then the eigenvalues of A ^{- 1} are 1 / ₁ , ..., 1 / _n .

Example 5: Eigenvalues and -vectors of an inverse

Proof. 1) If A has as an eigenvalue = 0 => det (A - 0· I) = 0 => det A = 0 => A^-1 => A^-1 only if every eigenvalue is 0.
If every eigenvalue is 0 => det(A - 0 · I) 0 => det A 0 => A^-1.
2) If is an eigenvalue of A and A^-1 exists, then and 0:

  

	A =	=>	A-1 (A) = A-1 ()
=>	= A-1	=>	A-1 = (1 / )

=>  1 / is an eigenvalue of A^-1 (with as a corresponding eigenvalue).

Proposition 3. The eigenvalues of A are the same as the eigenvalues of A^T.

Example 6: The eigenvalues and vectors of a transpose

Proof. is an eigenvalue of A <=> det (A - I ) = 0 <=> det (A - I )^T = 0 <=> det (A^T - I ) = 0 <=> is an eigenvalue of A^T.

Note. If is an eigenvector of A corresponding to an eigenvalue , then does not necessarily correspond to as an eigenvector of A^T. For example, the matrix

A =		1	0
		1	1

has = 1 as an eigenvalue with the corresponding eigenvectors = t (0,1), t C, t 0, but the eigenvectors of

AT =		1	1
		0	1

corresponding to = 1 are = t (1,0), t C, t 0.

Proposition 4. If is an eigenvector of A with as a corresponding eigenvector, then
1) ^k is an eigenvalue of A^k with a corresponding eigenvector ,
2) c is an eigenvalue of cA with a corresponding eigenvector ,
3) c_m ^m + c_{m - 1} ^{m - 1} + ··· + c₁ + c₀ is an eigenvalue of (c_m A^m + c_{m - 1} A^{m - 1} + ··· + c₁ A + c₀ I ) with a corresponding eigenvector .

Proof. 1) A¹ = ¹ and if A^k = ^k, then A^{k + 1} = A (A^k) = A (^k) = ^k A = ^{k + 1} => the statement.
2) (cA) = cA = c
3) (c_m A^m + ··· + c₁ A + c₀ I) = c_m A^m + ··· + c₁ A + c₀ = (c_m ^m + ··· + c₁ + c₀ ).

Since the determinant of an upper triangular matrix is the product of the diagonal elements, we have:

Proposition 5. The eigenvalues of an upper triangular matrix are the diagonal elements. The same is true also for a lower triangular matrix.

Proposition 6. If A has distinct eigenvalues ₁ , ..., _r , then the corresponding eigenvalues ₁ , ..., _r are linearly independent.

Proof. Let k r be the smallest positive integer such that {₁ , ..., _k } is linearly dependent. Let

	(1)	ai i =	=> ai Ai =
=>	(2)	ai i i =
	=>	ai k i	- ai i i =
	=>	ai (k - i )i	=
	=>	ai (k - i )i	=
	=>	a1 = a2 = ··· = ak - 1	= 0
	=>	{1 , ..., k }	linearly independent
		(since also ak = 0)	=>
	=>	{1 , ..., r }	linearly independent

Example 7: Linearly independent eigenvectors

Note. If the eigenvalues ₁ , ..., _n of A are not all distinct, it is still possible that A has n linearly independent eigenvectors.

Example 8: Linearly independent eigenvectors