characteristic values and vectors (of a matrix)

Over the spectrum σ(A) of a matrix A, its eigenvaluesMathworldPlanetmathPlanetmathPlanetmathPlanetmath λ1,λ2,,λs possess multiplicities n1,n2,,ns, respectively, with k=1snk=n. Its associated characteristic polynomialMathworldPlanetmathPlanetmath is then factored as

Δ(λ)|λI-A|=Πk=1s(λ-λk)nk. (1)

Let us set mult(λk)=nk for multiplicity of λk(k=1,,s). We will now prove the following theorem.

Theorem 1.

If σ(A)={λk}k=1s, mult(λk)=nk, and g(μ) is a scalar polynomialPlanetmathPlanetmath, then σ(g(A))={g(λk)}k=1s, mult(g(λk))=nk.


Let g(μ) be an arbitrary scalar polynomial. We want to find the characteristic values of g(A). For this purpose we split g(μ) into linear factors

g(μ)=a0Πi=1t(μ-μi)li,a00,i=1tli=l. (2)

On substitution μA, we have

g(A)=a0Πi=1t(A-μiI)li, (3)

being I the identity matrixMathworldPlanetmath. Let us compute the determinantDlmfMathworldPlanetmath of g(A). (Coefficient a0 will be powered to n, the order of the square matrixMathworldPlanetmath A).

|g(A)| =a0nΠi=1t|(-1)(μiI-A)|li=a0nΠi=1t(-1)nli|μiI-A|li

because on substitution λμi in (1). Next we commute the binomial by introducing (-1)nl into the product signs and also we note that a0n=a0k=1snk=Πk=1sa0nk, so that


and we may use (2) for μ=λk to obtain

|g(A)|=Πk=1sg(λk)nk. (4)

Finally we substitute the polynomial g(μ) by λ-g(μ), where λ is an arbitrary parameter, getting for (4)

Δ(g(A))|λI-g(A)|=Πk=1s[λ-g(λk)]nk. (5)

This proves the theorem. ∎

As an important particular case we have: σ(Am)={λkm}k=1s, (m=0,1,), mult(λk)=nk.

Connection between the characteristic polynomial Δ(λ) and the adjugate matrix B(λ) of A.
As it is well known, the adjugate matrix B of a matrix A there corresponds to the algebraic complement or cofactor matrix of the transposeMathworldPlanetmath of A. From this definition we have

B(λ)(λI-A)=Δ(λ)I  and  (λI-A)B(λ)=Δ(λ)I. (6)

Let us suppose Δ(λ) is given by

Δ(λ)=λn-k=1nckλn-k. (7)

It is clear that the difference Δ(λ)-Δ(μ) is divisible by λ-μ without remainder, hence

δ(λ,μ)Δ(λ)-Δ(μ)λ-μ=λn-1+(μ-c1)λn-2+(μ2-c1μ-c2)λn-3+ (8)

is a polynomial in λ,μ. If we replace in (8) (λ,μ) by the permutable matrices (λI,A) and recalling that from Cayley-Hamilton theoremMathworldPlanetmath Δ(A)=0, then

δ(λI,A)(λI-A)=Δ(λ)I, (9)

which by comparing it with (6)1 we conclude that

B(λ)=δ(λI,A) (10)

is the desired formula by virtue of the uniqueness of the quotientPlanetmathPlanetmath. Therefore (10) and (8) let to write the adjugatePlanetmathPlanetmath B(λ) as the matrix polynomial

B(λ)=Iλn-1+k=1n-1Bkλn-k-1, (11)

where (μA in (8))

Bk=Ak-i=1kciAk-i,(k=1,,n-1), (12)

which can also be obtained from the recurrence equation

Bk=ABk-1-ckI,(k=1,,n-1;B0=I). (13)

What is more,

ABn-1-cnI=0Bn. (14)

(13) as well as (14) follow inmediately from (6)2 if we equate the coefficients of equal powers of λ on both sides. Also, if we substitute Bn-1 from (12), into (14), we get Δ(A)=0 (Cayley-Hamilton), an implicit consequence of generalized Bézout theorem. On the other hand, by setting λ=0 in (7) we obtain cn=Δ(0)/(-1)=|-A|/(-1)=(-1)n-1|A|0, whenever A be non- singular. From this and from (14) follow that

A-1=1cnBn-1. (15)

Let now λc be a characteristic value of A, then Δ(λc)=0 and (6)2 becomes

(λcI-A)B(λc)=0. (16)

Let us assume that B(λc)0 and denote by 𝐛 an arbitrary non-zero column of this matrix. From (16) we have (λcI-A)𝐛=𝟎. That is,

A𝐛=λc𝐛. (17)

Therefore every non-zero column of B(λc) determines a characteristic vector corresponding to the characteristic value λc. Moreover, if to the characteristic value λc there correspond l linearly independentMathworldPlanetmath characteristic vectors, n-l will be the rank of λcI-A and so the rank of B(λc) does not exceed l. In particular, if only one characteristic vector there corresponds to λc, then in B(λc) the elements of any two columns will be proportional (In such a case l=1, hence the rank of λcI-A will be n-1).
In conclusion: If the coefficients of the characteristic polynomial are known, then the adjugate matrix may be found by (10). In addition, if the given matrix A is non-singular, then the inverse matrix A-1 can be found from (15). Also if λc is a characteristic value of A, the non-zero columns of B(λc) are characteristc vectors of A for λ=λc.

Example.  We find out the characteristic values and vectors from the matrix


From (1),


Comparing with (7), we have


Next we use (8),


so that from (11)


We will now evaluate B1 and B2 by using (12) and (13), respectively.


thus B(λ) is


Also |A|=16 and A-1 is obtained from (15), i.e.




We notice the eigenvalue λ=2 possesses multiplicity 2 and also that all the entries of the adjugate B(λ) are divisible by the binomial λ-2 (|B(2)|=0, i.e. λ=2 annihilates it), therefore it can be reduced which makes instructive this problem. Thus,


which for λ=2 it becomes


From this we get the charactreristic vectors (1,1,1) by multiplying the first colum by -1, and also (-3,1,3), both correponding to λ=2. Third column is a linear combinationMathworldPlanetmath of the first two (subtract it). Likewise we find for the another characteristic value λ=4


whence we get the eigenvectorMathworldPlanetmathPlanetmathPlanetmath (1,-1,-1), being the remaining two columns clearly proportional to the first one.

Title characteristic values and vectors (of a matrix)
Canonical name CharacteristicValuesAndVectorsofAMatrix
Date of creation 2013-03-22 17:43:58
Last modified on 2013-03-22 17:43:58
Owner perucho (2192)
Last modified by perucho (2192)
Numerical id 6
Author perucho (2192)
Entry type Topic
Classification msc 15A18
Synonym eigenvalues
Synonym eigenvectors