Begin2.DVI

given by col(1 , 0 ,0). Note that any nonzero constant times this vector is also an

eigenvector. Substituting the eigenvalue λ= 3 into the equation (10.32) gives the

equations −x 1 +2x 2 +2x 3 = 0,− 3 x 2 +4x 3 = 0 and − 3 x 2 +4x 3 = 0. These equations imply

that x 2 =^27 x 1 and x 3 = 143 x 1. This gives the eigenvector col(x 1 ,^27 x 1 , 143 x 1 ). Selecting

the value x 1 = 14 for convenience, one finds the eigenvector col(14, 4 ,3) corresponding

to the eigenvalue λ= 3. Note that any nonzero constant times this eigenvector is

also an eigenvector. Substituting the eigenvalue λ = 4 into the equation (10.32)

gives the equations − 2 x 1 + 2x 2 + 2x 3 = 0, − 4 x 2 + 4x 3 = 0 and − 3 x 2 + 3x 3 = 0. These

equations imply that x 3 =^12 x 1 and x 2 =^12 x 1 and so the eigenvector can be expressed

col(x 1 ,^12 x 1 ,^12 x 1 ). Selecting the value x 1 = 2 for convenience gives the eigenvector

col(2 , 1 ,1) corresponding to the eigenvalue λ= 4.

Properties of Eigenvalues and Eigenvectors

The following are some important properties concerning the eigenvalues

and eigenvectors associated with an n×nsquare matrix A.

Property 1: If Xis an eigenvector of A, then kX is also an eigenvector of

Afor any nonzero scalar k.

Assume that the vector Xis an eigenvector of A, so that it must satisfy

the equation AX =λX. If this equation is multiplied by a nonzero constant k

there results kAX =kλX which can be written A(kX ) = λ(kX )and given the

interpretation kX is an eigenvector of A.

Property 2: An eigenvector of a square matrix cannot correspond to two

different eigenvalues.

Let λ 1 , λ 2 with λ 1
=λ 2 be two different eigenvalues of A. Assume X 1 is an

eigenvector of Acorresponding to both λ 1 and λ 2 .Our assumption implies

that the equations

AX 1 =λ 1 X 1 and AX 1 =λ 2 X 1

must be satisfied simultaneously. Subtracting these equations shows us that

(λ 1 −λ 2 )X 1 = [0]. But, if λ 1 −λ 2
= 0 , then this equation would imply that

X 1 = [0],which contradicts the fact that X 1 must be a nonzero eigenvector.

Hence, the original assumption must be false.

Property 3: If a matrix Ahas one of its eigenvalues as zero and λ= 0 ,

then Ais a singular matrix.