Eigenvalues, eigenvectors, and eigenspaces of linear ...

1 Eigenvalues, eigenvectors , andeigenspaces of linear operatorsMath 130 linear AlgebraD Joyce, Fall 2015 Eigenvalues and re lookingat linear operators on a vector spaceV, that is, linear transformationsx7 T(x) from the vectorspaceVto finite dimensionnwith a specifiedbasis , thenTis described by a squaren nmatrixA= [T] .We re particularly interested in the study the ge-ometry of these transformations in a way that wecan t when the transformation goes from one vec-tor space to a different vector space, namely, we llcompare the original vectorxto its imageT(x).Some of these vectors will be sent to other vectorson the same line, that is, a vectorxwill be sent toa scalar multiple xof a given linear operatorT:V V, a nonzero vectorxand a constant scalar arecalled aneigenvectorand itseigenvalue, respec-tively, whenT(x) = x.

2 For a given eigenvalue , the set of allxsuch thatT(x) = xis calledthe -eigenspace. The set of all eigenvalues for atransformation is called the operatorTis described by a matrixA, then we ll associate the eigenvectors , eigenval-ues, eigenspaces , and spectrum toAas well. AsAdirectly describes a linear operator onFn, we lltake its eigenspaces to be subsets -eigenspace is a subspace thatxandyare - eigenvectors andcis a scalar. ThenT(x+cy) =T(x) +cT(y) = x+c y= (x+cy).Thereforex+cyis also a -eigenvector. Thus,the set of - eigenvectors form a subspace reason these eigenvalues and eigenspaces areimportant is that you can determine many of theproperties of the transformation from them, andthat those properties are the most important prop-erties of the are matrix that theeigenvalues, eigenvectors , and eigenspaces of a lin-ear transformation were defined in terms of thetransformation, not in terms of a matrix that de-scribes the transformation relative to a particu-lar basis.

3 That means that they are invariants ofsquare matrices under change of basis. Recall thatifAandBrepresent the transformation with re-spect to two different bases, thenAandBare con-jugate matrices, that is,B=P 1 APwherePisthe transition matrix between the two bases. Theeigenvalues are numbers, and they ll be the sameforAandB. The corresponding eigenspaces willbe isomorphic as subspaces ofFnunder the linearoperator of conjugation byP. Thus we have thefollowing eigenvalues of a square matrixAare the same as any conjugate matrixB=P 1 APofA. Furthermore, each -eigenspace forAis iso-morphic to the -eigenspace forB. In particular,the dimensions of each -eigenspace are the 0 is an s a special situa-tion when a transformation has 0 an an meansAx=0for some nontrivial general, a 0- eigenspaces is the solution space ofthe homogeneous equationAx=0, what we vebeen calling the null space ofA, and its dimensionwe ve been calling the nullity ofA.

4 Since a squarematrix is invertible if and only if it s nullity is 0, wecan conclude the following square matrix is invertible if andonly if 0 is not one of its eigenvalues. Put another1way, a square matrix is singular if and only if 0 isone of its example transformation that has 0 as aneigenvalue is a projection, like (x,y,z)7 (x,y,0)that maps space to thexy-plane. For this projec-tion, the 0-eigenspace is 1 is an is another im-portant situation. It means the transformation hasa subspace of fixed points. That s because vectorxis in the 1-eigenspace if and only ifAx= example transformation that has 1 as aneigenvalue is a reflection, like (x,y,z)7 (x,y, z)that reflects space across 1-eigenspace, that is, its subspace of fixed points, isthexy-plane. We ll look at reflections inR2in de-tail in a transformation with 1 as an eigenvalueis the shear transformation (x,y)7 (x+y,y).

5 Its1-eigenspace is of reflections in ve lookedat reflections across some lines in the plane. There sa general form for a reflection across the line of slopetan , that is, across the line that makes an angleof with thex-axis. Namely, the matrix transfor-mationx7 Ax, whereA=[cos 2 sin 2 sin 2 cos 2 ],describes a such a reflection has fixed points, namely, the pointson the line being reflected across. Therefore, 1 isan eigenvalue of a reflection, and the 1-eigenspaceis the line of to that line is a line passing throughthe origin and its points are reflected across theorigin, that is to say, they re negated. Therefore, 1 is an eigenvalue , and the orthogonal line is itseigenspace. Reflections have only these two eigen-values, characteristic polynomial, the main toolfor finding do you find whatvalues the eigenvalues can be?

6 In the finite di-mensional case, it comes down to finding the rootsof a particular polynomial, called the that is an eigenvalue ofA. That meansthere is a nontrivial vectorxsuch thatAx= ,Ax x=0, and we can rewrite thatas (A I)x=0, whereIis the identity a homogeneous equation like (A I)x=0hasa nontrivial solutionxif and only if the determinantofA Iis 0. We ve shown the following scalar is an eigenvalue ofAif andonly if det(A I) = 0. In other words, is a rootof the polynomial det(A I), which we call thecharacteristic polynomialoreigenpolynomial. Theequation det(A I) = 0 is called that the characteristic polynomial has de-green. That means that there are at mostneigen-values. Since some eigenvalues may be repeatedroots of the characteristic polynomial, there maybe fewer reason there may be fewer thannval-ues is that the roots of the eigenvalue may not liein the fieldF.

7 That won t be a problem ifFisthe field of complex numbersC, since the Funda-mental Theorem of Algebra guarantees that rootsof polynomials lie can use characteristic polynomials to givean alternate proof that conjugate matrices havethe same eigenvalues. Suppose thatB=P ll show that the characteristic polynomials ofAandBare the same, that is,det(A I) = det(B I).That will imply that they have the same (B I) = det(P 1AP I)= det(P 1AP P 1 IP)= det(P 1(A I)P)= det(P 1) det(A I) det(P)= det(A I)How to find eigenvalues and we know the eigenvalues are the roots of thecharacteristic polynomial. We ll illustrate this withan example. Here s the process to find all the eigen-values and their associated ). Form the characteristic polynomialdet(A I).2). Find all the roots of it.

8 Since it is annthde-gree polynomial, that can be hard to do by hand ifnis very large. Its roots are the eigenvalues 1, 2,..3). For each eigenvalue i, solve the matrix equa-tion (A iI)x=0to find the ll find the characteristic polyno-mial, the eigenvalues and their associated eigenvec-tors for this matrix:A= 10 0 3 3 032 2 The characteristic polynomial is|A I|= 1 00 33 0322 = (1 )(3 )(2 ).Fortunately, this polynomial is already in factoredform, so we can read off the three eigenvalues: 1=1, 2= 3, and 3= 2. (It doesn t matter the orderyou name them.) Thus, the spectrum of this matrixis the set{1,2,3}.Let s find the 1-eigenspace. We need to solveAx= 1x. That s the same as solving (A 1I)x=0. The matrixA 1 Iis 00 0 3 2 032 1 which row reduces to 1 0160 1140 0 0 and from that we can read off the general solution(x,y,z) = ( 16z, 14z,z)wherezis arbitrary.

9 That s the one-dimensional1-eigenspace (which consists of the fixed points ofthe transformation).Next, find the 2-eigenspace. The matrixA 2 Iis 2 00 3 0032 1 which row reduces to 1 000 1 120 00 and from that we can read off the general solution(x,y,z) = (0,12z,z)wherezis arbitrary. That s the , find the 3-eigenspace. The matrixA 3 Iis 1 0 0 3 1 032 0 which row reduces to 1 0 00 1 00 0 0 and from that we can read off the general solution(x,y,z) = (0,0,z)wherezis arbitrary. That s the 130 Home Page ~ma130/4