MATH 304Linear AlgebraLecture 36:Complex eigenvalues and eigenvectors.Symmetric and orthogonal matrices.Complex numbersC: complex numbers.Complex number:z = x + iy,where x, y ∈ R and i2= −1.i =√−1: imaginary unitAlternative notation: z = x + yi.x = real part of z,iy = imaginary part of zy = 0 =⇒ z = x (real number)x = 0 =⇒ z = iy (purely imaginary number)We add, subtract, and multiply complex numbers aspolynomials in i (but keep in mind that i2= −1).If z1= x1+ iy1and z2= x2+ iy2, thenz1+ z2= (x1+ x2) + i(y1+ y2),z1− z2= (x1− x2) + i(y1− y2),z1z2= (x1x2− y1y2) + i(x1y2+ x2y1).Given z = x + iy, the complex conjugate of z is¯z = x −iy. The modulus of z is |z| =px2+ y2.z¯z = (x + iy )(x −iy) = x2−(iy)2= x2+ y2= |z|2.z−1=¯z|z|2, (x + iy)−1=x −iyx2+ y2.Complex exponentialsDefinition. For any z ∈ C letez= 1 + z +z22!+ ··· +znn!+ ···Remark. A sequence of complex numbersz1= x1+ iy1, z2= x2+ iy2, . . . convergesto z = x + iy if xn→ x and yn→ y as n → ∞.Theorem 1 If z = x + iy, x, y ∈ R, thenez= ex(cos y + i sin y).In particular, eiφ= cos φ + i sin φ, φ ∈ R.Theorem 2 ez+w= ez· ewfor all z, w ∈ C.Proposition eiφ= cos φ + i sin φ for all φ ∈ R.Proof: eiφ= 1 + iφ +(iφ)22!+ ··· +(iφ)nn!+ ···The sequence 1, i, i2, i3, . . . , in, . . . is periodic:1, i , −1, −i|{z }, 1, i, −1, −i| {z }, . . .It follows thateiφ= 1 −φ22!+φ44!− ··· + (−1)kφ2k(2k)!+ ···+ iφ −φ33!+φ55!− ··· + (−1)kφ2k+1(2k + 1)!+ ···= cos φ + i sin φ.Geometric representationAny complex number z = x + iy is represented bythe vector/point (x, y ) ∈ R2.yx0rφ0x = r cos φ, y = r sin φ =⇒ z = r (cos φ + i sin φ) = reiφIf z1= r1eiφ1and z2= r2eiφ2, thenz1z2= r1r2ei(φ1+φ2), z1/z2= (r1/r2)ei(φ1−φ2).Fundamental Theorem of AlgebraAny polynomial of d egree n ≥ 1, with complexcoefficients, has exactly n roots (counting withmultiplicities).Equivalently, ifp(z) = anzn+ an−1zn−1+ ··· + a1z + a0,where ai∈ C and an6= 0, then there exist complexnumbers z1, z2, . . . , znsuch thatp(z) = an(z − z1)(z − z2) . . . (z − zn).Complex eigenvalues and eigenvectorsExample. A =0 −11 0. det(A − λI ) = λ2+ 1.Characteristic roots: λ1= i and λ2= −i.Associated e igenvectors: v1=1−iand v2=1i.0 −11 01−i=i1= i1−i,0 −11 01i=−i1= −i1i.v1, v2is a basis of eigenvectors. In which space?ComplexificationInstead of the real vector space R2, we consider acomplex vector space C2(all complex numbers areadmissible as scalars).The linear operator f : R2→ R2, f (x) = Ax isextended to a complex linear operatorF : C2→ C2, F (x) = Ax.The vectors v1= (1, −i ) and v2= (1, i ) form abasis for C2.C2is also a real vector space (of real dimension 4). Thestandard real basis for C2is e1= (1, 0), e2= (0, 1),ie1= (i, 0), ie2= (0, i). The matrix of the operator F withrespect to this bas is has block structureA OO A.Dot product of complex vectorsDot product of real vectorsx = (x1, . . . , xn), y = (y1, . . . , yn) ∈ Rn:x · y = x1y1+ x2y2+ ··· + xnyn.Dot product of c omplex vectorsx = (x1, . . . , xn), y = (y1, . . . , yn) ∈ Cn:x · y = x1y1+ x2y2+ ··· + xnyn.If z = r + it (t, s ∈ R) thenz = r − it,zz = r2+ t2= |z|2.Hence x · x = |x1|2+ |x2|2+ ··· + |xn|2≥ 0.Also, x · x = 0 if and only if x = 0.The norm is defined by kxk =√x · x.Normal matricesDefinition. An n×n matrix A is called• symmetric if AT= A;• orthogonal if AAT= ATA = I , i.e., AT= A−1;• normal if AAT= ATA.Theorem Let A be an n×n matrix with realentries. Then(a) A is normal ⇐⇒ there exists an orthonormalbasis for Cnconsisting of eigenvectors of A;(b) A is symmetric ⇐⇒ there exists an orthonormalbasis for Rnconsisting of eigenvectors of A.Example. A =1 0 10 3 01 0 1.• A is symmetric.• A has three eigenvalues: 0, 2, and 3.• Associated eigenvectors are v1= (−1, 0, 1),v2= (1, 0, 1), and v3= (0, 1, 0), respectively.• Vectors1√2v1,1√2v2, v3form an orthonormalbasis for R3.Theorem Suppose A is a normal matrix. Then forany x ∈ Cnand λ ∈ C one hasAx = λx ⇐⇒ ATx =λx.Thus any normal matrix A shares with ATall realeigenvalues and the corresponding eigenvect ors.Also, Ax = λx ⇐⇒ Ax = λ x for any matrix Awith real entries.Corollary All eigenvalues λ of a symmetric matrixare real (λ = λ). All eigenvalues λ of anorthogonal matrix satisfy λ = λ−1⇐⇒ |λ| = 1.Why are orthogonal matrices called so?Theorem Given an n×n matrix A, the followingconditions are equivalent:(i) A is orthogonal: AT= A−1;(ii) columns of A form an orthonormal b asis for Rn;(iii) rows of A form an orthonormal basis for Rn.Proof: Entries of the matrix ATA are dot products ofcolumns of A. Entries of AATare dot products of rows of A.Thus an orthogonal matrix is the transition matrixfrom one orthonormal basis to another.Example. Aφ=cos φ −sin φsin φ cos φ.• AφAψ= Aφ+ψ• A−1φ= A−φ= ATφ• Aφis orthogonal• det(Aφ− λI ) = (cos φ − λ)2+ sin2φ.• Eigenvalues: λ1= cos φ + i sin φ = eiφ,λ2= cos φ − i sin φ = e−iφ.• Associated eigenvectors: v1= (1, −i ),v2= (1, i ).• Vectors1√2v1and1√2v2form an orthonormalbasis for C2.Consider a linear operator L : Rn→ Rn, L(x) = Ax,where A is an n×n orthogonal matrix.Theorem There exists an orthonormal basis for Rnsuch that the matrix of L relative to this basis has adiagonal block structureD±1O . . . OO R1. . . O............O O . . . Rk,where D±1is a diagonal matrix whose diagonalentries are equal to 1 or −1, andRj=cos φj−sin φjsin φjcos φj, φj∈
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