Studocu is not sponsored or endorsed by any college or university Evalue-magic-tricks-handout Mathematics IA (The University of Adelaide) Studocu is not sponsored or endorsed by any college or university Evalue-magic-tricks-handout Mathematics IA (The University of Adelaide) Downloaded by Alice Yu ([email protected])lOMoARcPSD|32041740
Facts About Eigenvalues By Dr David Butler Definitions SupposeAis ann×nmatrix. •AneigenvalueofAis a numberλsuch thatAv=λvfor some nonzero vectorv. •AneigenvectorofAis a nonzero vectorvsuch thatAv=λvfor some numberλ. Terminology LetAbe ann×nmatrix. •The determinant|λI−A|(for unknownλ) is called thecharacteristic polynomialofA. (The zeros of this polynomial are the eigenvalues ofA.) •The equation|λI−A|= 0 is called thecharacteristic equationofA. (The solutions of this equation are the eigenvalues ofA.) •Ifλis an eigenvalue ofA, then the subspaceEλ ={v|Av=λv}is called theeigenspace ofA associated withλ. (This subspace contains all the eigenvectors with eigenvalueλ, and also the zero vector.) •An eigenvalueλ*ofAis said to havemultiplicitymif, when the characteristic polynomial is factorised into linear factors, the factor (λ−λ*) appearsmtimes. Theorems LetAbe ann×nmatrix. •The matrixAhasneigenvalues (including each according to its multiplicity). •The sum of theneigenvalues ofAis the same as the trace ofA(that is, the sum of the diagonal elements ofA). •The product of theneigenvalues ofAis the same as the determinant ofA. •Ifλis an eigenvalue ofA, then the dimension ofEλ is at most the multiplicity ofλ. •A set of eigenvectors ofA, each corresponding to a different eigenvalue ofA, is a linearly independent set. •Ifλn+cn-1λn-1+· · ·+c1λ+c0 is the characteristic polynomial ofA, thencn-1=−trace(A) andc0 = (−1)n |A|. •Ifλn+cn-1λn-1 +· · ·+c1λ+c0 is the characteristic polynomial ofA, thenAn+cn-1An-1+· · ·+c1A+c0 I=O. (The Cayley-Hamilton Theorem.) 1 Downloaded by Alice Yu ([email protected])lOMoARcPSD|32041740
Examples of Problems using Eigenvalues Problem: Ifλis an eigenvalue of the matrixA, prove thatλ2is an eigenvalue ofA2 . Solution: Sinceλis an eigenvalue ofA,Av=λvfor somevnegationslash =0. Multiplying both sides byAgives A(Av) =A(λv) A2 v=λAv =λλv =λ2 v Thereforeλ2is an eigenvalue ofA2 .squaresolid Problem: Prove that then×nmatrixAand its transposeAT have the same eigenvalues. Solution: Consider the characteristic polynomial ofAT:|λI−AT|=|(λI−A)T|=|λI−A|(since a matrix and its transpose have the same determinant). This result is the characteristic polynomial ofA, so ATandAhave the same characteristic polynomial, and hence they have the same eigenvalues.squaresolid Problem: The matrixAhas (1,2,1)Tand (1,1,0)Tas eigenvectors, both with eigenvalue 7, and its trace is 2. Find the determinant ofA. Solution: The matrixAis a 3×3 matrix, so it has 3 eigenvalues in total. The eigenspaceE7 contains the vectors (1,2,1)Tand (1,1,0)T, which are linearly independent. SoE7 must have dimension at least 2, which implies that the eigenvalue 7 has multiplicity at least 2. Let the other eigenvalue beλ, then from the traceλ+7+7 = 2, soλ=−12. So the three eigenvalues are 7, 7 and -12. Hence, the determinant ofAis 7×7× −12 =−588.squaresolid 2 Downloaded by Alice Yu ([email protected])lOMoARcPSD|32041740
The sum and product of eigenvalues Theorem:IfAis ann×nmatrix, then the sum of theneigenvalues ofAis the trace ofAand the product of theneigenvalues is the determinant ofA. Proof: WriteA=a11. . .a 1n .... .. . . . an1. . .a nn . Also let theneigenvalues ofAbeλ1, . . . ,λn . Finally, denote the characteristic polynomial ofAby p(λ) =|λI−A|=λn+cn-1λn-1 +· · ·+c1λ+c0 . Note that since the eigenvalues ofAare the zeros ofp(λ), this implies thatp(λ) can be factorised asp(λ) = (λ−λ1). . .(λ−λn ). Consider the constant term ofp(λ),c0 . The constant term ofp(λ) is given byp(0), which can be calculated in two ways: Firstly,p(0) = (0−λ1). . .(0−λn) = (−1)nλ1. . . λn . Secondly,p(0) =|0I−A|=| −A|= (−1)n|A|. Thereforec0= (−1)nλ1. . . λn= (−1)n |A|, and soλ1. . . λn =|A|.That is, the product of then eigenvalues ofAis the determinant ofA. Consider the coefficient ofλn-1,cn-1 . This is also calculated in two ways. Firstly, it can be calculated by expandingp(λ) = (λ−λ1). . .(λ−λn ). In order to get theλn-1 term, theλmust be chosen fromn−1 of the factors, and the constant from the other. Hence, theλn-1 term will be−λ1λn-1− · · · −λλn-1=−(λ1+· · ·+λn)λn-1. Thuscn-1=−(λ1+· · ·+λn ). Secondly, this coefficient can be calculated by expanding|λI−A|: |λI−A|= vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingleλ−a11−a12. . .−a 1n −a21λ−a22. . .−a 2n ....... .. . . . −an1−an2. . .λ−a nn vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle One way of calculating determinants is to multiply the elements in positions 1j1, 2j2, . . . ,njn , for each possible permutationj1. . . jn of 1. . . n.If the permutation is odd, then the product is also multiplied by−1. Then all of thesen! products are added together to produce the determinant. One of these products is (λ−a11). . .(λ−ann ). Every other possible product can contain at mostn−2 elements on the diagonal of the matrix, and so will contain at mostn−2λ's. Hence, when all of these other products are expanded, they will produce a polynomial inλof degree at mostn−2. Denote this polynomial byq(λ). Hence,p(λ) = (λ−a11). . .(λ−ann ) +q(λ). Sinceq(λ) has degree at mostn−2, it has noλn-1 term, and so theλn-1term ofp(λ) must be theλn-1 term from (λ−a11). . .(λ−ann ). However, the argument above for (λ−λ1). . .(λ−λn) shows that this term must be−(a11+· · ·+ann )λn-1. Thereforecn-1=−(λ1+· · ·+λn) =−(a11+· · ·+ann), and soλ1+· · ·+λn=a11+· · ·+ann . That is, the sum of theneigenvalues ofAis the trace ofA.squaresolid 3 Downloaded by Alice Yu ([email protected])lOMoARcPSD|32041740
The Cayley-Hamilton Theorem Theorem: LetAbe ann×nmatrix. Ifλn+cn-1λn-1+· · ·+c1λ+c0 is the characteristic polynomial ofA, thenAn+cn-1An-1 +· · ·+c1A+c0 I=O. Proof: Consider the matrixλI−A. If this matrix is multiplied by its adjoint matrix, the result will be its determinant multiplied by the identity matrix. That is, (λI−A)adj(λI−A) =|λI−A|I(1) Consider the matrix adj(λI−A). Each entry of this matrix is either the positive or the negative of the determinant of a smaller matrix produced by deleting one row and one column ofλI−A. The determinant of such a matrix is a polynomial inλof degree at mostn−1 (since removing one row and one column is guaranteed to remove at least oneλ). Let the polynomial in positionijof adj(λI−A) bebij0+bij1λ+· · ·+bij(n-1) λn-1. Then adj(λI−A) = b110+b111λ+· · ·+b11(n-1)λn-1. . .b1n0+b1n1λ+· · ·+b1n(n-1) λ n-1 .... .. . . . bn10+bn11λ+· · ·+bn1(n-1)λn-1. . .bnn0+bnn1 λ+· · ·+bnn(n-1) λn-1 =b110. . .b1n0.........bn10. . .bnn0+b111λ. . .b1n1λ.........bn11λ. . .bnn1λ+· · ·+ b11(n-1)λn-1. . .b1n(n-1) λn-1 .... ... . . bn1(n-1)λn-1. . .bnn(n-1) λ n-1 =b110. . .b1n0.........bn10. . .bnn0+λb111. . .b1n1.........bn11. . .bnn1+· · ·+λ n-1 b11(n-1). . .b 1n(n-1) .... ... . . bn1(n-1). . .b nn(n-1) Denote the matrices appearing in the above expression byB0,B1, . . . ,Bn-1 , respectively so that adj(λI−A) =B0+λB1+· · ·+λn-1 B n-1 Then (λI−A)adj(λI−A) = (λI−A)(B0+λB1+· · ·+λn-1 B n-1 =λB0+λ2B1+· · ·+λn B n-1 −AB0−λAB1− · · · −λn-1 AB n-1 =−AB0+λ(B0−AB1) +· · ·+λn-1(Bn-2−ABn-1) +λn B n-1 Next consider|λI−A|I. This is the characteristic polynomial ofA, multiplied by I. That is, |λI−A|I= (λn+cn-1λn-1 +· · ·+c1λ+c0 )I =λnI+cn-1λn-1 I+· · ·+c1λI+c0 I =c0I+λ(c1I) +· · ·+λ(cn-1 I) +λn I Substituting these two expressions into the equation (λI−A)adj(λI−A) =|λI−A|Igives −AB0+λ(B0−AB1) +. . .+λn-1(Bn-2−ABn-1) +λn B n-1 =c0I+λ(c1I)+. . .+λn-1(cn-1 I)+λn I 4 Downloaded by Alice Yu ([email protected])lOMoARcPSD|32041740
If the two sides of this equation were evaluated separately, each would be ann×nmatrix with each entry a polynomial inλ. Since these two resulting matrices are equal, the entries in each position are also equal. That is, for each choice ofiandj, the polynomials in positionijof the two matrices are equal. Since these two polynomials are equal, the coefficients of the matching powers ofλmust also be equal. That is, for each choice ofi,jandk, the coefficient ofλkin positionijof one matrix is equal to the coefficient ofλk in positionijof the other matrix. Hence, when each matrix is rewritten as sum of coefficient matrices multiplied by powers ofλ(as was done above for adj(λI−A) ), then for everyk, the matrix multiplied byλk in one expression must be the same as the matrix multiplied byλkin the other. In other words, we can equate the matrix coeffiicients of the powers ofλin each expression. This results in the following equations: c0I=−AB 0 c1I=B0−AB 1 . . . cn-1I=Bn-2−AB n-1 I=B n-1 Now right-multiply each equation by successive powers ofA(that is, the first is multiplied byI, the second is multiplied byA, the third is multiplied byA2, and so on until the last is multiplied byAn ). This produces the following equations: c0I=−AB 0 c1A=AB0−A2 B 1 . . . cn-1An-1=An-1Bn-2−An B n-1 AnI=An B n-1 Adding all of these equations together produces: c0I+c1A+· · ·+cn-1An-1+An-1=−AB0+AB0−A2B1+· · ·+An-1Bn-2−AnBn-1+An B n-1 c0I+c1A+· · ·+cn-1 An-1+An-1 =Osquaresolid 5 Downloaded by Alice Yu ([email protected])lOMoARcPSD|32041740
Polynomials acted upon matrices Theorem: LetAbe ann×nmatrix with eigenvaluesλ1, . . . ,λn(including multiplicity). Letg(x) =a0+a1 x+ · · ·+akxkbe a polynomial, and letg(A) =a0I+a1A+· · ·+ak Ak . Then the eigenvalues ofg(A) areg(λ1), . . . ,g(λn ) (including multiplicity). Proof: We will begin by showing that the determinant ofg(A) isg(λ1). . . g(λn ). By the fundamental theorem of algebra, the polynomialg(x) can be factorised intoklinear factors over the complex numbers.Hence we can writeg(x) =ak(x−c1). . .(x−ck ) for some complex numbersc1, . . . ,ck . Now a matrix commutes with all its powers, and with the identity, so it is also possible to writeg(A) asg(A) =ak(A−c1I). . .(A−ck I). Also, denote the characteristic polynomial ofAbyp(λ) =|λI−A|. Since the eigenvalues ofAare λ1, . . . , λn, the characteristic polynomial can be factorised asp(λ) = (λ−λ1). . .(λ−λn ). Consider the determinant ofg(A): |g(A)|=|ak(A−c1I). . .(A−ck I)| = (ak)n |A−c1I|. . .|A−ck I| = (ak)n | −(c1I−A)|. . .| −(ck I−A)| = (ak)n(−1)n|c1I−A|. . .(−1)n |ck I−A| = (ak)n(−1)nk |c1I−A|. . .|ck I−A| Now|ciI−A|is|λI−A|withλreplaced byci , that is, it is the characteristic polynomial ofA evaluated atλ=ci. Thus|ciI−A|=p(ci) = (ci−λ1). . .(ci−λn ). So,|g(A)|= (ak)n(−1)nk p(c1). . . p(ck ) = (ak)n(−1)nk ×(c1−λ1). . .(c1−λn ) ×. . . ×(ck−λ1). . .(ck−λn ) = (ak)n ×(λ1−c1). . .(λn−c1 ) ×. . . ×(λ1−ck). . .(λn−ck ) = (ak)n ×(λ1−c1). . .(λ1−ck ) ×. . . ×(λn−c1). . .(λn−ck ) =ak(λ1−c1). . .(λ1−ck ) ×. . . ×ak(λn−c1). . .(λn−ck ) =g(λ1)× · · · ×g(λn ) The above argument shows that ifg(x) is any polynomial, then|g(A)|=g(λ1). . . g(λn ). 6 Downloaded by Alice Yu ([email protected])lOMoARcPSD|32041740
Now we will show that the eigenvalues ofg(A) areg(λ1), . . . ,g(λn ). Letabe any number and consider the polynomialh(x) =a−g(x). Thenh(A) =aI−g(A), and the argument above shows that|h(A)|=h(λ1). . . h(λn ). Substituting the formulas forh(x) andh(A) into this equation gives that|aI−g(A)|= (a−g(λ1)). . .(a−g(λn )). Since this is true for all possiblea, it can be concluded that as polynomials,|λI−g(A)|= (λ−g(λ1)). . .(λ−g(λn )). But|λI−g(A)|is the characteristic polynomial ofg(A), which has been fully factorised here, so this implies that the eigenvalues ofg(A) areg(λ1), . . . ,g(λn ).squaresolid Some corollaries: LetAbe ann×nmatrix with eigenvaluesλ1, . . . ,λn . Then: •2Ahas eigenvalues 2λ1, . . . , 2λn . •A2has eigenvaluesλ21, . . . ,λ2n . •A+ 2Ihas eigenvaluesλ1+ 2, . . . ,λn + 2. •Ifp(λ) is the characteristic polynomial ofA, thenp(A) has eigenvalues 0, . . . , 0. 7 Downloaded by Alice Yu ([email protected])lOMoARcPSD|32041740
Similar matrices Definition: Two matricesAandBare calledsimilarif there exists an invertible matrixXsuch thatA=X-1BX. Theorem: SupposeAandBare similar matrices. ThenAandBhave the same characteristic polynomial and hence the same eigenvalues. Proof: Consider the characteristic polynomial ofA: |λI−A|=|λI−X-1 BX| =|λX-1IX−X-1 BX| =|X-1 (λI−B)X| =|X-1 ||λI−B||X| =1 |X| |λI−B||X| =|λI−B| This is the characteristic polynomial ofB, soAandBhave the same characteristic polynomial. HenceAandBhave the same eigenvaluessquaresolid. 8 Downloaded by Alice Yu ([email protected])lOMoARcPSD|32041740
Multiplicity and the dimension of an eigenspace Theorem: Ifλ*is an eigenvalue ofA, then the multiplicity ofλ*is at least the dimension of the eigenspaceEλ* . Proof: Suppose the dimension ofEλ*ismand letv1, . . . ,vmform a basis forEλ * . It is possible to findn−mother vectorsum+1, . . . ,unso thatv1, . . . ,vm,um+1, . . . ,un form a basis forRn . LetXbe then×nmatrix with thesenbasis vectors as its columns. This matrixX is invertible since its columns are linearly independent. Consider the matrixB=X-1AX. This matrix is similar toAand so it has the same characteristic polynomial asA. In order to describe the entries ofB, we will first investigateAX. AX=A[v1| · · · |vm|um+1| · · · |un ] = [Av1| · · · |Avm|Aum+1| · · · |Aun ] = [λ*v1| · · · |λ* vm|Aum+1| · · · |Aun](sincev1, . . . ,vm are eigenvalues ofA) B=X-1 AX =X-1[λ*v1| · · · |λ* vm|Aum+1| · · · |Aun ] = [X-1(λ*v1)| · · · |X-1(λ*vm)|X-1Aum+1| · · · |X-1 Aun ] = [λ*X-1v1| · · · |λ*X-1vm|X-1Aum+1| · · · |X-1 Aun ] Now considerX-1vi . This isX-1multiplied by thei'th column ofX, and so it is thei'th column ofX-1X. HoweverX-1 X=I, so itsi'th column is thei'th standard basis vectorei . Thus: B= [λ*X-1v1| · · · |λ*X-1vm|X-1Aum+1| · · · |X-1 Aun ] = [λ*e1| · · · |λ*em|X-1Aum+1| · · · |X-1 Aun ] =λ* 0. . .0b1(m+1). . .b 1n 0λ*. . .0b2(m+1). . .b 2n.................. . . . 00. . .λ* bm(m+1). . .bmn 00. . .0b(m+1)(m+1). . .b (m+1)n ................ .. . .. 00. . .0bn(m+1). . .b nn So,λI−B=λ−λ* 0. . .0−b1(m+1). . .−b 1n 0λ−λ*. . .0−b2(m+1). . .−b 2n................ .. . . . 00. . .λ−λ* −bm(m+1). . .−bmn 00. . .0λ−b(m+1)(m+1). . .−b (m+1)n ................ .. . .. 00. . .0−bn(m+1). . .λ−b nn If the determinant of this matrixλI−Bis expanded progressively along the firstmcolumns, it 9 Downloaded by Alice Yu ([email protected])lOMoARcPSD|32041740
results in the following: |λI−B|= (λ−λ*). . .(λ−λ*) vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingleλ−b(m+1)(m+1). . .−b (m+1)n .... .. . . . −bn(m+1). . .λ−b nn vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = (λ−λ*)m vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingleλ−b(m+1)(m+1). . .−b (m+1)n .... .. . . . −bn(m+1). . .λ−b nn vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle Hence, the factor (λ−λ*) appears at leastmtimes in the characteristic polynomial ofB(it may appear more times because of the part of the determinant that is as yet uncalculated).SinceA andBhave the same characteristic polynomial, the factor (λ−λ*) appears at leastmtimes in the characteristic polynomial ofA. That is, the multiplicity of the eigenvalueλ*is at leastm.squaresolid 10 Downloaded by Alice Yu ([email protected])lOMoARcPSD|32041740
The product of two matrices Theorem: LetAbe anm×nmatrix and letBbe ann×mmatrix, withn≥m. Then theneigenvalues of BAare themeigenvalues ofABwith the extra eigenvalues being 0. Proof: Consider the (m+n)×(m+n) matrices: M=parenleftbiggOn×nOn×mAABparenrightbigg,N= parenleftbiggBAO n×m AO m×m parenrightbigg Also letX= parenleftbiggIn×n B Om×nI m×m parenrightbigg Then ThenXM= parenleftbiggIBOIparenrightbigg parenleftbiggOO AAB parenrightbigg = parenleftbiggBABAB AAB parenrightbigg AndNX= parenleftbiggBAOAOparenrightbigg parenleftbiggIB OI parenrightbigg = parenleftbiggBABAB AAB parenrightbigg SoXM=NX. NowXis an upper triangular matrix with every entry on the diagonal equal to 1.Therefore it is invertible.Hence we can multiply both sides of this equation byX-1 to get M=X-1NX. ThusMandNare similar and so have the same characteristic polynomial. Consider the characteristic polynomial of each: |λI−M|=vextendsingle vextendsinglevextendsinglevextendsingleλI− parenleftbiggOn×nO n×m AAB parenrightbiggvextendsingle vextendsingle vextendsingle vextendsingle = vextendsinglevextendsinglevextendsinglevextendsingleparenleftbiggλIn×nO n×m −AλIm×m −AB parenrightbiggvextendsingle vextendsingle vextendsingle vextendsingle =|λIn×n||λIm×m −AB| =λn |λIm×m −AB| |λI−N|=vextendsinglevextendsinglevextendsinglevextendsingleλI− parenleftbiggBAO n×m AO m×m parenrightbiggvextendsingle vextendsingle vextendsingle vextendsingle = vextendsinglevextendsinglevextendsinglevextendsingleparenleftbiggλIn×n−ABO n×m −AλI m×m parenrightbiggvextendsingle vextendsingle vextendsingle vextendsingle =|λIn×n−BA||λIm×m | =λm |λIm×m −BA| 11 Downloaded by Alice Yu ([email protected])lOMoARcPSD|32041740
SinceMandNhave the same characteristic polynomial, |λI−M|=|λI−N| λn|λIm×m−AB|=λm |λIm×m −BA| λn-m |λIm×m−AB|=|λIm×m −BA| So the characteristic polynomial ofBAis the same as the characteristic polynomial ofAB, but multiplied byλn-m . HenceBAhas all of the eigenvalues ofAB, but withn−mextra zeros.squaresolid 12 Downloaded by Alice Yu ([email protected])lOMoARcPSD|32041740
A proof in finite geometry with a surprising use of eigenvalues Preliminaries: •Afinite projective planeis a collection of finitely many points and finitely many lines such that -Every two points are contained in precisely one line. -Every two lines share precisely one point. -There are at least three points on every line. -Not all the points are on the same line. •For a finite projective plane, there is a numberqcalled theordersuch that: -There areq+ 1 points on every line. -There areq+ 1 lines through every point. -There areq2 +q+ 1 points in total. -There areq2 +q+ 1 lines in total. •Apolarityof a finite projective plane is a one-to-one mapσwhich maps points to lines and lines to points, so that if the pointPis on the lineℓ, then the pointσ(ℓ) is on the lineσ(P), and also for any point or lineX,σ(σ(X)) =X. •Anabsolute pointwith respect to a polarityσof a projective plane is a pointPsuch that Pis on the lineσ(P). Theorem:A polarity of a finite projective plane plane must have an absolute point. Proof:Letσbe a polarity of a finite projective plane of orderq. Denote the points in the projective plane byP1,P2, . . . ,Pq2+q+1, and denote the lineσ(Pi) byℓi for eachi= 1, . . . ,q2 +q+ 1. Note that sinceσis a polarity, thenσ(ℓi) =σ(σ(Pi)) =Pi for anyi. Create a (q2+q+ 1)×(q2+q+ 1) matrixAas follows: if the pointPiis on the lineℓj , then put a 1 in entryijof the matrixA, otherwise, put a 0 in positionij. The matrixAis called an incidence matrix of the projective plane. Sinceσis a polarity, ifPiis onℓj, thenσ(Pi) is onσ(ℓj) =σ(σ(Pj)) =Pj . Hence if there is a 1 in positionij, then there is also a 1 in positionji. Thus the matrixAis symmetric andAT=A. Now an abolute point is a pointPisuch thatPiis onσ(Pi ). That is, an absolute point is a point such thatPiis onℓi . Hence, ifσhas an absolute point, then there is a 1 on the diagonal ofA. Therefore, the number of absolute points ofσis the sum of the diagonal elements ofA. That is, it is the trace ofA. To find the trace ofA, we may instead find the sum of the eigenvalues ofA. Firstly note that every row ofAcontains preciselyq+ 1 entries that are 1 since each point lies on q+ 1 lines. Hence the rows ofAall add toq+ 1. Therefore whenAis multiplied by (1,1, . . . ,1)T , 13 Downloaded by Alice Yu ([email protected])lOMoARcPSD|32041740
the result is (q+ 1, . . . , q+ 1)T. This means that (1, . . . ,1)T is an eigenvector ofAwith eigenvalue q+ 1. Consider the matrixA2. IfAhas eigenvalues ofλ1, . . . ,λq2 +q+1 , thenA2will have eigenvalues of λ21 , . . . ,λ2 q2+q+1 . Hence information about the eigenvalues ofAcan be obtained from the eigenvalues ofA2 . SinceA=AT, we have thatA2=AA=ATA. Theijelement ofATAis the dot product of the ith row ofAT with thejth column ofA, which is the dot product of theith column ofAwith the jth column ofA. Now each column ofArepresents a line of the projective plane and has a 1 in the position of each of the points on this line. Two lines share precisely one point, and so two columns ofAare both 1 in precisely one position. Hence the dot product of two distinct columns ofAis 1. One the other hand, any line containsq+ 1 points, and therefore any column ofAhasq+ 1 entries equal to 1. Hence the dot product of a column ofAwith itself isq+ 1. Therefore, the diagonal entries ofA2areq+1 and the other entries ofA2are 1. That isA2=J+qI, whereJis a matrix with all of its entries equal to 1. Now the matrixJis equal to the product of two vectors (1, . . . ,1)T (1, . . . ,1).Multiplying these vectors in the opposite order gives a 1×1 matrix: (1, . . . ,1)(1, . . . ,1)T= [q2+q+ 1]. This 1×1 matrix has eigenvalueq2 +q+ 1. Now for any two matrices such thatABandBAare both defined, the eigenvalues of the larger matrix are the same as the eigenvalues of the smaller matrix with the extra eigenvalues all being zero. Thus theq2+q+ 1 eigenvalues ofJmust beq2 +q+ 1, 0, . . . , 0. Consider the polynomialg(x) =x+q. The matrixg(J) =J+qI=A2 . Therefore the eigenvalues ofA2areg(q2+q+ 1),g(0) . . . ,g(0). That is, the eigenvalues ofA2areq2+ 2q+ 1,q, . . . ,q. One of the eigenvalues ofAisq+ 1, and so the remaining eigenvalues ofAmust all be√qor−√ q Suppose there arekeigenvalues equal to−√qandq2+q−kequal to√ q. Then the trace ofAis equal to−k√q+ (q2+q−k)√q+q+ 1 =√ q(q2+q−2k) +q+ 1 Ifq=p2for some natural numberp, then the trace isp(q2+q−2k) +p2 + 1, which is one more than a multiple ofp, and so it cannot be 0. Ifqis not a square number, then√ qis irrational and so ifq2+q−2knegationslash = 0 then the trace ofA must be irrational also. However, the entries ofAare all 0 or 1, so its trace is an integer. Hence q2+q−2k= 0 and the trace isq+ 1. In either case, the trace is not 0, so the polarity must have an absolute point.squaresolid 14 Downloaded by Alice Yu ([email protected])lOMoARcPSD|32041740
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