Christian Parkinson UCLA Basic Exam Solutions: Linear ...

1 Christian ParkinsonUCLA Basic Exam Solutions: Linear Algebra1 Problem a Linear operator on a finite dimensional complex inner prod-uct spaceVsuch thatT T=TT . Show that there is an orthonormal basis ofVconsistingof eigenvectors T=TT , we prove this by induction on the dimension of the space thatToperates on. IfTisoperating on a 1-dimensional space, the claim is the claim holds for any normalToperating on ann 1 dimensional space(n 2). By the fundamental theorem of algebra, the characteristic polynomial ofT has aroot which is an eigenvalue ofT . Letvbe the corresponding non-zero eigenvector (wlog||v||= 1). Thenv ={x V: (x,v) = 0}has dimensionn 1. Also, ifx v , then(Tx,v) = (x,T v) = (x,v) = isT-invariant. Then the restriction ofTtov is a normal operator on ann 1 dimen-sional space.

2 Then by our inductive hypothesis, there is an orthonormal basis{v2,..,vn}forv consisting of eigenvectors ofT. Then{v,v2,..,vn}is an orthonormal set withnelements and is thus a basis forV. It remains to prove thatvis also an eigenvector ofTandthen we will have the required basis. SinceTis normal, so isT and thusT cIfor everyc C. Also, for any normal operatorSand any vectorx, we see||Sx||2= (Sx,Sx) = (x,S Sx) = (x,SS x) = (S x,S x) =||S x|| sincevis an eigenvector ofT , we have (T I)v= =||(T I)v||2=||(T I) v||2= (T I)v vsovis also an eigenvector ofT. Thus{v,v2,..,vn}is a basis ofVconsistingof eigenvectors ofT. This completes the induction and the :V WandS:W Xbe Linear transformations of real finitedimensional vector spaces.

3 Prove thatrank(T) + rank(S) dim(W) rank(S T) max{rank(T),rank(S)}. the Rank-Nullity Theorem,rank(T) + dim(ker(T)) = dim(V),(1)rank(S) + dim(ker(S)) = dim(W),(2)rank(S T) + dim(ker(S T)) = dim(V).(3)Adding the (1), (2) and then subtracting (3) givesrank(T) + rank(S) rank(S T) + dim(ker(T)) + dim(ker(S)) dim(ker(S T)) = dim(W).Let{v1,..,v`}be a basis for ker(T). Then (S T)(vi) = 0 for eachiso ker(T) ker(S T).Thus we can extend this to a basis{v1,..,v`,y1,..,yk}for ker(S T). Then for eachj= 1,..,k, we have0 = (S T)(yj) =S(T(yj)). Christian ParkinsonUCLA Basic Exam Solutions: Linear Algebra2 HenceT(yj) ker(S) for eachj. Further ifa1,..,ak Care such thata1T(y1) + +akT(yk) = 0,ThenT(a1y1+ +akyk) = 0soa1y1+ +akyk ker(T) so there areb1.

4 ,b` Csuch thata1y1+ +akyk=b1v1+ b`v`= a1y1+ +akyk b1v1 b`v`= these vectors form a basis for ker(S T) so in particular,a1= =ak= 0. Thus{T(y1),..,T(yk)}is a linearly independent subset of ker(S) and so dim(ker(S)) k. Hencedim(ker(T)) + dim(ker(S)) dim(ker(S T))and so the equation above yieldsrank(T) + rank(S) rank(S T) dim(W)orrank(T) + rank(S) dim(W) rank(S T)which is the first half of the suppose that{x1,..,xm}is a basis for im(S T). Then there areu1,..,um Vsuch that (S T)(ui) =xi,i= 1,.., (T(ui)) =xifor eachi, and so in particularxi im(S) for eachiand so we havemlinearly independent vectors in im(S). This givesrank(S T) rank(S) max{rank(S),rank(T)}.This is the second half of the a finite dimensional complex inner product space andf:V Ca Linear functional.

5 Show that there exists a vectorw Vsuch thatf(v) = (v,w)for allv ,..,vnbe an orthonormal basis forV. Givenf V , putf(vi) = i setw= 1v1+ + anyv V, there are 1,.., n Csuch thatv= 1v1+ + (v) = 1f(v1) + + nf(vn) = 1 1+ + n nand(v,w) =(n i=1 ivi,n j=1 jvj)=n i=1n j=1( ivi, jvj) =n i=1n j=1 i j(vi,vj).Thus by orthonormality,(v,w) =n i=1 i i=f(v). Christian ParkinsonUCLA Basic Exam Solutions: Linear Algebra3 Sincevwas arbitrary,f(v) = (v,w) for allv a finite dimensional complex inner product space andT:V Va Linear transformation. Prove that there exists an orthonormal ordered basisforVsuch that the matrix representation ofTin this basis is upper prove this by induction on the dimension of the spaceTacts upon. IfTisacting on a 1-dimensional space, the claim is the claim holds for Linear maps acting onn 1 dimensional spaces.

6 Let dim(V) =n. By the fundamental theorem of algebra, there is an eigenvalue Cand correspondingnon-zero eigenvector 06=v1 VofT ; wlog||v1||= 1. Thenv 1={x V: (x,v1) = 0}isann 1-dimensional space. Also forx v 1, we have(T(x),v1) = (x,T (v1)) = (x, v1) = (x,v1) = 1isT-invariant. ThusT v 1is an operator acting on ann 1-dimensional our inductive hypothesis, there is an orthonormal basis{v2,..,vn}forv such thatthe matrix ofTis upper-triangular with respect to this basis. Then{v1,v2,..,vn}is anorthonormal basis forV. Further,T(v1) = nj=1a1jvjfor somea1j Cand by assumptionT(vi) = nj=iaijxj. Thus the matrix ofTwith respect to this basis isA= a11a12a13 a1n0a22a23 a2n00a33 ann .Problem ann-dimensional complex vector space andT:V Valinear operator.

7 Suppose that the characteristic polynomial ofThasndistinct roots. Showthat there is a basisBofVsuch that the matrix representation ofTin the basisBis each root of the characteristic polynomial (and thus each eigenvalue ofT)is distinct and since eigenvectors corresponding to different eigenvalues are linearly indepen-dent, each eigenspaceE is a one-dimensionalT-invariant subspace. Let 1,.., nbe thedistinct eigenvalues ofTwith corresponding eigenvectorsv1,..,vn. We know that eigen-vectors corresponding to distinct eigenvalues are linearly independent, thusE i E j={0}wheneveri6=j. Further, since we haven-linearly independent vectors,{v1,..,vn}is a basisforV. The matrix ofTwith respect to this basis is[T] = 1 n , Christian ParkinsonUCLA Basic Exam Solutions: Linear Algebra4sinceT(vi) = ivi,i= 1.

8 , M3(R) satisfy det(A) = 1 andAtA=I=AAtwhereIis theidentity matrix. Prove that the characteristic polynomial ofAhas 1 as a the characteristic polynomial ofAhas a real root since it has odd be a real root of the characteristic polynomial. Then is an eigenvalue ofA. Suppose06=v R3is a normalized eivengector corresponding to . Then 2= 2(v,v) = ( v, v) = (Av,Av) = (v,AtAv) = (v,v) = = 1. If = 1, then we are done. If = 1, suppose , Care the othereigenvalues ofA. Then 1 = det(A) = = . If , are not real, they must be aconjugate pair sinceAis real. But this is impossible, because then 0. Thus both , are real. By the same reasoning as above, , = 1. Then = 1 forces = 1, = 1(or vice versa). ThusAhas 1 as an eigenvalue and so the characteristic polynomial ofAhas1 as a a finite dimensional real inner product space andT:V Va hermitian Linear operator.

9 Suppose the matrix representation ofT2in the standard basishas trace zero. Prove thatTis the zero dim(V) =nand letAbe the matrix ofTin the standard basis. SinceTis hermitian, so isAand thus by the spectral theorem, there is an orthonormal basis{v1,..,vn}forRnconsisting of eigenvectors ofA. Let 1,.., n Rbe the correspondingeigenvalues (repeats are allowed and the eigenvalues are real sinceAis hermitian). ThenAvi= ivi= A2vi= iAvi= ( 2i,vi) is an eigenpair forA2which is the matrix ofT2. We are given that the traceof the matrix is zero, but the trace is the sum of the eigenvalues. Hencen i=1 2i= 0= 1= = n= 0;again this holds since all eigenvalues of a hermitian operator are real. ThenAvi= 0,i=1,.., this meansAsends a basis forRnto zero.

10 This is only possible ifAis the zeromatrix. ThusTis the zero a 3 3 real symmetric matrix with determinant 6. Assume that(1,2,3) and (0,3, 2) are eigenvectors with corresponding eigenvalues 1 and 2 respectively.(a) Give an eigenvector of the form (1,x,y) which is linearly independent from the twovectors above.(b) What is the eigenvalue of this eigenvector? Christian ParkinsonUCLA Basic Exam Solutions: Linear answer the questions in the reverse order. The product of the eigenvaluesequals the determinant, so the third eigenvalue is 3. This answers (b).By the spectral theorem, the eigenspaces corresponding to distinct eigenvalues will beorthogonal. Here all eigenvalues are distinct. Since the first two eigenvectors span a twodimensional space, any vector orthogonal to both will necessarily be a third the cross product of the two vectors gives a vector which is orthogonal to both.