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.

2 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. 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.

3 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. 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).

4 (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.

5 ,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).

6 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. 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.

7 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. 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.

8 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. Suppose that the characteristic polynomial ofThasndistinct roots.

9 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.}

10 ,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,.., 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.