Linear algebra II Homework #1 solutions 1.

1 Linear algebra IIHomework #1 the eigenvalues and the eigenvectors of the matrixA= 4 62 5 .Since trA= 9 and detA= 20 12 = 8, the characteristic polynomial isf( ) = 2 (trA) + detA= 2 9 + 8 = ( 1)( 8).The eigenvectors with eigenvalue = 1 satisfy the systemAv=v, namely(A I)v= 0 = 3 62 4 xy = 00 = x+ 2y= 0 = x= means that every eigenvector with eigenvalue = 1 must have the formv= 2yy =y 21 ,y6= , the eigenvectors with eigenvalue = 8 are solutions ofAv= 8v, so(A 8I)v= 0 = 4 62 3 xy = 00 = 2x 3y= 0 = x= 3y/2and every eigenvector with eigenvalue = 8 must have the formv= 3y/2y =y 3/21 ,y6= the eigenvalues and the eigenvectors of the matrixA= 3 3 11 1 13 9 5 .The eigenvalues ofAare the roots of the characteristic polynomialf( ) = det(A I) = 3+ 7 2 16 + 12 = ( 2)2( 3).The eigenvectors ofAare nonzero vectors in the null spacesN(A 2I) = Span 310 , 101 ,N(A 3I) = Span 113.

2 Following matrix has eigenvalues = 1,1,1,4. Is it diagonalisable? 3 1 3 32 2 3 32 1 2 33 0 3 4 .When it comes to the eigenvalue = 1, row reduction ofA IgivesA I= 2 1 3 32 1 3 32 1 3 33 0 3 3 1 0 1 12 1 3 30 0 0 00 0 0 0 1 0 1 10 1 1 10 0 0 00 0 0 0 ,so there are 2 pivots and 2 linearly independent eigenvectors. When = 4, we similarly getA I= 1 1 3 32 2 3 32 1 6 33 0 3 0 .. 1 0 0 10 1 0 10 0 1 10 0 0 0 ,so there are 3 pivots and 1 linearly independent eigenvector. This gives a total of 3 linearlyindependent eigenvectors, soAis not diagonalisable. One does not really need to find theeigenvectors in this case, but those are nonzero elements ofthe null spacesN(A I) = Span 1110 , 1 101 ,N(A 4I) = Span 1111 . a square matrix which is both diagonalisable and invertible. Show thatevery eigenvector ofAis an eigenvector ofA 1and thatA 1is thatvis an eigenvector ofAwith eigenvalue.

3 Then is nonzero because = 0 = Av= v= 0 = A 1Av= 0 = v= 0and eigenvectors are nonzero. To see thatvis also an eigenvector ofA 1, we note that v=Av= A 1v=v= A 1v= proves the first part. To prove the second, assume thatAis ann nmatrix. It mustthen havenlinearly independent eigenvectors which form a matrixBsuch thatB 1 ABisdiagonal. These vectors are also eigenvectors ofA 1, soB 1A 1 Bis diagonal as algebra IIHomework #2 a basis for both the null space and the column space of thematrixA= 1 2 0 53 1 5 04 2 6 24 3 5 5 .The reduced row echelon form ofAis easily found to beR= 1 0 2 10 1 1 30 0 0 00 0 0 0 .Since the pivots appear in the first and the second columns, this implies thatC(A) = Span{Ae1, Ae2}= Span 1344 , 2123 .The null space ofAis the same as the null space ofR. It can be expressed in the formN(A) = 2x3+x4x3 3x4x3x4 :x3, x4 R = Span 2110 , 1 301.

4 That the following matrix is 7 1 33 5 35 1 1 .The eigenvalues ofAare the roots of the characteristic polynomialf( ) = det(A I) = 3+ 11 2 38 + 40 = ( 2)( 4)( 5).Since the eigenvalues ofAare distinct, we conclude thatAis the eigenvalues and the generalised eigenvectors of the matrixA= 1 1 2 3 1 3 5 1 6 .The eigenvalues ofAare the roots of the characteristic polynomialf( ) = det(A I) = 3+ 6 2 9 + 4 = (4 )( 1) it comes to the eigenvalue = 4, one can easily check thatN(A 4I) = Span 112 ,N(A 4I)2=N(A 4I).This implies thatN(A 4I)j=N(A 4I) for allj 1, so we have found all generalisedeigenvectors with = 4. When it comes to the eigenvalue = 1, one similarly hasN(A I) = Span 101 ,N(A I)2=N(A I)3= Span 010 , 101 .In view of the general theory, we must thus haveN(A I)j=N(A I)2for allj thatv1,v2, .. ,vkform a basis for the null space of a square matrixAandthatC=B 1 ABfor some invertible matrixB.

5 Find a basis for the null space note that a vectorvlies in the null space ofC=B 1 ABif and only ifB 1 ABv= 0 ABv= 0 Bv N(A) Bv=kXi=1civifor some scalar coefficientsci. In particular, the vectorsB 1vispan the null space ofC. Toshow that these vectors are also linearly independent, we note thatkXi=1ciB 1vi= 0 kXi=1civi= 0 ci= 0 for algebra IIHomework #3 following matrix has = 4 as its only eigenvalue. What is its Jordan form?A= 1 5 1 2 7 1 1 2 4 .In this case, the null space ofA 4 Iis one-dimensional, as row reduction givesA 4I= 3 5 1 2 3 1 1 2 0 1 2 00 1 10 1 1 1 0 20 1 10 0 0 .Similarly, the null space of (A 4I)2is two-dimensional because(A 4I)2= 2 2 2 1 1 1 1 1 1 1 1 10 0 00 0 0 ,while (A 4I)3is the zero matrix, so its null space is three-dimensional. The diagramofJordan chains is then and there is a single 3 3 Jordan block, namelyJ= 1 1 = 41 41 4.

6 Following matrix has = 4 as its only eigenvalue. What is its Jordan form?A= 1 3 3 2 6 2 1 1 5 .In this case, the null space ofA 4 Iis two-dimensional, as row reduction givesA 4I= 3 3 3 2 2 2 1 1 1 1 1 10 0 00 0 0 .On the other hand, (A 4I)2is the zero matrix, so its null space is three-dimensional. Thus,the diagram of Jordan chains is and there is a Jordan chain of length 2 as well as aJordan chain of length 1. These Jordan chains give a 2 2 block and an 1 1 block, soJ= 41 44 . a Jordan chain of length 2 for the matrixA= 5 22 1 .The eigenvalues of the given matrix are the roots of the characteristic polynomialf( ) = 2 (trA) + detA= 2 6 + 9 = ( 3)2,so = 3 is the only eigenvalue. Using row reduction, we now getA 3I= 2 22 2 1 10 0 ,so the null space ofA 3 Iis one-dimensional. On the other hand, (A 3I)2is the zeromatrix, so its null space is two-dimensional.

7 To find a Jordan chain of length 2, we pick avectorv1that lies in the latter null space, but not in the former. We can alwaystakev1=e1= 10 = v2= (A 3I)v1= 22 ,but there are obviously infinitely many choices. Another possible choice would bev1=e2= 01 = v2= (A 3I)v1= 2 2 . a 3 3 matrix that hasv1,v2,v3as a Jordan chain of length 3 and letBbe the matrix whose columns arev3,v2,v1(in the order listed). ComputeB the vectorsv1,v2,v3form a Jordan chain with eigenvalue , in which case(A I)v1=v2,(A I)v2=v3,(A I)v3= find the columns ofB 1AB, we need to find the coefficients that one needs in order toexpress the vectorsAv3, Av2, Av1in terms of the given basis. By above, we haveAv3= v3,Av2=v3+ v2,Av1=v2+ the coefficients of the vectorsv3,v2,v1in the given order, we conclude thatB 1AB= 1 1 . Linear algebra IIHomework #4 the Jordan form and a Jordan basis for the matrixA= 3 1 26 2 62 1 3.

8 The characteristic polynomial of the given matrix isf( ) = det(A I) = 3+ 4 2 5 + 2 = (2 )( 1)2,so its eigenvalues are = 1,1,2. The corresponding null spaces are easily found to beN(A I) = Span 120 , 101 ,N(A 2I) = Span 131 .These contain 3 linearly independent eigenvectors, soAis diagonalisable andB= 1 1 12 0 30 1 1 = J=B 1AB= 112 . the Jordan form and a Jordan basis for the matrixA= 2 4 21 4 12 0 2 .The characteristic polynomial of the given matrix isf( ) = det(A I) = 3+ 8 2 20 + 16 = (4 )( 2)2,so its eigenvalues are = 4,2,2. The corresponding null spaces are easily found to beN(A 4I) = Span 111 ,N(A 2I) = Span 012 .This implies thatAis not diagonalisable and that its Jordan form isJ=B 1AB= 421 2 .To find a Jordan basis, we need to find vectorsv1,v2,v3such thatv1is an eigenvector witheigenvalue = 4 andv2,v3is a Jordan chain with eigenvalue = 2. In our case, we haveN(A 2I)2= Span 100 , 012 ,so it easily follows that a Jordan basis is provided by the vectorsv1= 111 ,v2= 100 ,v3= (A 2I)v2= 012.

9 ThatAis a 2 2 matrix withA2=A. Show thatAis thatAis not diagonalisable. Then its eigenvalues are not distinct, so there is adouble eigenvalue and the Jordan form isJ= 1 = J2= 1 1 = 22 2 .WriteJ=B 1 ABfor some invertible matrixB. SinceA2=A, we must also haveJ2=B 1AB B 1AB=B 1A2B=B 1AB=J= 1 .Comparing the last two equations now gives 2= and 2 = 1, a 5 5 matrixAhas characteristic polynomialf( ) = 4(2 ) and its columnspace is two-dimensional. Find the dimension of the column space eigenvalue = 0 has multiplicity 4 and the number of Jordan chains isdimN(A) = 5 dimC(A) = particular, the Jordan chain diagram is and we have dimN(A2) = 4, sodimC(A2) = 5 dimN(A2) = algebra IIHomework #5 1 andy0= 2. Suppose the sequencesxn, ynare such thatxn= 4xn 1 yn 1,yn=xn 1+ 2yn 1for eachn 1. Determine each ofxnandynexplicitly in terms can express this problem in terms of matrices by writingun= xnyn = 4 11 2 xn 1yn 1 =Aun 1= un= us now focus on computingAn.

10 The characteristic polynomial ofAisf( ) = 2 (trA) + detA= 2 6 + 9 = ( 3)2,so the only eigenvalue is = 3. As one can easily check, the corresponding null space isN(A 3I) = Span 11 ,while (A 3I)2is the zero matrix. Thus, a Jordan basis is provided by the vectorsv1= 10 ,v2= (A 3I)v1= 11 .LettingBbe the matrix whose columns arev1andv2, we must thus haveJ=B 1AB= 31 3 = Jn=B 1 AnB= 3nn3n 13n .Once we now solve this equation forAn, we find thatAn=BJnB 1= 1 10 1 3nn3n 13n 1 10 1 = (n+ 3)3n 1 n3n 1n3n 1(3 n)3n 1 .In particular, the sequencesxn, ynare given explicitly by xnyn =un=Anu0= (n+ 1)3n(n 2)3n . of the following matrices are similar? 2 0 01 2 00 1 2 ,B= 2 1 00 2 10 0 2 ,C= 2 0 10 2 00 0 2 .Two matrices are similar if and only if they have the same Jordan this case,Ais already in Jordan form, all matrices have = 2 as their only eigenvalue, whiledimN(B 2I) = 1,dimN(C 2I) = means that the Jordan form ofBconsists of one block, soBis similar toA.