Transcription of SOLUTION: ASSIGNMENT 5 - UCB Mathematics | Department …
1 SOLUTION: ASSIGNMENT the subset of P2given below a subspace? Find a basis if it is.{p(t) :p (1)=p(2)}(p is the derivative).S . LetS={p(t) :p (1)=p(2)}. For anyp1,p2 S,k1,k2 R, by(k1 p1+k2 p2) |t=1=k1p 1(1)+k2p 2(1)=k1p1(2)+k2p2(2)=(k1 p1+k2 p2)|t=2we checkedk1 p1+k2 p2 S. ThereforeSis a (t)=a+bt+ct2 S,p (1)=p(2) b+2c=a+2b+4c a+b+2c=0 (a,b,c)=( 2 , , )where , R. Thenp(t)= 2 + t+ t2= (t 1)+ (t2 2), {t 1,t2 2}. Butt 1,t2 2 Sare obviously independent. Therefore (t 1,t2 2) is a basis as in {p(t) :1 0p(t)dt=0}S.
2 LetS={p(t) : 10p(t)dt=0}. For anyp1,p2 S,k1,k2 R, by1 0(k1 p1+k2 p2)dt=k11 0p1(t)dt+k21 0p2(t)dt=k10+k20=0we checkedk1 p1+k2 p2 S. ThereforeSis a (t)=a+bt+ct2 S,1 0p(t)dt=0 a+b2+c3=0 (a,b,c)=( 2 3, , )where , R. Thenp(t)= 2 3+ t+ t2= (t 12)+ (t2 13), {t 12,t2 13. Butt 12,t 13 Sare obviously independent. Therefore (t 12,t 13) is a basis a basis for the space of all diagonal2 2matrices, and determine its . Any diagonal 2 2 matrix looks like(a00d)=a(1 00 0)+d(0 00 1)This tells us that ((1 00 0),(0 00 1)) is a basis because these two matrices are already independent as inR2 2.}
3 The dimension is a basis for the space of all diagonal n n matrices, and determine its . Any diagonaln nmatrix looks like =a1E11+ +anEnnwhereEiiis the matrix with entries all 0 except a 1 at the i th diagonal entry. This tells us that (E11, ,Enn)is a basis because thesenmatrices are already independent as inRn n. The dimension a basis of the space of all upper triangular3 3matrices and determine its .Any upper triangular 3 3 matrix looks like a11a12a13a22a23a33 =a11E11+a12E12+a13E13+a22E22+a23E23+a33E 33whereEi jdenotes the matrix with entries all 0 except for a 1 at the (i,j)-th place.
4 This tells us that(E11,E12,E13,E22,E23,E33) is a basis because these 6 matrices are already independent as inR3 3. Thedimension is a basis of the space of all2 2matrices S such that(1 11 1)S=S(2 00 0)S . LetS=(a bc d), the condition becomes(1 11 1) (a bc d)=(a bc d) (2 00 0)which is simplified to be(a+c b+da+c b+d)=(2a02c0)By comparing the entries we get a+c=2ab+d=0a+c=2cb+d=0so (a,b,c,d)=( , , , ). ThereforeS=( )= (1 01 0)+ (010 1), ((1 01 0),(010 1))is a basis of the space of allS s.
5 The dimension is all solutions of the differential equation f (x)+8f (x) 20f(x)= . Solving the characteristic equation 2+8 20=0 we have two eigenvalues 1=2, 2= all solutions aref(x)=c1e2t+c2e 10t, wherec1,c2 . Review how to solve a homogeneous linear ODE in Math 1B. Another way is to follow what wedid in Example 18 of Section this exercise we will show that the functioncosx andsinx span the solution space V of thedifferential equation f (x)= f(x).2a. Show that if g(x)is in V , then the function g(x)2+g (x)2is Show that if g(x)is in V , with g(0)=g (0)=0, then g(x)=0for all If f(x)is in V , then g(x)=f(x) f(0) cosx f (0) sinx is in V as well.
6 Show then f(x)=f(0) cosx f (0) . a. Taking derivative we have(g2+(g )2) =(g2) +((g )2) =2gg +2g g =2g(g+g )Butg Vmeansg = g, so the right handed side is zero. The derivative being always zero meansg2+(g )2is Becauseg(x)2+g (x)2=Cis constant, plugging ing(0)=g (0)=0 we know the constantC= (x)2+g (x)2=0, and since the summands are nonnegative, they must both be zero. In particular,g(x)= (x) is inVbecause it s a linear combination off, cosx, sinxwhich are all inV. Butg(0)=f(0) f(0) cos 0 f (0) sin 0=0,g (0)=f (0)+f(0) sin 0 f (0) cos 0=0.
7 By b. we knowg(x)=0, (x)=f(0) cosx+f (0) out if the transformation it is linear, and when linear, if it is (M)=(1 23 6)fromR2 2toR2 . Denote(1 23 6)asA. For anyM1,M2 R2 2andk1,k2 R,T(k1M1+k2M2)=(k1M1+k2M2)A=k1M1A+k2M2A= k1T(M1)+k2T(M2)ThereforeTis not an isomorphism becauseAis not invertible. In this caseTmaps any matrixto a non-invertible matrix so imTcannot beR2 2( it avoids all invertible matrices). as in , T(c)=c(2 34 5)fromRtoR2 . Denote(2 34 5)asA. For anyc1,c2 Randk1,k2 R,T(k1c1+k2c2)=(k1c1+k2c2)A=k1c1A+k2c2A= k1T(c1)+k2T(c2)ThereforeTis an isomorphism because the dimensions of target space and the source spaceare as in , T(M)=(2 35 7)M M(2 35 7)fromR2 2toR2.
8 Denote(2 35 7)asA. For anyM1,M2 R2 2andk1,k2 R,T(k1M1+k2M2)=A(k1M1+k2M2) (k1M1+k2M2)A=k1(AM1 M1A)+k2(AM2 M2A)=k1T(M1)+k2T(M2)ThereforeTis linear. Observe thatT(I2)=AI2 I2A=A A=0 whereI2is the unit matrix, and henceTis not an isomorphism because kerT,{0}( it containsI2). as in , T(f(t))=f( t)from P2to . For anyf1,f2 P2andk1,k2 R,T(k1f1+k2f2)(t)=(k1f1+k2f2)( t)=k1f1( t)+k2f2( t)=k1T(f1)(t)+k2T(f2)(t)ThereforeTis linear. Observe thatT2is identity, (T(f(t)))=T(f( t))=f(t), soTis the inverse invertible and hence is as in , T(f(t))=t f (t)from P2to.
9 For anyf1,f2 P2andk1,k2 R,T(k1f1+k2f2)(t)=t(k1f1(t)+k2f2(t)) =k1t f 1(t)+k2t f 2(t)=k1T(f1)(t)+k2T(f2)(t)ThereforeTis linear. Observe thatT(c)=t0=0 wherecis any constant polynomial, and henceTis not anisomorphism because kerT,{0}( it contains constant polynomials). as in , T(f(t))=f (t)from P to . For anyf1,f2 Pandk1,k2 R,T(k1f1+k2f2)(t)=(k1f1(t)+k2f2(t)) =k1f 1(t)+k2f 2(t)=k1T(f1)(t)+k2T(f2)(t)ThereforeTis linear. Observe thatT(c)=t0=0 wherecis any constant polynomial, and henceTis not anisomorphism because kerT,{0}( it contains constant polynomials).
10 R . But in this case, imT=P, it is surjective. The explanation is thatPis infinite image,rank, kernel and nullity of the transformation in . Letf(t)=a+bt+ct2, thenT(f)(t)=t(a+bt+ct2) =bt+2ct2. Thus imT=span{t,2t2}.Considerbt+2ct2=0 as a polynomial, all coefficients are zero, so (a,b,c)=( ,0,0) where kerT=span{1}. The rank is dim imT=2, and the nullity is dim kerT= which constants k is the linear transformationT(M)=(2 30 4)M M(3 00k)an isomorphism fromR2 2toR2 2?S . The most standard way to understand a linear transformation is to find out its matrix under a favor-able basis.
