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Solution to Linear Algebra Hoffman & Kunze Chapter 3.7


Exercise 3.7.1

Let F be a field and let f be the linear functional on F2 defined by f(x1,x2)=ax1+bx2. For each of the following linear operators T, let g=ftf, and find g(x1,x2).

(a) T(x1,x2)=(x1,0);
(b) T(x1,x2)=(x2,x1);
(c) T(x1,x2)=(x1x2,x1+x2).

Solution:

(a) g(x1,x2)=Ttf(x1,x2)=f(T(x1,x2))=f(x1,0)=ax1.

(b) g(x1,x2)=Ttf(x1,x2)=f(T(x1,x2))=f(x2,x1)=axs+bx1.

(c) We have g(x1,x2)=Ttf(x1,x2)=f(T(x1,x2))=f(x1x2,x1+x2)=a(x1x2)+b(x1+x2)=(a+b)x1+(ba)x2.


Exercise 3.7.2

Let V be the vector space of all polynomial functions over the field of real numbers. Let a and b be fixed real numbers and let f be the linear functional on V defined by
f(p)=abp(x)dx. If D is the differentiation operator on V, what is Dtf?

Solution: Let p(x)=c0+c1x++cnxn. Then
Dtf(p)=f(D(p))=f(c1+2c2x+3c3x2++ncnxn1)=c1+c2x2+cnxn0b=p(b)p(a).


Exercise 3.7.3

Let V be the space of all n×n matrices over a field F and let B be a fixed n×n matrix. If T is the linear operator on V defined by T(A)=ABBA, and if f is the trace function, what is Ttf?

Solution: By exercise 3 in section 3.5, we know trace(AB) = trace(BA). Thus Ttf(A)=f(T(A))=trace(ABBA)=trace(AB)trace(BA)=0.


Exercise 3.7.4

Let V be a finite-dimensional vector space over the field F and let T be a linear operator on V. Let c be a scalar and suppose there is a non-zero vector α in V such that Tα=cα. Prove that there is a non-zero linear functional f on V such that Ttf=cf.

Solution: Consider the operator U=TcI. Then U(α)=0 so rank(U)<n. Therefore rank(Ut)<n as an operator on V. It follows that there’s a fV such that Ut(f)=0. Now Ut=TtcI, thus Tt(f)=cf.


Exercise 3.7.5

Let A be an m×n matrix with real entries. Prove that A=0 if and only if trace(AtA)=0.

Solution: Suppose A is the m×n matrix with entries aij and B is the n×k matrix with entries bij. Then the i,j entry of AB is
k=1naikbkj.Substituting At for A and A for B we get the i,j entry of AtA is (note that AtA has dimension n×n)
k=1makiakj.Thus the diagonal entries of AtA are
k=1makiakifor i=1,,n. Thus the trace is
i=1nk=1maki2.If all aijR then this sum is zero if and only if each aij=0 because every one of them appears in this sum.


Exercise 3.7.6

Let n be a positive integer and let V be the space of all polynomial functions over the field of real numbers which have degree at most n, i.e., functions of the form
f(x)=c0+c1x++cnxn.Let D be the differentiation operator on V. Find a basis for the null space of the transpose operator Dt.

Solution: The null space of Dt consists of all linear functionals g:VR such that Dtg=0, or equivalently gD(f)=0 fV. Now g(a0+a1x++anxn)=c0a0+c1a1++cnan for some consants c0,,cnR. So gD(a0+a1x++anxn)=g(a1+2a2x++nanxn1)=i=0n1(i+1)ciai+1.

This sum i=0n1(i+1)ciai+1 equals zero for all vectors (a0,a1,,an)Rn if and only if c0=c1==cn1=0. While cn can be anything. Therefore the null space has dimension one and a basis is given by taking cn=1, which gives the function g:aixian, the projection onto the xn coordinate.


Exercise 3.7.7

Let V be a finite-dimensional vector space over the field F. Show that TTt is an isomorphism of L(V,V) onto L(V,V).

Solution: Choose a basis B for V. This gives an isomorphism L(V,V)Mn the space of n×n matrices. And the dual basis B gives an isomorphism L(V,V)Mn. Now we know by Theorem 23 that the following diagram of functions commutes. This means if we start at L(V,V) and follow two functinos to the Mn in the bottom left, it doesn’t matter which way around the diagram we go, we end up at the same place.
L(V,V)MntransposetransposeL(V,V)MnBoth horizontal arrows are isomorphisms (by Theorem 12, page 88). Now clearly transpose on matrices is a one-to-one and onto function from the set of matrices to itself. Also (rA)t=rAt and (A+B)t=At+Bt for any two n×n matrices A and B. Thus transpose is also a linear transformation. Thus transpose is an isomorphism. Therefore three of the arrows in this diagram are isomorphisms and it follows that the fourth arrow must also be an isomorphism.


Exercise 3.7.8

Let V be the vector space of n×n matrices over the field F.

(a) If B is a fixed n×n matrix, define a function fB on V by FB(A)=trace(BtA). Show that FB is a linear functional on V.
(b) Show that every linear functional on V is of the above form, i.e., is fB for some B.
(c) Show that BfB is an isomorphism of V onto V.

Solution:

(a) This follows from the fact that the trace function is a linear functional and left multiplication by a matrix is a linear transformation from V to V. In other words FB(cA1+A2)=trace(Bt(cA1+A2))=trace(cBtA1+BtA2)=ctrace(BtA1)+trace(BtA2)=cFB(A1)+FB(A2).

(b) Let f:VF be a linear functional. Let A=(aij)V. Then
(1)f(A)=i,j=1ncijaijfor some fixed constants cijF.

Now let B=(bij)V be any matrix. Then the i,j element of BtA is k=1nbkiakj. Thus
(2)trace(BtA)=i=1nk=1nbkiaki.Comparing (1) and (2) we see each aij apperas exactly once in each sum. So setting bki=cki for all i,k=1,,n we get the appropriate matrix B such that trace(BtA)=f.

(c) Let F be the function F:VV such that F(B)=fB. Part (a) shows this function is into V. Part (b) shows it is onto V. We must show it is linear and one-to-one. Let rF and B1,B2V. Then (rB1+B2)tA=(rB1t+B2t)A=rB1tA+B2tA. We know the trace function itself is linear. Thus F is linear. Now suppose trace(BtA)=0 AV. Fix i,j{1,2,,n}. Let A be the matrix with a one in the i,j position and zeros elsewhere. Then by the proof of part (b) we know that trace(BtA)=bij. So if F(A)=0 then bij=0. Thus B must be the zero matrix. Thus F is one-to-one. It follows that F is an isomorphism.

From http://greggrant.org

Linearity

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