MAT 342 Linear Algebra October 29, 2003
Commented sample solutions for selected problems of test 3 ♥
1.a. Form the matrix S= (A|B|F) whose columns are A, B and F. Since rref(S) = I3,βis linearly
independent. And because |β|= 3, βalso spans R3, i.e. βis a basis for R3.
(MATLAB: A=[1 1 1]’,B=[-1 1 1]’,F=[-1 1 -1]’,E=[1 1 -1]’,ABF=[A B F], rref(ABF))
b. Note that S=ε[id]βis the transition matrix that transforms coordinates w.r.t. βinto coord.’s
w.r.t. the standard basis (e.g. [A]β= (1,0,0)T, and S·(1,0,0)T=A= [A]ε, i.e. S·[A]β= [A]ε).
Hence [E]β=S−1E= (1,−1,1)T. (MATLAB: ABF\E. Geometrically note: A+F=B+E)
c. Form the matrix Ywhose columns are (the coord.’s w.r.t. the standard basis εof ) the images of
the vectors in βunder the map L, i.e., Y= (L(A)|L(B)|L(F)) = (B|E|E) [LONG SENTENCE!].
Note that this is the matrix representation for Lw.r.t. the basis βfor the input and w.r.t. the
standard basis εfor the output, e.g. B=Y·(1,0,0)T
and [A]β= (1,0,0)T, or [B]ε=ε[L]β·[A]β. Hence
ε[L]ε=ε[L]β·β[id]ε=ε[L]β·(ε[id]β)−1=Y S−1,and
β[L]β=β[id]ε·ε[L]β= (ε[id]β)−1·ε[L]β=S−1Y.
d. L(C) = [L(C)]ε=ε[L]ε·[C]ε= (1,−1,−1)T=H.
ε[L]ε=Y·S−1=
−1 0 0
0 1 0
1 0 0
β[L]β=S−1·Y=
011
1−1−1
011
(In MATLAB: eLb=[B,E,E], eLe=eLb/ABF, bLb=ABF\eLb, LC=eLe*C)
Bonus: Because {A, D, F }are linearly dependent (compare a.) the given equations are overdeter-
mined. These may be consistent or inconsistent. Geometrically, since F=−D,T(F) must equal
T(−D) = −T(D) = −E=C. Algebraically, with the usual abuse of notation, we want a matrix
Xsuch that X·(A|D|F) = (A|E|F), i.e. (A|D|F)T·XT= (A|E|F)T.
Thus (in)consistency and possible solutions can be read off of the rref(³(A|D|F)T|(A|E|F)T´.
If e.g. T(F) = Cwas specified, we may pick any vector vthat is linearly independent from {A, D},
and completely freely specify T(v)∈R3. Then ε[T]γ= (A|E|T(v)) is the matrix representation
for the desired map Tw.r.t. the basis γ={A, D, v}for the input and εfor the output.
2. The rank of the composition can be no larger than the minimum of the ranks of the individual
maps, i.e. at most 2. On the other hand the dimension of the kernel (nullity) of the composition
can be no larger than the sum of the dimensions of the kernels of the individual maps, i.e. at most
2. Hence the rank of the composition must be at least 3 −2 = 1.
For example, T1: (x1, x2, x3)7→ (x1, x2,0) and T2: (x1, x2, x3)7→ (x1,0, x3) have both rank 2, while
T1◦T1=T1and T2◦T1: (x1, x2, x3)7→ (x1,0,0) have rank 2 and 1, respectively
4.a. The Wronskian matrix of the functions {e−3x, xe−3x, x2e−3x}evaluated at x= 0 is lower tri-
angular with nonzero entries along the diagonal. Hence the Wronskian has full rank. Thus βis
linearly independent. Since βspans by defn., it is a basis for V.b. [(2 −5x2)e−3x]β= (2,0,−5)T
c. Suppose c∈Rand y1, y2∈V. Then L(cy1+y2) = (cy1+y2)0=cy0
1+y0
2=cL(y1) + L(y2).
d. For vi∈βcalculate the coordinates of L(vi) w.r.t. β.
They are the columns of the desired matrix.
e. Since β[L]βhas full rank, ker(L) = {0}and a basis is the
empty set { }.f. Calculate β[L]−1
β[f(x)]β= (−8
9,10
9,5
3)T.
Hence ³−8
9+10
9x+5
3x2´e−3xis the desired antiderivative.
Bonus. Compare 4.d. This interesting pair of linear maps
whose commutator ε[X◦D−D◦X]ε=I3is the identity is
useful e.g. as a mathematical model in quantum mechanics.
β[L]β=
−3 1 0
0−3 2
0 0 −3
ε[D◦X]ε=
300
020
001
ε[X◦D]ε=
200
010
000
.