Math 415 HW #6 Sol’n page 1of 2
Section3.5 Exercise5:
a) The transition matrix from [e1, e2, e3] to [u1, u2, u3] is
A=
112
123
124
−1
=
2 0 −1
−1 2 −1
0−1 1
.
b) The coordinates wrt [u1, u2, u3] are:
i)A(3,2,5)T= (1,−4,3)T;ii)A(1,1,2)T= (0,−1,1)T;iii)A(3,3,2)T= (2,2,−1)T.
Section3.6 Exercise10:
Recall that dimension of row space equals to the dimension of column space and that
row equivalent matrices have the same rank, we have rank of A = rank of U = 2. Also
notice that a1and a2are linear independent, hence a3and a4are linear combinations
of a1and a2. Thus we can write a3=xa1+ya2,a4=za1+wa2for some x, y, z , w ∈R:
A=
−3 4 −3x+ 4y−3z+ 4w
5−3 5x−3y5z−3w
2 7 2x+ 7y2z+ 7w
1−1x−y z −w
Now since A and U are row equivalent, they have the same row space. Thus (−3,4,−3x+
4y, −3z+4w),(5,−3,5x−3y , 5z−3w) are linear combinations of (1,0,2,1) and (0,1,1,4).
We have (−3,4,−3x+ 4y, −3z+ 4w) = a(1,0,2,1) + b(0,1,1,4) and (5,−3,5x−3y, 5z−
3w) = c(1,0,2,1) + d(0,1,1,4) for some a, b, c, d ∈R. Comparing the first two coordi-
nates in these two equations, we see immediately that a=−3, b = 4, c = 5, d =−3.
Substituting them back and looking the rest two coordinates, we have
−3x+ 4y= 2 ·(−3) + 1 ·(4
−3z+ 4w= 1 ·(−3) + 4 ·(4)
5x−3y= 2 ·(5) + 1 ·(−3)
5z−3w= 1 ·(5) + 4 ·(−3).
Solving these equations, we have x= 2, y = 1, z = 1, w = 4. Therefore a3=
(−2,7,11,1)Tand a4= (13,−7,30,−3)T.
Section3.6 Exercise22:
a)To show the column space of C=AB is a subspace of the column space of A, it
suffices to show that each column vector of Cis a linear combination of column vectors
of A. Now let bibe the i−th column vector of B. Then we know the i−th column
vector of C=AB is Abi. (See textbook page 38). Now we write A= (a1, ..., an) where
