Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 185
Five.II.2.16 Yes, ct is diagonalizable by the final theorem of this subsection.
No, t+s need not be diagonalizable. Intuitively, the problem arises when the two maps diagonalize
with respect to different bases (that is, when they are not simultaneously diagonalizable). Specifically,
these two are diagonalizable but their sum is not:
( ) ( )
1 1 −1 0
0 0 0 0
(the second is already diagonal; for the first, see Exercise 15). The sum is not diagonalizable because
its square is the zero matrix.
The same intuition suggests that t ◦ s is not be diagonalizable. These two are diagonalizable but
their product is not: ( ) ( )
1 0 0 1
0 0 1 0
(for the second, see Exercise 15).
Five.II.2.17
then
If
P
P
( ) 1 c
P −1 =
0 1
( )
1 c
=
0 1
( ) a 0
0 b
( )
a 0
P
0 b
so
( ) ( ) ( ) ( )
p q 1 c a 0 p q
=
r s 0 1 0 b r s
( ) ( )
p cp + q ap aq
=
r cr + s br bs
The 1, 1 entries show that a = 1 and the 1, 2 entries then show that pc = 0. Since c ≠ 0 this means
that p = 0. The 2, 1 entries show that b = 1 and the 2, 2 entries then show that rc = 0. Since c ≠ 0
this means that r = 0. But if both p and r are 0 then P is not invertible.
Five.II.2.18 (a) Using the formula for the inverse of a 2×2 matrix gives this.
( ) ( ) ( ) ( )
a b 1 2 1 d −b
·
c d 2 1 ad − bc ·
1 ad + 2bd − 2ac − bc −ab − 2b
=
2 + 2a 2 + ab
−c a ad − bc cd + 2d 2 − 2c 2 − cd −bc − 2bd + 2ac + ad
Now pick scalars a, . . . , d so that ad − bc ≠ 0 and 2d 2 − 2c 2 = 0 and 2a 2 − 2b 2 = 0. For example,
these will do. ( ) ( ) ( )
1 1 1 2 1 −1 −1
·
1 −1 2 1 −2 ·
= 1 ( ) −6 0
−1 1 −2 0 2
(b) As above,
( ) ( ) ( ) ( )
a b x y 1 d −b
·
c d y z ad − bc ·
1 adx + bdy − acy − bcz −abx − b
=
2 y + a 2 y + abz
−c a ad − bc cdx + d 2 y − c 2 y − cdz −bcx − bdy + acy + adz
we are looking for scalars a, . . . , d so that ad − bc ≠ 0 and −abx − b 2 y + a 2 y + abz = 0 and
cdx + d 2 y − c 2 y − cdz = 0, no matter what values x, y, and z have.
For starters, we assume that y ≠ 0, else the given matrix is already diagonal. We shall use that
assumption because if we (arbitrarily) let a = 1 then we get
and the quadratic formula gives
−bx − b 2 y + y + bz = 0
(−y)b 2 + (z − x)b + y = 0
b = −(z − x) ± √ (z − x) 2 − 4(−y)(y)
y ≠ 0
−2y
(note that if x, y, and z are real then these two b’s are real as the discriminant is positive). By the
same token, if we (arbitrarily) let c = 1 then
and we get here
dx + d 2 y − y − dz = 0
(y)d 2 + (x − z)d − y = 0
d = −(x − z) ± √ (x − z) 2 − 4(y)(−y)
2y
y ≠ 0