Linear Algebra Exercises-n-Answers.pdf

Chapter Five: Similarity
Subsection Five.II.1: Definition and Examples
Five.II.1.4 One way to proceed is left to right.
( ) ( ) ( ) ( ) ( )
P SP −1 4 2 1 3 2/14 −2/14 0 0 2/14 −2/14
=
=
=
−3 2 −2 −6 3/14 4/14 −7 −21 3/14 4/14
( )
0 0
−11/2 −5
Five.II.1.5 (a) Because the matrix (2) is 1×1, the matrices P and P −1 are also 1×1 and so where
P = (p) the inverse is P −1 = (1/p). Thus P (2)P −1 = (p)(2)(1/p) = (2).
(b) Yes: recall that scalar multiples can be brought out of a matrix P (cI)P −1 = cP IP −1 = cI. By
the way, the zero and identity matrices are the special cases c = 0 and c = 1.
(c) No, as this example shows.
( ) ( ) ( ) ( )
1 −2 −1 0 −1 −2 −5 −4
=
−1 1 0 −3 −1 −1 2 1
Five.II.1.6 Gauss’ method shows that the first matrix represents maps of rank two while the second
matrix represents maps of rank three.
Five.II.1.7
easy:
gives this.
(a) Because t is described with the members of B, finding the matrix representation is
⎛
⎞
Rep B (t(x 2 )) = ⎝ 0 1⎠
1
⎝ 3 0⎠
0
D
B
⎛
⎞
Rep B (t(x)) = ⎝ 1 0 ⎠
−1
B
⎛
Rep B,B (t) ⎝ 0 1 0
⎞
1 0 0⎠
1 −1 3
⎛
⎞
Rep B (t(1)) = ⎝ 0 0⎠
3
(b) We will find t(1), t(1 + x), and t(1 + x + x 2 , to find how each is represented with respect
to D. We are given that t(1) = 3, and the other two are easy to see: t(1 + x) = x 2 + 2 and
t(1 + x + x 2 ) = x 2 + x + 3. By eye, we get the representation of each vector
⎛ ⎞
⎛
Rep D (t(1)) = Rep D (t(1 + x)) = ⎝ 2
⎞
⎛ ⎞
⎠ Rep D (t(1 + x + x 2 )) = ⎠
and thus the representation of the map.
⎛ ⎞
3 2 2
Rep D,D (t) = ⎝0 −1 0⎠
0 1 1
(c) The diagram, adapted for this T and S,
shows that P = Rep D,B (id).
V w.r.t. D
⏐
id
↓P
V w.r.t. B
−1
1
t
−−−−→
S
t
−−−−→
T
D
V w.r.t. D
⏐
id
↓P
V w.r.t. B
⎛
P = ⎝ 0 0 1
⎞
0 1 1⎠
1 1 1
B
⎝ 2 0
1
D