Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 181
Five.II.1.12 A matrix is similar to itself; take P to be the identity matrix: IP I −1 = IP I = P .
If T is similar to S then T = P SP −1 and so P −1 T P = S. Rewrite this as S = (P −1 )T (P −1 ) −1 to
conclude that S is similar to T .
If T is similar to S and S is similar to U then T = P SP −1 and S = QUQ −1 . Then T =
P QUQ −1 P −1 = (P Q)U(P Q) −1 , showing that T is similar to U.
Five.II.1.13 Let f x and f y be the reflection maps (sometimes called ‘flip’s). For any bases B and D,
the matrices Rep B,B (f x ) and Rep D,D (f y ) are similar. First note that
( )
( )
1 0
−1 0
S = Rep E2,E 2
(f x ) =
T = Rep
0 −1
E2,E 2
(f y ) =
0 1
are similar because the second matrix is the representation of f x with respect to the basis A = 〈⃗e 2 , ⃗e 1 〉:
( ) ( )
1 0 −1 0
= P P −1
0 −1 0 1
where P = Rep A,E2 (id).
R 2 w.r.t. A
⏐
id
↓P
f x
−−−−→
T
V R 2 w.r.t. A
⏐
id↓P
R 2 f x
w.r.t. E 2 −−−−→ R 2 w.r.t. E S 2
Now the conclusion follows from the transitivity part of Exercise 12.
To finish without relying on that exercise, write Rep B,B (f x ) = QT Q −1 = QRep E2 ,E 2
(f x )Q −1 and
Rep D,D (f y ) = RSR −1 = RRep E2 ,E 2
(f y )R −1 . Using the equation in the first paragraph, the first of
these two becomes Rep B,B (f x ) = QP Rep E2 ,E 2
(f y )P −1 Q −1 and rewriting the second of these two as
R −1 · Rep D,D (f y ) · R = Rep E2,E 2
(f y ) and substituting gives the desired relationship
Rep B,B (f x ) = QP Rep E2,E 2
(f y )P −1 Q −1
= QP R −1 · Rep D,D (f y ) · RP −1 Q −1 = (QP R −1 ) · Rep D,D (f y ) · (QP R −1 ) −1
Thus the matrices Rep B,B (f x ) and Rep D,D (f y ) are similar.
Five.II.1.14 We must show that if two matrices are similar then they have the same determinant and
the same rank. Both determinant and rank are properties of matrices that we have already shown to
be preserved by matrix equivalence. They are therefore preserved by similarity (which is a special case
of matrix equivalence: if two matrices are similar then they are matrix equivalent).
To prove the statement without quoting the results about matrix equivalence, note first that rank
is a property of the map (it is the dimension of the rangespace) and since we’ve shown that the rank of
a map is the rank of a representation, it must be the same for all representations. As for determinants,
|P SP −1 | = |P | · |S| · |P −1 | = |P | · |S| · |P | −1 = |S|.
The converse of the statement does not hold; for instance, there are matrices with the same determinant
that are not similar. To check this, consider a nonzero matrix with a determinant of zero.
It is not similar to the zero matrix, the zero matrix is similar only to itself, but they have they same
determinant. The argument for rank is much the same.
Five.II.1.15 The matrix equivalence class containing all n×n rank zero matrices contains only a single
matrix, the zero matrix. Therefore it has as a subset only one similarity class.
In contrast, the matrix equivalence class of 1 × 1 matrices of rank one consists of those 1 × 1
matrices (k) where k ≠ 0. For any basis B, the representation of multiplication by the scalar k is
Rep B,B (t k ) = (k), so each such matrix is alone in its similarity class. So this is a case where a matrix
equivalence class splits into infinitely many similarity classes.
Five.II.1.16 Yes, these are similar ( ) ( )
1 0 3 0
0 3 0 1
since, where the first matrix is Rep B,B (t) for B = 〈 ⃗ β 1 , ⃗ β 2 〉, the second matrix is Rep D,D (t) for D =
〈 ⃗ β 2 , ⃗ β 1 〉.
Five.II.1.17 The k-th powers are similar because, where each matrix represents the map t, the k-
th powers represent t k , the composition of k-many t’s. (For instance, if T = reptB, B then T 2 =
Rep B,B (t ◦ t).)