Linear Algebra Exercises-n-Answers.pdf

126 Linear Algebra, by Hefferon
As in the prior item, a check provides some confidence that this calculation was performed without
mistakes. We can for instance, fix the vector
( )
−1
⃗v =
2
(this is selected for no reason, out of thin air). Now we have
( ) ( ) ( )
−1 1 2 −1
Rep B (⃗v) =
2 3 4 2
and so t(⃗v) is this vector.
B,D
( ( (
1 1 8
3 · + 5 · =
1)
−1)
−2)
With respect to ˆB, ˆD we first calculate
( ) ( ) ( )
1 −28/3 −8/3 1
Rep ˆB(⃗v) =
−2 38/3 10/3 ˆB, ˆD −2
and, sure enough, that is the same result for t(⃗v).
( ( ( 1 2 8
−4 · + 6 · =
2)
1)
−2)
Three.V.2.13
B
(
3
=
5)
ˆB
=
D
( )
−4
6 ˆD
Where H and Ĥ are m×n, the matrix P is m×m while Q is n×n.
Three.V.2.14 Any n×n matrix is nonsingular if and only if it has rank n, that is, by Theorem 2.6, if
and only if it is matrix equivalent to the n×n matrix whose diagonal is all ones.
Three.V.2.15 If P AQ = I then QP AQ = Q, so QP A = I, and so QP = A −1 .
Three.V.2.16 By the definition following Example 2.2, a matrix M is diagonalizable if it represents
M = Rep B,D (t) a transformation with the property that there is some basis ˆB such that Rep ˆB, ˆB(t)
is a diagonal matrix — the starting and ending bases must be equal. But Theorem 2.6 says only that
there are ˆB and ˆD such that we can change to a representation Rep ˆB, ˆD(t) and get a diagonal matrix.
We have no reason to suspect that we could pick the two ˆB and ˆD so that they are equal.
Three.V.2.17 Yes. Row rank equals column rank, so the rank of the transpose equals the rank of the
matrix. Same-sized matrices with equal ranks are matrix equivalent.
Three.V.2.18
Only a zero matrix has rank zero.
Three.V.2.19 For reflexivity, to show that any matrix is matrix equivalent to itself, take P and Q to
be identity matrices. For symmetry, if H 1 = P H 2 Q then H 2 = P −1 H 1 Q −1 (inverses exist because P
and Q are nonsingular). Finally, for transitivity, assume that H 1 = P 2 H 2 Q 2 and that H 2 = P 3 H 3 Q 3 .
Then substitution gives H 1 = P 2 (P 3 H 3 Q 3 )Q 2 = (P 2 P 3 )H 3 (Q 3 Q 2 ). A product of nonsingular matrices
is nonsingular (we’ve shown that the product of invertible matrices is invertible; in fact, we’ve shown
how to calculate the inverse) and so H 1 is therefore matrix equivalent to H 3 .
Three.V.2.20 By Theorem 2.6, a zero matrix is alone in its class because it is the only m×n of rank
zero. No other matrix is alone in its class; any nonzero scalar product of a matrix has the same rank
as that matrix.
Three.V.2.21 There are two matrix-equivalence classes of 1×1 matrices — those of rank zero and those
of rank one. The 3×3 matrices fall into four matrix equivalence classes.
Three.V.2.22 For m×n matrices there are classes for each possible rank: where k is the minimum
of m and n there are classes for the matrices of rank 0, 1, . . . , k. That’s k + 1 classes. (Of course,
totaling over all sizes of matrices we get infinitely many classes.)
Three.V.2.23 They are closed under nonzero scalar multiplication, since a nonzero scalar multiple of
a matrix has the same rank as does the matrix. They are not closed under addition, for instance,
H + (−H) has rank zero.
Three.V.2.24 (a) We have
( )
1 −1
Rep B,E2 (id) =
Rep
2 −1
E2,B(id) = Rep B,E2 (id) −1 =
and thus the answer is this.
( ) ( ) ( )
1 −1 1 1 −1 1
Rep B,B (t) =
=
2 −1 3 −1 −2 1
( ) −1
1 −1
=
2 −1
(
−2
)
0
−5 2
(
−1
)
1
−2 1