Linear Algebra Exercises-n-Answers.pdf

196 Linear Algebra, by Hefferon
that space, and a transformation that acts on that basis in this way.
⃗β 1 ↦→ β ⃗ 2 ↦→ β ⃗ 3 ↦→ β ⃗ 4 ↦→ ⃗0
⃗β 5 ↦→ β ⃗ 6 ↦→ β ⃗ 7 ↦→ β ⃗ 8 ↦→ ⃗0
.
.
⃗β 4k−3 ↦→ β ⃗ 4k−2 ↦→ β ⃗ 4k−1 ↦→ β ⃗ 4k ↦→ ⃗0
.
.
–possibly other, shorter, strings–
So the dimension of the rangespace of T 3 can be as large as desired. The smallest that it can be is
one — there must be at least one string or else the map’s index of nilpotency would not be four.
Five.III.2.32 These two have only zero for
(
eigenvalues
) ( )
0 0 0 0
0 0 1 0
but are not similar (they have different canonical representatives, namely, themselves).
Five.III.2.33 A simple reordering of the string basis will do. For instance, a map that is assoicated
with this string basis
⃗β 1 ↦→ β ⃗ 2 ↦→ ⃗0
is represented with respect to B = 〈 β ⃗ 1 , β ⃗ 2 〉 by this matrix
( )
0 0
1 0
but is represented with respect to B = 〈 β ⃗ 2 , β ⃗ 1 〉 in this way.
( )
0 1
0 0
Five.III.2.34 Let t: V → V be the transformation. If rank(t) = nullity(t) then the equation rank(t) +
nullity(t) = dim(V ) shows that dim(V ) is even.
Five.III.2.35 For the matrices to be nilpotent they must be square. For them to commute they must
be the same size. Thus their product and sum are defined.
Call the matrices A and B. To see that AB is nilpotent, multiply (AB) 2 = ABAB = AABB =
A 2 B 2 , and (AB) 3 = A 3 B 3 , etc., and, as A is nilpotent, that product is eventually zero.
The sum is similar; use the Binomial Theorem.
Five.III.2.36 Some experimentation gives the idea for the proof. Expansion of the second power
t 2 S(T ) = S(ST − T S) − (ST − T S)S = S 2 − 2ST S + T S 2
the third power
t 3 S(T ) = S(S 2 − 2ST S + T S 2 ) − (S 2 − 2ST S + T S 2 )S
= S 3 T − 3S 2 T S + 3ST S 2 − T S 3
and the fourth power
t 4 S(T ) = S(S 3 T − 3S 2 T S + 3ST S 2 − T S 3 ) − (S 3 T − 3S 2 T S + 3ST S 2 − T S 3 )S
= S 4 T − 4S 3 T S + 6S 2 T S 2 − 4ST S 3 + T S 4
suggest that the expansions follow the Binomial Theorem. Verifying this by induction on the power
of t S is routine. This answers the question because, where the index of nilpotency of S is k, in the
expansion of t 2k
S
t 2k
S (T ) =
∑
0≤i≤2k
(−1) i ( 2k
i
)
S i T S 2k−i
for any i at least one of the S i and S 2k−i has a power higher than k, and so the term gives the zero
matrix.
Five.III.2.37 Use the geometric series: I − N k+1 = (I − N)(N k + N k−1 + · · · + I). If N k+1 is the zero
matrix then we have a right inverse for I − N. It is also a left inverse.
This statement is not ‘only if’ since (1 ) ( )
0 −1 0
−
0 1 0 −1
is invertible.