Linear Algebra, Theory And Applications, 2012a

490 ANSWERS TO SELECTED EXERCISES
6
⎛
⎝
⎞
1
2 e−t 1
0
2 e−t
1
2 cos t + 1 2 sin t − sin t 1
2 sin t − 1 2 cos t ⎠
1
2 sin t − 1 2 cos t cos t − 1 2 cos t − 1 2 sin t
8 det(λI − A) =det ( λI − S −1 BS )
=det ( λS −1 S − S −1 BS )
=det ( S −1 (λI − B) S )
=det ( S −1) det (λI − B)det(S)
=det ( S −1 S ) det (λI − B) =det(λI − B)
9 From the Cayley Hamilton theorem,A n +a n−1 A n−1 +
···+ a 1 A + a 0 I =0. Also the characteristic polynomial
is det (tI − A) and the constant term is
(−1) n det (A) . Thus a 0 ≠0ifandonlyifdet(A) ≠
0 if and only if A −1 has an inverse. Thus if A −1
exists, it follows that
a 0 I = − ( A n + a n−1 A n−1 + ···+ a 1 A )
= A ( −A n−1 − a n−1 A n−2 −···−a 1 I ) and also
a 0 I = ( −A n−1 − a n−1 A n−2 −···−a 1 I ) A Therefore,
the inverse is
(
1
a 0
−A n−1 − a n−1 A n−2 −···−a 1 I )
11 Say the characteristic polynomial is q (t) whichisof
degree 3. Then if n ≥ 3,t n = q (t) l (t)+r (t) where
the degree of r (t) is either less than 3 or it equals
zero. Thus A n = q (A) l (A) +r (A) =r (A) and
so all the terms A n for n ≥ 3 can be replaced with
some r (A) where the degree of r (t) is no more than
2. Thus, assuming there are no convergence issues,
the infinite sum must be of the form ∑ 2
k=0 b kA k .
G.9 Exercises
4.6
1 A typical thing in {Ax : x ∈ P (u 1 , ··· , u n )} is
∑ n
k=1 t kAu k : t k ∈ [0, 1] and so it is just
P (Au 1 , ··· ,Au n ) .
( )
1 1
2 E =
0 1
P (e 1 , e 2 ) E(P (e 1 , e 2 ))
5Heretheyare.
⎛
1 0 0
⎞ ⎛
⎝ 0 1 0 ⎠ , ⎝
⎛
0 0 1
0 1 0
⎞ ⎛
⎝ 1 0 0 ⎠ , ⎝
0 0 1
0 0 1
1 0 0
0 1 0
1 0 0
0 0 1
0 1 0
⎞ ⎛
⎠ , ⎝
⎞ ⎛
⎠ , ⎝
0 1 0
0 0 1
1 0 0
0 0 1
0 1 0
1 0 0
⎞
⎠
⎞
⎠
So what is the dimension of the span of these? One
way to systematically accomplish this is to unravel
them and then use the row reduced echelon form.
Unraveling these yields the column vectors
⎛
⎜
⎝
1
0
0
0
1
0
0
0
1
⎞ ⎛
⎟ ⎜
⎠ ⎝
0
0
1
1
0
0
0
1
0
⎞ ⎛
⎟ ⎜
⎠ ⎝
0
1
0
0
0
1
1
0
0
⎞ ⎛
⎟ ⎜
⎠ ⎝
0
1
0
1
0
0
0
0
1
⎞ ⎛
⎟ ⎜
⎠ ⎝
1
0
0
0
0
1
0
1
0
⎞ ⎛
⎟ ⎜
⎠ ⎝
0
0
1
0
1
0
1
0
0
⎞
⎟
⎠
Then arranging these as the columns of a matrix
yields the following along with its row reduced echelon
form.
⎛
⎞
1 0 0 0 1 0
0 0 1 1 0 0
0 1 0 0 0 1
0 1 0 1 0 0
1 0 0 0 0 1
,rowechelonform:
0 0 1 0 1 0
⎜ 0 0 1 0 0 1
⎟
⎝ 0 1 0 0 1 0 ⎠
1 0 0 1 0 0
⎛
⎜
⎝
1 0 0 0 0 1
0 1 0 0 0 1
0 0 1 0 0 1
0 0 0 1 0 −1
0 0 0 0 1 −1
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
The dimension is 5.
⎞
⎟
⎠
10 It is because you cannot have more than min (m, n)
nonzero rows in the row reduced echelon form. Recall
that the number of pivot columns is the same
as the number of nonzero rows from the description
of this row reduced echelon form.