Linear Algebra Exercises-n-Answers.pdf

142 Linear Algebra, by Hefferon
and this second description exactly says this.
⎛
⎞
N (f) ⊥ = [{ ⎝ 1 2⎠}]
3
(b) The generalization is that for any f : R n → R there is a vector ⃗ h so that
⎛ ⎞
v 1
⎜
⎝
⎟
. ⎠ ↦−→ f
h 1 v 1 + · · · + h n v n
v n
and ⃗ h ∈ N (f) ⊥ . We can prove this by, as in the prior item, representing f with respect to the
standard bases and taking ⃗ h to be the column vector gotten by transposing the one row of that
matrix representation.
(c) Of course,
( ) 1 2 3
Rep E3 ,E 2
(f) =
4 5 6
and so the nullspace is this set.
⎛
N (f){ ⎝ v ⎞
1
v 2
⎠ ∣ v 3
That description makes clear that
⎛
⎞
⎝ 1 2⎠ ,
3
( ) ⎛
1 2 3 ⎝ v ⎞
(
1
v
4 5 6 2
⎠ 0
= }
0)
v 3
⎛
⎝ 4 5
6
⎞
⎠ ∈ N (f) ⊥
and since N (f) ⊥ is a subspace of R n , the span of the two vectors is a subspace of the perp of the
nullspace. To see that this containment is an equality, take
⎛ ⎞
⎛ ⎞
M = [{ ⎝ 1 2⎠}] N = [{ ⎝ 4 5⎠}]
3
6
in the third item of Exercise 23, as suggested in the hint.
(d) As above, generalizing from the specific case is easy: for any f : R n → R m the matrix H representing
the map with respect to the standard bases describes the action
⎛ ⎛
⎜
⎝
⎞
v 1
. ⎟
. ⎠
v n
f
↦−→
⎜
⎝
⎞
h 1,1 v 1 + h 1,2 v 2 + · · · + h 1,n v n
.
⎟
.
⎠
h m,1 v 1 + h m,2 v 2 + · · · + h m,n v n
and the description of the nullspace gives that on transposing the m rows of H
⎛ ⎞ ⎛ ⎞
h 1,1
h m,1
⃗ h 1,2
h1 = ⎜ ⎟
⎝ . ⎠ , . . .⃗ h m,2
h m = ⎜ ⎟
⎝ . ⎠
h 1,n h m,n
we have N (f) ⊥ = [{ ⃗ h 1 , . . . , ⃗ h m }]. (In [Strang 93], this space is described as the transpose of the
row space of H.)
Three.VI.3.25 (a) First note that if a vector ⃗v is already in the line then the orthogonal projection
gives ⃗v itself. One way to verify this is to apply the formula for projection into the line spanned by
a vector ⃗s, namely (⃗v ⃗s/⃗s ⃗s) · ⃗s. Taking the line as {k · ⃗v ∣ k ∈ R} (the ⃗v = ⃗0 case is separate but
easy) gives (⃗v ⃗v/⃗v ⃗v) · ⃗v, which simplifies to ⃗v, as required.
Now, that answers the question because after once projecting into the line, the result proj l (⃗v) is
in that line. The prior paragraph says that projecting into the same line again will have no effect.
(b) The argument here is similar to the one in the prior item. With V = M ⊕ N, the projection of
⃗v = ⃗m+⃗n is proj M,N (⃗v ) = ⃗m. Now repeating the projection will give proj M,N (⃗m) = ⃗m, as required,
because the decomposition of a member of M into the sum of a member of M and a member of N
is ⃗m = ⃗m + ⃗0. Thus, projecting twice into M along N has the same effect as projecting once.