Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 141
Three.VI.3.14 No, a decomposition of vectors ⃗v = ⃗m + ⃗n into ⃗m ∈ M and ⃗n ∈ N does not depend on
the bases chosen for the subspaces — this was shown in the Direct Sum subsection.
Three.VI.3.15 The orthogonal projection of a vector into a subspace is a member of that subspace.
Since a trivial subspace has only one member, ⃗0, the projection of any vector must equal ⃗0.
Three.VI.3.16 The projection into M along N of a ⃗v ∈ M is ⃗v. Decomposing ⃗v = ⃗m + ⃗n gives ⃗m = ⃗v
and ⃗n = ⃗0, and dropping the N part but retaining the M part results in a projection of ⃗m = ⃗v.
Three.VI.3.17 The proof of Lemma 3.7 shows that each vector ⃗v ∈ R n is the sum of its orthogonal
projections onto the lines spanned by the basis vectors.
⃗v = proj [⃗κ1](⃗v ) + · · · + proj [⃗κn](⃗v ) = ⃗v ⃗κ 1
· ⃗κ 1 + · · · + ⃗v ⃗κ n
· ⃗κ n
⃗κ 1 ⃗κ 1 ⃗κ n ⃗κ n
Since the basis is orthonormal, the bottom of each fraction has ⃗κ i ⃗κ i = 1.
Three.VI.3.18 If V = M ⊕ N then every vector can be decomposed uniquely as ⃗v = ⃗m + ⃗n. For all ⃗v
the map p gives p(⃗v) = ⃗m if and only if ⃗v − p(⃗v) = ⃗n, as required.
Three.VI.3.19 Let ⃗v be perpendicular to every ⃗w ∈ S. Then ⃗v (c 1 ⃗w 1 + · · · + c n ⃗w n ) = ⃗v (c 1 ⃗w 1 ) +
· · · + ⃗v (c n ⃗w n ) = c 1 (⃗v ⃗w 1 ) + · · · + c n (⃗v ⃗w n ) = c 1 · 0 + · · · + c n · 0 = 0.
Three.VI.3.20
Three.VI.3.21
of the two.
True; the only vector orthogonal to itself is the zero vector.
This is immediate from the statement in Lemma 3.7 that the space is the direct sum
Three.VI.3.22 The two must be equal, even only under the seemingly weaker condition that they yield
the same result on all orthogonal projections. Consider the subspace M spanned by the set {⃗v 1 , ⃗v 2 }.
Since each is in M, the orthogonal projection of ⃗v 1 into M is ⃗v 1 and the orthogonal projection of ⃗v 2
into M is ⃗v 2 . For their projections into M to be equal, they must be equal.
Three.VI.3.23 (a) We will show that the sets are mutually inclusive, M ⊆ (M ⊥ ) ⊥ and (M ⊥ ) ⊥ ⊆ M.
For the first, if ⃗m ∈ M then by the definition of the perp operation, ⃗m is perpendicular to every
⃗v ∈ M ⊥ , and therefore (again by the definition of the perp operation) ⃗m ∈ (M ⊥ ) ⊥ . For the other
direction, consider ⃗v ∈ (M ⊥ ) ⊥ . Lemma 3.7’s proof shows that R n = M ⊕ M ⊥ and that we can give
an orthogonal basis for the space 〈⃗κ 1 , . . . , ⃗κ k , ⃗κ k+1 , . . . , ⃗κ n 〉 such that the first half 〈⃗κ 1 , . . . , ⃗κ k 〉 is a
basis for M and the second half is a basis for M ⊥ . The proof also checks that each vector in the
space is the sum of its orthogonal projections onto the lines spanned by these basis vectors.
⃗v = proj [⃗κ1](⃗v ) + · · · + proj [⃗κn](⃗v )
Because ⃗v ∈ (M ⊥ ) ⊥ , it is perpendicular to every vector in M ⊥ , and so the projections in the second
half are all zero. Thus ⃗v = proj [⃗κ1 ](⃗v ) + · · · + proj [⃗κk ](⃗v ), which is a linear combination of vectors
from M, and so ⃗v ∈ M. (Remark. Here is a slicker way to do the second half: write the space both
as M ⊕ M ⊥ and as M ⊥ ⊕ (M ⊥ ) ⊥ . Because the first half showed that M ⊆ (M ⊥ ) ⊥ and the prior
sentence shows that the dimension of the two subspaces M and (M ⊥ ) ⊥ are equal, we can conclude
that M equals (M ⊥ ) ⊥ .)
(b) Because M ⊆ N, any ⃗v that is perpendicular to every vector in N is also perpendicular to every
vector in M. But that sentence simply says that N ⊥ ⊆ M ⊥ .
(c) We will again show that the sets are equal by mutual inclusion. The first direction is easy; any ⃗v
perpendicular to every vector in M +N = {⃗m + ⃗n ∣ ∣ ⃗m ∈ M, ⃗n ∈ N} is perpendicular to every vector
of the form ⃗m + ⃗0 (that is, every vector in M) and every vector of the form ⃗0 + ⃗n (every vector in
N), and so (M + N) ⊥ ⊆ M ⊥ ∩ N ⊥ . The second direction is also routine; any vector ⃗v ∈ M ⊥ ∩ N ⊥
is perpendicular to any vector of the form c⃗m + d⃗n because ⃗v (c⃗m + d⃗n) = c · (⃗v ⃗m) + d · (⃗v ⃗n) =
c · 0 + d · 0 = 0.
Three.VI.3.24
(a) The representation of
⎛
⎝ v 1
⎞
v 2
⎠
v 3
f
↦−→ 1v 1 + 2v 2 + 3v 3
is this.
Rep E3 ,E 1
(f) = ( 1 2 3 )
By the definition of f
⎛
N (f) = { ⎝ v ⎞
⎛
1
v 2
⎠ ∣ 1v 1 + 2v 2 + 3v 3 = 0} = { ⎝ v ⎞ ⎛
1
v 2
⎠ ∣ ⎝ 1 ⎞
2⎠
v 3 v 3 3
⎛
⎝ v ⎞
1
v 2
⎠ = 0}
v 3