Download Lecture 3: Vector subspaces, sums, and direct sums (1)

Lecture 3: Vector subspaces, sums, and direct
sums (1)
Travis Schedler
Thurs, Sep 15, 2011 (version: Thurs, Sep 15, 1:00 PM)
Goals (2)
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Understand vector subspaces and examples
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Go over a model proof
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Understand intersections, sums, and direct sums
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Preview bases
Warm-up exercise 1 (3)
Which of the following are subspaces of R3 ?
(a) The plane x = y
(b) The line (1 + t, 2t, 3t)
(c) The locus x 2 + y 2 + z 2 = 0
(d) The locus x 2 + y 2 − z 2 = 0
Answer: (a): yes: this is a plane through the origin.
(b): no: this does not contain the origin (it is a line though) so it
is not a subspace.
(c): yes: this is just the zero point, so it is a subspace.
(d): no: this is a conic which is closed under scaling but not under
addition, so is not a subspace.
Warm-up exercise 2 (4)
Which of the following are subspaces of the vector space of all
functions {0, 1, . . . , n − 1} → C?
(a) All functions such that f (3) = 2f (1)
(b) All functions such that f (5) = f (6) = 0
(c) All functions such that f (2) − f (0) = 1
(d) All functions such that f (2)f (5) = 0
Answer: (a): yes: if f and g satisfy these properties, so do f + g
and af for all a ∈ C: f (3) = 2f (1) and g (3) = 2g (1) imply
(f + g )(3) = 2(f + g )(1) and similarly af (3) = 2af (1).
(b): yes: The same argument applies.
(c): This is not a subspace: the zero function doesn’t satisfy this
condition.
(d): No: we can take f (2) = 1 and f (k) = 0 for all k 6= 0, and
g (5) = 1 and g (k) = 0 for all k 6= 5, then
f (2)f (5) = g (2)g (5) = 0 but (f + g )(2)(f + g )(5) = 1 6= 0.
Warm-up exercise 3 (5)
Which of the following sums are direct?
(a) {(x = y )-axis} + {yz-plane}
(b) {x-axis} + {xy -plane}
(c) {0} + V for V a vector space.
(d) {Constant functions R → R} +
{Functions f : R → R such that f (0) = 0}
Here, a constant function f : X → F is a function such that
f (x) = f (y ) for all x, y ∈ X .
Answer: (a): Yes: this sum is R3 , and every (a, b, c) can be
written uniquely as (a, a, 0) + (0, b − a, c) of this form.
(b): No: the sum is the xy -plane and, for example,
(1, 0) = (1, 0) + (0, 0) = (0, 0) + (1, 0), so there is no unique
expression as a sum of a vector in the x-axis and one in the
xy -plane.
(c): Yes: for every v ∈ V , we can uniquely write v = 0 + v .
(d): Yes: for every function f : R → R, let g be the constant
function g (x) = f (0) for all x. Then (f − g )(0) = 0, and
f = g + (f − g ) is the unique decomposition (why?).
Easier solution (6)
Which of the following sums are direct?
(a) {(x = y )-axis} + {yz-plane}
(b) {x-axis} + {xy -plane}
(c) {0} + V for V a vector space.
(d) {Constant functions R → R} +
{Functions f : R → R such that f (0) = 0}
Let us use what you should have read in the book:
Proposition (Proposition 1.9)
Suppose that U and W are subspaces of V . Then V = U ⊕ W if
and only if V = U + W and U ∩ W = {0}.
We can apply this simply to V = U + W , so then the sums are
direct if and only if U ∩ W = {0}.
Then, this property is true for (a), (c), and (d), but not for (b).
Caution: this proposition does not generalize to sums of three or
more subspaces (Ui ∩ Uj = 0 for all i, j does not imply
U1 + · · · + Um is direct)! Can you give an example?
Ex: x-axis + y -axis + (x = y line) = R2 is not direct!
Subspaces of vector spaces (recap) (7)
Definition
A subspace U of a vector space V is a subset containing 0 ∈ V
such that, for all u1 , u2 ∈ U and all a ∈ F,
u1 + u2 ∈ U,
au1 ∈ U.
We write U ⊆ V to denote that U is a subspace [or subset] of V .
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Caution: ⊆ can be used to denote either subspace or subset.
The meaning should be clear by context.
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A subspace is the same thing as a subset which is also a
vector space, using the addition and scalar multiplication.
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Note that the condition above that a subspace U contains 0 is
equivalent to the condition that it be nonempty, by the
following result:
Claim. If u ∈ U, then 0 · u = 0.
Proof of claim (8)
Claim. If u ∈ U, then 0 · u = 0.
Proof.
First,
0 · u = (0 + 0) · u = 0 · u + 0 · u.
(0.2)
Here, we used that 0 + 0 = 0 (additive identity of a field, or fact
for F = R and C); then distributivity. Next,
0 = 0 · u + (−(0 · u)) = (0 · u + 0 · u) + (−(0 · u))
= (0 · u) + (0 · u + (−(0 · u))) = 0 · u + 0 = 0 · u.
Here, we used the existence of an additive inverse −(0 · u) of 0 · u,
then (0.2), then associativity, then the additive inverse property,
then the additive identity property.
Examples of subspaces (9)
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For every vector space V , {0} ⊆ V .
{0} ⊆ x-axis ⊆ x, y -plane ⊆ R3 . More generally, we can take
any line or plane through the origin in R3 , which is a subspace.
More generally, for m ≤ n, there is a subspace
{(a1 , . . . , am , 0, . . . , 0) : a1 , . . . , am ∈ F} ⊆ Fn .
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Informally, this says that Fm ⊆ Fn for m ≤ n (we will make
this precise later).
Caution: there are other ways to realize Fm inside Fn , e.g.,
the subspace such that the first n − m coordinates are zero,
rather than the last n − m coordinates. (Correspondingly, in
the previous example, we could have taken the z-axis and the
y , z-plane).
Similarly, the vector space of functions f : {1, 2, . . . , n} → F,
and the subspace of functions f such that
0 = f (m + 1) = f (m + 2) = · · · = f (n). Informally, this says
that functions on {1, . . . , m} are a subspace of functions on
{1, . . . , n}.
More examples of subspaces (10)
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For F = R, we have:
P(R) ⊆ Continuous functions R → R ⊆ All functions R → R.
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We already observed that, for all F, P(F) ⊆ F∞ . This is the
subspace of lists which terminate in an infinite sequence of
zeros.
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For any set X and any F, let V be the vector space of all
functions X → F. Then, for every subset Y ⊆ X , we can
consider the vector subspace U ⊆ V of all functions vanishing
on Y , i.e., all f such that f (y ) = 0 for all y ∈ Y . (This
generalizes the last example from the last slide).
Intersections, unions, and sums (11)
We can form intersections of vector spaces:
Proposition
If U, W ⊆ V are subspaces, so is the intersection U ∩ W .
Idea of proof: Each defining property of a subspace (containing
zero, being closed under addition, being closed under scalar
multiplication) is preserved by taking intersections.
However, this is not true of unions. Why?
Example: x-axis ∪ y -axis is not a subspace (not closed under
addition).
Instead, we can perform sum operations:
Definition
U + W = {u + w | u ∈ U, w ∈ W }.
Proposition
If U, W ⊆ V are subspaces, so is U + W .
Idea of proof: (u1 + w1 ) + (u2 + w2 ) = (u1 + u2 ) + (w1 + w2 ), and
a(u1 + w1 ) = au1 + aw1 .
Direct sums (12)
Suppose U1 , . . . , Um ⊆ V .
Definition
The sum U1 + · · · + Um is direct if every v ∈ U1 + · · · + Um has a
unique expression as v = u1 + · · · + um for
u1 ∈ U1 , u2 ∈ U2 , . . . , um ∈ Um .
Example: x-axis + y -axis + z-axis = R3 .
Definition
Let U1 ⊕ · · · ⊕ Um denote U1 + · · · + Um in the case that the sum
is direct (otherwise we may not use ⊕).
So, we may write x-axis ⊕ y -axis ⊕ z-axis, but we may not write
x-axis ⊕ xy -plane.
Example: R3 (13)
Take V = R3 . We can describe all subspaces:
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{0} and R3 themselves;
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All lines and planes through the origin.
(*) Given any two distinct lines U1 , U2 through the origin, we can
take the plane U1 + U2 that they span.
(**) Given a plane U through the origin, and a line W not in that
plane, we can take U + W = R3 : they span everything.
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On the other hand, if U1 = U2 , then U1 + U2 = U1 = U2 .
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Similarly, if U1 ⊆ U2 , then U1 + U2 = U2 .
In the cases of (*) and (**), every vector v ∈ R3 is a unique sum
of a vector of U and one of W . So these are direct sums. The last
two examples are not direct (unless U1 = {0}).
Preview: bases (14)
We want to understand bases: in the case of the vector space of all
greyscale 800 × 600-images, there are at least two important ones:
the basis of pixel coordinates, and the basis of frequency
coordinates.
Definition (cf. Proposition 2.8)
A basis of a vector space V is a list of vectors (v1 , . . . , vn ) in V
such that, for all v ∈ V , there is a unique expression
v = a1 v1 + · · · + an vn ,
for a1 , . . . , an ∈ F.
Notice the similarity with direct sums: here we have a unique
expression using vectors, whereas a direct sum is a unique
expression involving subspaces. (Exercise: Make a precise
connection between the two notions!)
In the case of images, the pixel basis is v0 , . . . , v479999 where vi is
the image with only the pixel i in white, and all other pixels black
(black is the value 0).
Bases preview continued (15)
The frequency basis is w0 , . . . , w479999 where wi is the image
corresponding to a sine wave with frequency i: so w0 is solid white,
and w479999 alternates every pixel between black and white; the
halfway w240000 would be white, grey, black, grey, repeated over an
over, etc.
We are going to explain bases from a theoretical point of view, as
lists having two weaker properties at the same time: linear
independence, and spanning. Roughly, linear independence says
that there is at most one expression v = a1 v1 + · · · + an vn , and
spanning says there is at least one such expression.
For next time: Read Section 2 through Proposition 2.7, paying
particular attention to the proof of Theorem 2.6. Come prepared
with questions! (Otherwise, you won’t understand this proof.)
Reminder: PS1 due tomorrow (Friday) by 5:10 PM, in envelope
outside room 2-172 (or email PDF to instructor).