Download Lecture 7: Examples of linear operators, null space and range, and

Lecture 7: Examples of linear operators, null
space and range, and the rank-nullity theorem (1)
Travis Schedler
Thurs, Sep 29, 2011 (version: Thurs, Sep 29, 1:00 PM)
Goals (2)
• Understand dimension and infinite-dimensionality
• Dimension formula, finish Chapter 2
• Introduce linear operators
• Null space and range of linear operators
Reading for next time: Finish Chapter 3. Pay attention to isomorphisms and matrices.
Warm-up exercise (3)
Suppose V = V1 ⊕ V2 , and W is another vs.
(a) Given any linear maps T1 : V1 → W and T2 : V2 → W , prove that there
is a unique linear map T : V → W such that, for all v1 ∈ V1 and v2 ∈ V2 ,
T (v1 + v2 ) = T1 (v1 ) + T2 (v2 ).
Call this linear map T1 ⊕ T2 .
(b) For every linear map T : V → W , let T1 : V1 → W and T2 : V2 → W be
the restrictions. Prove that T = T1 ⊕ T2 .
Conclusion: Linear maps T : V → W are the same as pairs (T1 , T2 ) of linear
maps T1 : V1 → W and T2 : V2 → W .
Solution to warm-up exercise (4)
(a) Well-definition: For every v ∈ V , there exist unique v1 ∈ V1 and v2 ∈ V2
such that v = v1 + v2 , so we can define T by T (v) = T (v1 ) + T (v2 ).
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Linearity: We first show that T (u + v) = T (u) + T (v). Write u = u1 + u2
and v = v1 + v2 for u1 , v1 ∈ V1 and u2 , v2 ∈ V2 . Then,
T (u + v) = T ((u1 + v1 ) + (u2 + v2 )) = T1 (u1 + v1 ) + T2 (u2 + v2 )
= (T1 (u1 ) + T1 (v1 )) + (T2 (u2 ) + T2 (v2 ))
= (T1 (u1 ) + T2 (u2 )) + (T1 (v1 ) + T2 (v2 )) = T (u) + T (v).
Similarly, T (av) = T (a(v1 + v2 )) = T1 (av1 ) + T2 (av2 ) = a(T1 (v1 ) +
T2 (v2 )) = aT (v).
(b) For every v ∈ V , write v = v1 + v2 for vi ∈ Vi . Then T (v) = T (v1 + v2 ) =
T (v1 ) + T (v2 ) = T1 (v1 ) + T2 (v2 ) = (T1 ⊕ T2 )(v).
Linear transformations (review) (5)
Definition 1. Given two vector spaces V and W over F, a linear transformation
T : V → W is a function satisfying:
• T (u + v) = T (u) + T (v), ∀u, v ∈ V ;
• T (av) = aT (v), ∀a ∈ F, v ∈ V .
Examples:
• The identity map I : V → V is a linear transformation. So is the zero
map V → {0} (or the zero map V → V ).
• Scalar multiplication: λI : V → V : λI(v) = λv for all v.
• The inclusion T : Fm → Fn for m ≤ n as the first coordinates: T (a1 , . . . , am ) =
(a1 , . . . , am , 0, . . . , 0).
• The evaluation map: let V = the vs. of functions {1, . . . , n} → F. Then
for any 1 ≤ j ≤ n, we have evj : V → F, evj (f ) = f (j).
More sophisticated examples (6)
• Differentiation of differentiable functions (R to R).
– E.g., xn 7→ nxn−1 . Gives P(R) → P(R), linear.
• Integration of continuous real functions (denoted C(R)):
Rb
– Definite integration: given a ≤ b, f 7→ a f (y) dy ∈ R. Gives
C(R) → R.
Rx
– Indefinite integration: f 7→ a f (y) dy, gives C(R) → C(R).
• Multiplication by a function: Fix a function f . This defines Mf : functions →
functions, Mf (g) := f g.
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– For example, multiplication by 2x3 : M2x3 (1 + x) = 2x3 + 2x4 .
• Backward shift: T : F∞ → F∞ , T (x1 , x2 , x3 , . . .) = (x2 , x3 , . . .).
• Projections: Given V = U ⊕W , we obtain a map V U called projection:
(u + w) 7→ u.
– Since all v can uniquely be written as v = u + w, the map v 7→ w is
well-defined!
Examples of linear transformations R2 → R2 (7)
• Rotations Rθ (counterclockwise). E.g., the 90◦ rotation, R90◦ (x, y) =
(−y, x).
• Reflections. E.g., about the x-axis, (x, y) 7→ (x, −y).
• Question: What is the reflection about the x = y axis? Answer: (x, y) 7→
(y, x).
Matrices (8)
given any two-by-two matrix,
A=
a
c
b
,
d
(0.1)
this defines a linear map TA : F2 → F2 :
x
• View v = (x, y) as the column vector
.
y
• Then, TA (v) is defined by the product Av, i.e.,
a b
x
ax + by
=
.
c d
y
cx + dy
(0.2)
• Thus, TA (v) = (ax + by, cx + dy).
Theorem 2 (Exercise: generalizes Chap 3, problem 1 on PSet). All linear maps
F2 →F2 are
of the form TA for a unique matrix A. This matrix is defined by
a b
A=
where TA (1, 0) = (a, c) and TA (0, 1) = (b, d).
c d
Null space (9)
Definition 3. The null space of a linear transformation T : V → W is null T :=
{v ∈ V | T (v) = 0} ⊆ V .
Exercise: The null space is a subspace.
Examples:
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• The null space of the backwards shift, T : F∞ → F∞ , given by T (a0 , a1 , a2 , . . .) =
(a1 , a2 , . . .) is: {(a, 0, 0, 0, . . .) | a ∈ F}.
• The null space of the identity, I : V → V , is: {0}.
• The null space of the zero map, 0 : V → V , is: V .
• The null space of a rotation of R2 is: {0}.
• The null space of the orthogonal (=perpendicular) projection R2 → x −
axis is: y-axis.
Proposition 3.2: Injectivity and null space (10)
Recall:
Definition 4. A function f : X → Y is injective if, for every pair x1 6= x2 of
elements of X, f (x1 ) 6= f (x2 ).
Proposition 0.3 (Proposition 3.2). A linear transformation T : V → W is
injective if and only if null T = {0}.
Proof (on board). Suppose T is injective. Since T (0) = 0, if T (v) = 0, then
v = 0. Hence, v ∈ null T implies v = 0, i.e., null T = {0}.
Conversely, suppose that null T = {0}. If T (v) = T (w), then T (v − w) = 0,
and hence v − w = 0. So v = w. Hence, T (v) = T (w) implies v = w, i.e., T is
injective.
Range (11)
Definition 5. The range of a function f : X → Y is the subset range f :=
{f (x) | x ∈ X} ⊆ Y .
Definition 6. A function f : X → Y is surjective if (range f ) = Y .
Exercise: If T : V → W is a linear map, then (range T ) ⊆ W is a subspace.
Rank-nullity theorem (12)
Theorem 7 (Theorem 3.4; “Rank-nullity theorem”). If V is f.d. and T : V →
W is linear, then
dim V = dim null(T ) + dim range(T ).
Definition: the rank of T is rk T := dim range(T ).
Corollary 8. rk T ≤ min{dim V, dim W }. Moreover, rk T = dim V if and only
if T is injective, and rk T = dim W if and only if T is surjective.
Corollary 9 (Corollaries 3.5 and 3.6). Let T ∈ L(V, W ) with V and W finitedimensional.
(i) If dim V > dim W , then T is not injective.
(ii) If dim V < dim W , then T is not surjective.
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Proof of rank-nullity theorem (on board) (13)
Recall: T : V → W is linear, and V is f.d.
• Let (u1 , . . . , um ) be a basis of null T .
• Extend (u1 , . . . , um ) to a basis (u1 , . . . , um , v1 , . . . , vn ) of V .
• Then, dim V = m + n = dim null T + n. We need: dim range T = n.
• Claim: (T (v1 ), . . . , T (vn )) is a basis for range T . This proves the theorem.
Proof of Claim (on board) (14)
Claim: (T (v1 ), . . . , T (vn )) is a basis for range T .
• Spanning: First, Span(T (u1 ), . . . , T (um ), T (v1 ), . . . , T (vn )) = T (Span(u1 , . . . , um , v1 , . . . , vn )) =
T (V ), by linearity of T .
• But, T (u1 ) = · · · = T (um ) = 0 since u1 , . . . , um ∈ null(T ).
• Hence, (T (v1 ), . . . , T (vn )) span T (V ) = range(T ).
• Lin. ind.: Suppose a1 T (v1 ) + · · · + an T (vn ) = 0.
• Then, T (a1 v1 + · · · + an vn ) = 0.
• So a1 v1 + · · · + an vn ∈ null T . We can therefore write
a1 v1 + · · · + an vn = b1 u1 + · · · + bm um .
• By linear independence of (u1 , . . . , um , v1 , . . . , vn ), this implies that a1 =
· · · = an = 0 = b1 = · · · = bm . So, (T (v1 ), . . . , T (vn )) is lin. ind.
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