Sums, Scalar Multiples and Products of Linear Maps
I Key FormulasKey Insight: L ( V , W ) \mathcal{L}(V, W) L ( V , W ) W W W v v v T T T ( − 1 ) T (-1)T ( − 1 ) T
II Sums and Scalar Multiples, One Vector at a TimeFor S ( x , y ) = ( 2 x , y ) S(x, y) = (2x, y) S ( x , y ) = ( 2 x , y ) T ( x , y ) = ( − y , x ) T(x, y) = (-y, x) T ( x , y ) = ( − y , x ) S + λ T S + \lambda T S + λ T v v v S v + λ ( T v ) Sv + \lambda(Tv) S v + λ ( T v )
III T D TD T D D T DT D T With D p = p ′ Dp = p' D p = p ′ ( T p ) ( x ) = x 2 p ( x ) (Tp)(x) = x^2 p(x) ( Tp ) ( x ) = x 2 p ( x ) 2 x p ( x ) 2x\,p(x) 2 x p ( x )
IV Why 2 x + 1 2x + 1 2 x + 1 For f ( x ) = m x + b f(x) = mx + b f ( x ) = m x + b b b b f f f f ( 0 ) = b = 0 f(0) = b = 0 f ( 0 ) = b = 0
💡 The same test rules out cosine: cos 0 = 1 ≠ 0 \cos 0 = 1 \neq 0 cos 0 = 1 = 0 T ( 0 ) = 0 T(0) = 0 T ( 0 ) = 0
V Quiz Questions
Problem 1 · Evaluate a Linear Combination of Maps
Given: S , T ∈ L ( R 2 , R 2 ) S, T \in \mathcal{L}(\mathbf{R}^2, \mathbf{R}^2) S , T ∈ L ( R 2 , R 2 ) S ( x , y ) = ( x + 2 y , 3 x ) S(x, y) = (x + 2y,\ 3x) S ( x , y ) = ( x + 2 y , 3 x ) T ( x , y ) = ( y , x − y ) T(x, y) = (y,\ x - y) T ( x , y ) = ( y , x − y ) Find ( S + 2 T ) ( 1 , 1 ) (S + 2T)(1, 1) ( S + 2 T ) ( 1 , 1 )
a ( 4 , 3 ) (4, 3) ( 4 , 3 ) b ( 7 , 6 ) (7, 6) ( 7 , 6 ) c ( 8 , 6 ) (8, 6) ( 8 , 6 ) d ( 5 , 3 ) (5, 3) ( 5 , 3 )
✅ Correct! S ( 1 , 1 ) = ( 3 , 3 ) S(1, 1) = (3, 3) S ( 1 , 1 ) = ( 3 , 3 ) and
2 T ( 1 , 1 ) = 2 ( 1 , 0 ) = ( 2 , 0 ) 2\,T(1, 1) = 2(1, 0) = (2, 0) 2 T ( 1 , 1 ) = 2 ( 1 , 0 ) = ( 2 , 0 ) , so the sum is
( 5 , 3 ) (5, 3) ( 5 , 3 ) .
❌ Not quite. ( 4 , 3 ) = S ( 1 , 1 ) + T ( 1 , 1 ) (4, 3) = S(1, 1) + T(1, 1) ( 4 , 3 ) = S ( 1 , 1 ) + T ( 1 , 1 ) is
( S + T ) ( 1 , 1 ) (S + T)(1, 1) ( S + T ) ( 1 , 1 ) ; the scalar
2 2 2 multiplies
T ( 1 , 1 ) T(1, 1) T ( 1 , 1 ) .
❌ Not quite. ( 7 , 6 ) = 2 S ( 1 , 1 ) + T ( 1 , 1 ) (7, 6) = 2\,S(1, 1) + T(1, 1) ( 7 , 6 ) = 2 S ( 1 , 1 ) + T ( 1 , 1 ) is
( 2 S + T ) ( 1 , 1 ) (2S + T)(1, 1) ( 2 S + T ) ( 1 , 1 ) ; in
S + 2 T S + 2T S + 2 T the
2 2 2 belongs to
T T T .
❌ Not quite. ( 8 , 6 ) = 2 ( S + T ) ( 1 , 1 ) (8, 6) = 2(S + T)(1, 1) ( 8 , 6 ) = 2 ( S + T ) ( 1 , 1 ) : the
2 2 2 scales only
T T T , not
S S S .
Show solution
By the definitions of sum and scalar multiple, ( S + 2 T ) ( v ) = S v + 2 ( T v ) (S + 2T)(v) = Sv + 2(Tv) ( S + 2 T ) ( v ) = S v + 2 ( T v ) v = ( 1 , 1 ) v = (1, 1) v = ( 1 , 1 )
S ( 1 , 1 ) = ( 1 + 2 ⋅ 1 , 3 ⋅ 1 ) = ( 3 , 3 ) , T ( 1 , 1 ) = ( 1 , 1 − 1 ) = ( 1 , 0 ) S(1, 1) = (1 + 2 \cdot 1,\ 3 \cdot 1) = (3, 3), \qquad T(1, 1) = (1,\ 1 - 1) = (1, 0) S ( 1 , 1 ) = ( 1 + 2 ⋅ 1 , 3 ⋅ 1 ) = ( 3 , 3 ) , T ( 1 , 1 ) = ( 1 , 1 − 1 ) = ( 1 , 0 ) ( S + 2 T ) ( 1 , 1 ) = ( 3 , 3 ) + 2 ( 1 , 0 ) = ( 3 , 3 ) + ( 2 , 0 ) = ( 5 , 3 ) (S + 2T)(1, 1) = (3, 3) + 2(1, 0) = (3, 3) + (2, 0) = (5, 3) ( S + 2 T ) ( 1 , 1 ) = ( 3 , 3 ) + 2 ( 1 , 0 ) = ( 3 , 3 ) + ( 2 , 0 ) = ( 5 , 3 )
Problem 2 · Which Map Acts First?
Given: D , T ∈ L ( P ( R ) ) D, T \in \mathcal{L}(\mathcal{P}(\mathbf{R})) D , T ∈ L ( P ( R )) D p = p ′ Dp = p' D p = p ′ ( T p ) ( x ) = x 2 p ( x ) (Tp)(x) = x^2 p(x) ( Tp ) ( x ) = x 2 p ( x ) p ( x ) = x 2 + 1 p(x) = x^2 + 1 p ( x ) = x 2 + 1 Find ( D T ) p (DT)p ( D T ) p
a 2 x 3 2x^3 2 x 3 b 4 x 3 + 2 x 4x^3 + 2x 4 x 3 + 2 x c 4 x 3 4x^3 4 x 3 d 2 x 2x 2 x
✅ Correct! T p = x 4 + x 2 Tp = x^4 + x^2 Tp = x 4 + x 2 , and differentiating gives
4 x 3 + 2 x 4x^3 + 2x 4 x 3 + 2 x .
❌ Close, but… 2 x 3 = x 2 ⋅ ( 2 x ) 2x^3 = x^2 \cdot (2x) 2 x 3 = x 2 ⋅ ( 2 x ) is
( T D ) p (TD)p ( T D ) p . In
D T DT D T the right-hand factor
T T T acts first: multiply by
x 2 x^2 x 2 , then differentiate.
❌ Not quite. T T T multiplies the constant term too:
x 2 ⋅ 1 = x 2 x^2 \cdot 1 = x^2 x 2 ⋅ 1 = x 2 , whose derivative
2 x 2x 2 x is missing.
❌ Not quite. 2 x = D p 2x = Dp 2 x = D p leaves out
T T T ; first form
T p = x 2 ( x 2 + 1 ) Tp = x^2(x^2 + 1) Tp = x 2 ( x 2 + 1 ) .
Show solution
( D T ) p = D ( T p ) (DT)p = D(Tp) ( D T ) p = D ( Tp ) T T T
( T p ) ( x ) = x 2 ( x 2 + 1 ) = x 4 + x 2 , ( D T ) p = ( x 4 + x 2 ) ′ = 4 x 3 + 2 x (Tp)(x) = x^2(x^2 + 1) = x^4 + x^2, \qquad (DT)p = (x^4 + x^2)' = 4x^3 + 2x ( Tp ) ( x ) = x 2 ( x 2 + 1 ) = x 4 + x 2 , ( D T ) p = ( x 4 + x 2 ) ′ = 4 x 3 + 2 x Check with the product rule: x 2 p ′ ( x ) + 2 x p ( x ) = x 2 ( 2 x ) + 2 x ( x 2 + 1 ) = 2 x 3 + 2 x 3 + 2 x = 4 x 3 + 2 x x^2 p'(x) + 2x\,p(x) = x^2(2x) + 2x(x^2 + 1) = 2x^3 + 2x^3 + 2x = 4x^3 + 2x x 2 p ′ ( x ) + 2 x p ( x ) = x 2 ( 2 x ) + 2 x ( x 2 + 1 ) = 2 x 3 + 2 x 3 + 2 x = 4 x 3 + 2 x ( T D ) p = x 2 ⋅ 2 x = 2 x 3 (TD)p = x^2 \cdot 2x = 2x^3 ( T D ) p = x 2 ⋅ 2 x = 2 x 3 2 x p ( x ) = 2 x 3 + 2 x 2x\,p(x) = 2x^3 + 2x 2 x p ( x ) = 2 x 3 + 2 x
Problem 3 · Both Orders on R²
Given: S , T ∈ L ( R 2 , R 2 ) S, T \in \mathcal{L}(\mathbf{R}^2, \mathbf{R}^2) S , T ∈ L ( R 2 , R 2 ) S ( x , y ) = ( x + y , y ) S(x, y) = (x + y,\ y) S ( x , y ) = ( x + y , y ) T ( x , y ) = ( y , x ) T(x, y) = (y,\ x) T ( x , y ) = ( y , x ) Find ( S T ) ( 1 , 2 ) (ST)(1, 2) ( ST ) ( 1 , 2 ) ( T S ) ( 1 , 2 ) (TS)(1, 2) ( TS ) ( 1 , 2 )
What is (ST)(1, 2)?
a ( 2 , 3 ) (2, 3) ( 2 , 3 ) b ( 3 , 2 ) (3, 2) ( 3 , 2 ) c ( 3 , 1 ) (3, 1) ( 3 , 1 ) d ( 2 , 1 ) (2, 1) ( 2 , 1 )
What is (TS)(1, 2)?
a ( 2 , 3 ) (2, 3) ( 2 , 3 ) b ( 3 , 1 ) (3, 1) ( 3 , 1 ) c ( 3 , 2 ) (3, 2) ( 3 , 2 ) d ( 2 , 1 ) (2, 1) ( 2 , 1 )
✅ Correct! ( S T ) ( 1 , 2 ) = ( 3 , 1 ) (ST)(1, 2) = (3, 1) ( ST ) ( 1 , 2 ) = ( 3 , 1 ) but
( T S ) ( 1 , 2 ) = ( 2 , 3 ) (TS)(1, 2) = (2, 3) ( TS ) ( 1 , 2 ) = ( 2 , 3 ) , so
S T ≠ T S ST \neq TS ST = TS even though both products are defined.
❌ Close, but… ( 2 , 3 ) (2, 3) ( 2 , 3 ) is
( T S ) ( 1 , 2 ) (TS)(1, 2) ( TS ) ( 1 , 2 ) . In
S T ST ST the map
T T T acts first:
T ( 1 , 2 ) = ( 2 , 1 ) T(1, 2) = (2, 1) T ( 1 , 2 ) = ( 2 , 1 ) , then apply
S S S .
❌ Not quite. Apply both maps,
T T T first:
( S T ) ( 1 , 2 ) = S ( T ( 1 , 2 ) ) = S ( 2 , 1 ) (ST)(1, 2) = S\big(T(1, 2)\big) = S(2, 1) ( ST ) ( 1 , 2 ) = S ( T ( 1 , 2 ) ) = S ( 2 , 1 ) .
❌ Close, but… ( 3 , 1 ) (3, 1) ( 3 , 1 ) is
( S T ) ( 1 , 2 ) (ST)(1, 2) ( ST ) ( 1 , 2 ) . In
T S TS TS the map
S S S acts first:
S ( 1 , 2 ) = ( 3 , 2 ) S(1, 2) = (3, 2) S ( 1 , 2 ) = ( 3 , 2 ) , then apply
T T T .
❌ Not quite. Apply both maps,
S S S first:
( T S ) ( 1 , 2 ) = T ( S ( 1 , 2 ) ) = T ( 3 , 2 ) (TS)(1, 2) = T\big(S(1, 2)\big) = T(3, 2) ( TS ) ( 1 , 2 ) = T ( S ( 1 , 2 ) ) = T ( 3 , 2 ) .
Show solution
The product is composition, with the right-hand factor acting first.
( S T ) ( 1 , 2 ) = S ( T ( 1 , 2 ) ) = S ( 2 , 1 ) = ( 2 + 1 , 1 ) = ( 3 , 1 ) (ST)(1, 2) = S\big(T(1, 2)\big) = S(2, 1) = (2 + 1,\ 1) = (3, 1) ( ST ) ( 1 , 2 ) = S ( T ( 1 , 2 ) ) = S ( 2 , 1 ) = ( 2 + 1 , 1 ) = ( 3 , 1 ) ( T S ) ( 1 , 2 ) = T ( S ( 1 , 2 ) ) = T ( 1 + 2 , 2 ) = T ( 3 , 2 ) = ( 2 , 3 ) (TS)(1, 2) = T\big(S(1, 2)\big) = T(1 + 2,\ 2) = T(3, 2) = (2, 3) ( TS ) ( 1 , 2 ) = T ( S ( 1 , 2 ) ) = T ( 1 + 2 , 2 ) = T ( 3 , 2 ) = ( 2 , 3 ) In general ( S T ) ( x , y ) = S ( y , x ) = ( x + y , x ) (ST)(x, y) = S(y, x) = (x + y,\ x) ( ST ) ( x , y ) = S ( y , x ) = ( x + y , x ) ( T S ) ( x , y ) = T ( x + y , y ) = ( y , x + y ) (TS)(x, y) = T(x + y,\ y) = (y,\ x + y) ( TS ) ( x , y ) = T ( x + y , y ) = ( y , x + y )
Problem 4 · Spot the Linear Map
Exactly one of these functions is a linear map. Which one?
a h : R 2 → R 2 h: \mathbf{R}^2 \to \mathbf{R}^2 h : R 2 → R 2 h ( x , y ) = ( 2 x − y , 5 x ) \ h(x, y) = (2x - y,\ 5x) h ( x , y ) = ( 2 x − y , 5 x ) b f : R → R f: \mathbf{R} \to \mathbf{R} f : R → R f ( x ) = 2 x + 1 \ f(x) = 2x + 1 f ( x ) = 2 x + 1 c g : R 2 → R 2 g: \mathbf{R}^2 \to \mathbf{R}^2 g : R 2 → R 2 g ( x , y ) = ( x + y , 1 ) \ g(x, y) = (x + y,\ 1) g ( x , y ) = ( x + y , 1 ) d k : R 2 → R 2 k: \mathbf{R}^2 \to \mathbf{R}^2 k : R 2 → R 2 k ( x , y ) = ( x y , x ) \ k(x, y) = (xy,\ x) k ( x , y ) = ( x y , x )
✅ Correct! Each coordinate of
h h h is a fixed combination
a x + b y ax + by a x + b y of the inputs, so
h h h is additive and homogeneous.
❌ Not quite. f ( 0 ) = 1 ≠ 0 f(0) = 1 \neq 0 f ( 0 ) = 1 = 0 , and every linear map sends
0 0 0 to
0 0 0 .
❌ Not quite. g ( 0 , 0 ) = ( 0 , 1 ) ≠ ( 0 , 0 ) g(0, 0) = (0, 1) \neq (0, 0) g ( 0 , 0 ) = ( 0 , 1 ) = ( 0 , 0 ) , so
g g g fails the
T ( 0 ) = 0 T(0) = 0 T ( 0 ) = 0 test.
❌ Close, but… k ( 0 , 0 ) = ( 0 , 0 ) k(0, 0) = (0, 0) k ( 0 , 0 ) = ( 0 , 0 ) , yet
k ( 2 , 2 ) = ( 4 , 2 ) k(2, 2) = (4, 2) k ( 2 , 2 ) = ( 4 , 2 ) while
2 k ( 1 , 1 ) = ( 2 , 2 ) 2\,k(1, 1) = (2, 2) 2 k ( 1 , 1 ) = ( 2 , 2 ) . Sending
0 0 0 to
0 0 0 is necessary for linearity, not sufficient.
Show solution
(a) is linear. For u = ( x 1 , y 1 ) u = (x_1, y_1) u = ( x 1 , y 1 ) w = ( x 2 , y 2 ) w = (x_2, y_2) w = ( x 2 , y 2 ) λ ∈ R \lambda \in \mathbf{R} λ ∈ R
h ( u + w ) = ( 2 ( x 1 + x 2 ) − ( y 1 + y 2 ) , 5 ( x 1 + x 2 ) ) = h ( u ) + h ( w ) h(u + w) = \big(2(x_1 + x_2) - (y_1 + y_2),\ 5(x_1 + x_2)\big) = h(u) + h(w) h ( u + w ) = ( 2 ( x 1 + x 2 ) − ( y 1 + y 2 ) , 5 ( x 1 + x 2 ) ) = h ( u ) + h ( w ) h ( λ u ) = ( 2 λ x 1 − λ y 1 , 5 λ x 1 ) = λ h ( u ) h(\lambda u) = (2\lambda x_1 - \lambda y_1,\ 5\lambda x_1) = \lambda\, h(u) h ( λ u ) = ( 2 λ x 1 − λ y 1 , 5 λ x 1 ) = λ h ( u ) (b) and (c) fail the quick test: f ( 0 ) = 1 f(0) = 1 f ( 0 ) = 1 g ( 0 , 0 ) = ( 0 , 1 ) g(0, 0) = (0, 1) g ( 0 , 0 ) = ( 0 , 1 )
(d) passes the quick test, k ( 0 , 0 ) = ( 0 , 0 ) k(0, 0) = (0, 0) k ( 0 , 0 ) = ( 0 , 0 ) k ( 2 ( 1 , 1 ) ) = k ( 2 , 2 ) = ( 4 , 2 ) ≠ ( 2 , 2 ) = 2 k ( 1 , 1 ) k\big(2(1, 1)\big) = k(2, 2) = (4, 2) \neq (2, 2) = 2\,k(1, 1) k ( 2 ( 1 , 1 ) ) = k ( 2 , 2 ) = ( 4 , 2 ) = ( 2 , 2 ) = 2 k ( 1 , 1 )
Solved: 0 / 4