Applicazioni lineari 1

Verificare che det di teta non e lineare

Quindi:

T(A) = det(A)

Dove A ∈ R^2x2

Se fisse linee

T(A) + T(B) = T(A + B)

∀ A, B ∈ N^1x1

Se A = (a₁₁ a₁₂ a₂₁ a₂₂) B = (b₁₁ b₁₂ b₂₁ b₂₂)

det(A) = T(A) = a₁₁a₂₂ - a₁₂a₂₁

det(B) = T(B) = b₁₁b₂₂ - b₁₂b₂₁

det(A + B) = T(A + B) = (a₁₁+b₁₁ a₁₂+b₁₂ a₂₁+b₂₁ a₂₂+b₂₂)

= (a₁₁+b₁₁)(a₂₂+b₂₂) - (a₂₁+b₂₁)(a₁₂+b₁₂)

≠ T(A) + T(B) = a₁₁a₂₂ - a₁₂a₂₁ + b₁₁b₂₂ - b₁₂b₂₁

Quindi non é lineare

3)

Stabilire se α : ℝ² → ℝ² è lin

T(1,2) = (3,0)

T(2,7) = (4,5) T(1,5) = (1,4)

Se entrambi

T(1,2) + T(2,7) = (1+2,2+7) = T(3,9)

T(1,2) + T(1,5) = T(2,2)

T(1,2) - T(1,5) = (3,0)*(1,4) = (4,0)

Dato che T(2,7) ≠ (4,4)

Non é lineare

7)

T: ℝ² → ℝ³

T(2) = (1, 2, 1)

T(2) = (1, 0, -1)

a) T(x, y)

b) A (app...

d) (3, 4, 1) ∈ Im(T)

v = x q₁ + y q₂

T(ᵁ) = xT(q₁) + yT(q₂) = x (1, 2, 1) + y (1, 0, -1)

Q = (x + y, 2x, x - y)

q₁ A = (

• 1
• 2
• 0

)

c)

{

x + y = 3

2x + = 4

x - y = 1

}

(

x = 2

y = 1

)

=>

(3, 4, 1) ∈ Im(T) ⊇ (1, 2, 1)

B_min(T) = { ( 1, 0, 0, 0 ), ( 0, 2, 0, 0 ), ( 1, 1, 0, 0 ) }

b_i i.e. v_i(C)

⇒ b_i(T) =

B_ker(T) = { ( 0, 0, 0, t ) }

1: R⁴ → R⁴

1(x₁, x₂, x₃, x₄) = (x₂ + 3x₃, -2x₃, x₄, x₁, 0, x₁ - x₄)

• 1 nulla (alge)
• 1 nulla (alge nulla = dim ker 1)
• Im (1)

1(e₁) = (0, 0, 0, 1)

1(e₂) = (1, 0, 0, -1)

1(e₃) = (3, -2, 0, 0)

1(e₄) = (0, 1, 0, 1)

A = | 0 1 3 0 | | 0 0 -2 0 | | 0 0 0 0 | | 1 -1 0 1 |

(RA ↔ R)

p(A) = 3 => dim (Im (1)) = 3 → None 2 soluzioni

(dim ker (1)) = 1

S := { (t, -3t, b, 2t) / t ∈ R² }

B = (1, 1, 3, 1, 2)

1) T(w) = Aw = ^{(2 0 0)}₍₁₎ ^{(0 0 1)}_(x) = (2, 2)

5) 2x = 0 ⇒ S = { (y, z, -t) | y, z, t ∈ ℝ }Passa sempre per (0, 1, -1) = 0 ≠ Im(A) ⇒ dim(Im(A)) = 1Im(T) = [(2, 2) ] ≠ linee di R²Essendo un gerarch intorno al S = < x₂, x₃ > ∈ ℝ²

(x₁x₂x₃)→ y - x₂ + z₃ + (0)(0) k. (1)= (x₁ = x₃)

→ x₂ = t→ x = x₃ → t = x₂, x₁ = x₂ - x₁= x₃ → P = (x₁, t, k )_B

W = {0, 1, 0 }

7) T(t₂) = (1, 1), T(t₃) = (0, 1), T(t₅) = (2, 1)

M_{C, B}(A|β) = ^{(4 0 2)}_{(0 1)}

8) F = M_{B, C} (I_q) (2, 1, 0) = (x, y, z)_y,t(0 2 ) (x₁ = 7)(0 0)→ x₂ = t Ch 5 f t = x₅ (x, y, z, x₅)_β3

P = ^{(0 2 0)}_{(0 0)}|P| = 2

9) N_{B, C}(T*) = P^t p^t ⇒ P₁₁ = (-1)² = z⁸2K₁ = ^-2_t = (0,1)

|P| = -t |(C_S) = (-t )= 1/2 ^{(1 0)}_{(0 0)} (0 0)(-1 1 2)= V₁ - V₂ 0V₂ = (V^t)_tV₂ = (V)1/2 ) - 1 V₂ (0 |V|)1) = (2, 2)(1 | (0)(-1)

21

T: R⁴ → R³

T (x, y, z, w) = (-x - y + z + w; x + 2y - z; -x + y + 2z - 3w)

1. A regola che T non è sur.
2. T non è on van che rg (T) ≠ L (T)

T (e₁) = (-1, -1, 1) T (e₂) = (-1, +2, 1) T (e₃) = (1, -1, 2) T (e₄) = (1, 0, -3)

A =

( -1 -1 1 1 )

( -1 2 1 0 )

( 1 -1 2 -3 )

R₂ → R₂ - R₁

R₃ → R₃ - R₁

( -1 -1 1 1 )

( 0 2 2 -1 )

( 0 0 3 -2 ) 3

R₃ → R₃ - 2R₂ / 3

( -1 -1 1 1 )

( 0 3 2 1 )

( 0 0 1 -1/3)

( -1 -1 1 1 ) =

( 0 3 2 1 )

( 0 0 10/3 16/3 )

⇒ δ (A) = 3 ⇒ dim (Im (A)) = 3

dim (ker(A)) = 1

B_Im(A) = {(1, 0, 0), (1, 3, 0), (1, -2, 1)}

- x - y - z + w = 0

3y - z - 2x = 0

10z + w = 0

{ 2 = 7/3 w

{ 3/7 16/5 w - 1 w = 0

{ -x - y - 2v + w = 5

T: ℝ³ → ℝ³

T(x) = A x

A = ⎛ ⎜ ⎝1 1 10 1 01 0 0⎞ ⎟ ⎠

B_s = {(1,1,1), v₂ = (1,0,0), v₃ = (0,0,1)}

B = {(1,1,1), v₂, v₃}

(a) M_B(T) = ?

(b) Base del ker(T)

T(v₁) = ⎛ ⎜ ⎝1 1 10 1 01 0 0⎞ ⎟ ⎠⎛ ⎜ ⎝111⎞ ⎟ ⎠= (3, 2, 1)

T(v₂) = ⎛ ⎜ ⎝1 1 10 1 01 0 0⎞ ⎟ ⎠⎛ ⎜ ⎝100⎞ ⎟ ⎠= (1, 0, 1)

T(v₃) = ⎛ ⎜ ⎝1 1 10 1 01 0 0⎞ ⎟ ⎠⎛ ⎜ ⎝001⎞ ⎟ ⎠= (1, 1, 0)

T(v_B) = x v₁ + y v₂ + z v₃ = x ⎛ ⎜ ⎝111⎞ ⎟ ⎠+ y ⎛ ⎜ ⎝100⎞ ⎟ ⎠+ z ⎛ ⎜ ⎝010⎞ ⎟ ⎠

⎧⎨⎩x + y = 3x = 2x + z = 1⎫⎬⎭⟹ ⎧⎨⎩x = 2y = 1z = -1⎫⎬⎭⟹T(v_B) = (2,1,-1)_B

T(v_B) = x v₁ + y v₂ + z v₃ = x ⎛ ⎜ ⎝111⎞ ⎟ ⎠y ⎛ ⎜ ⎝100⎞ ⎟ ⎠+ z ⎛ ⎜ ⎝010⎞ ⎟ ⎠= ⎛ ⎜ ⎝100⎞ ⎟ ⎠

⎧⎨⎩x + y = 1x = 0x + z = 1⎫⎬⎭⟹ ⎧⎨⎩x = 0y = 1z = 1⎫⎬⎭⟹T(v₂) = (0,1,1)_B