Esercizi di Algebra lineare e geometria sui matriciali

6.4.1

mercoledi 21 gennaio 2026     10:34

A = ( -5   0   6         k   1   -3         -3   0   10 ), k in R

H = ( -5 - l   0   6         k   1 - l   -3         -3   0   10 - l )

P(l) = (-5 - l)(1 - l)(10 - l) + 54(1 - l)

(1 - l)((-5 - l)(10 - l) + 54)

(1 - l)(-50 + 5l - 10l + l² + 54)

(1 - l)(l² - 5l + 4) = 0

l_1,2 = (5 +/- 3)/2 < 1

l₁ = 1 l₂ = 1 } Ma = 2    Mg = 2? l₃ = 4 } Ma = 1

H(1) = ( -6   0   6         k   0   -3         -3   0   9 ) -> ( 0   -6   6         0   k   -3         0   -3   3 )

18 - 6k = 0 => k = 3 <=> z = 2 per l₁, DIAGONALIZZABILE

ALTRIMENTI z = 1

H(1) = ( -6   0   6         3   0   -3         -3   0   3 )

-x + z = 0 <=> x = z

S: { x, beta, -x } => (1, 0, -1) (0, 1, 0)

H(4) = ( -3   0   6         3   -3   -3         -3   0   6 )

{ -3x + 2z = 0 x - y - z = 0 }

-> { x = 2z/3 2z/3 - y - z = 0 }

-> { x = -2y z = -3y }

-> S: { -2a, a, -3a }

{ 2, -1, 3 }

P = ( 1   0   2       0   1   -1       -1   0   3 )

D = diag(1, 1, 4)

6.4.2

mercoledi 21 gennaio 2026     11:12

A = ( -8   3   -3       -6   7   -6       0   0   1 )

B = ( 1   0   0       0   h+3   0       0   0   h )

P_B(x) = P_A(x)

H(A) = ( -8-x   3   -3       -6   7-x   -6       0   0   1-x )

P_A(x) = ((-8-x)(7-x) + 54)(1-x)

= -56 + 8x - 7x + x² + 54

= (x² + x - 2)(1-x) = 0

x_1,2 = (-1 ± 3)/2 < -2, 1

x₃ = 1

x = -2    m_a = 1

x = 1    m_a = 2

AUTOVALORE SEMPICE

m_a = m_g = 1

H_A(1) = ( -3   3   -3       -6   6   -6       0   0   0 )

-x + y - z = 0  → m_g = 2 = m_a

P_B(x) = P_A(x)

{1, h+3, h} = {1, 1, -2}

h+3 = 1  ⇐> h = -2

h = -2

h = -2  ⇐> A ≈ B

6.4.11

A = ( 0 -3h h )( 2 3 -1 )( 0 -6 2 )

|A - 2I| = ( -2 -3h h )( 2 1 -1 )( 0 -6 0 )

P(lambda) = -12h + 12 <=> h = 1 per lambda = 2

|A - lambda I|h = 1

( -lambda -3 1 )( 2 3-lambda -1 )( 0 -6 2-lambda )

P(lambda) = 6(lambda - 2) + (2 - lambda)((-lambda)(3-lambda) + 6)

= -6(2 - lambda) + (2 - lambda)(-3lambda + lambda² + 6)

= (2 - lambda)(lambda² - 3lambda)

= (2 - lambda)(lambda(lambda - 3)) = 0

lambda₁ = 2     m_p = 1

lambda₂ = 0     "

lambda₃ = 3     "

H(2) = ( -2 -3 1 )( 2 1 -1 )( 0 -6 0 )

{ -2x - 3y + z = 0 2x + y - z y = 0 }

-> { -2x + z = 0 2x - z = 0 } { x = z/2 }

m_g = 1

S: { alpha, 0, 2alpha }

(1, 0, 2)

H(0) = ( 0 -3 1 )( 2 3 -1 )( 0 -6 2 )

{ -3y + z = 0 2x + 3y - z = 0 }

{ z = 3y x = 0 }

S: { 0, alpha, 3alpha }(0, 1, 3)

H(3) = ( -3 -3 1 )( 2 0 -1 )( 0 -6 -1 )

{ -3x - 3y + z = 0 2x - z = 0 -6y - z = 0 }

-> { y = -x/3 z = 2x }

S: { alpha, -alpha/3, 2alpha }

(1, -1/3, 2)

(3, -1, 6)

P = ( 0 1 3 )( 1 0 -1 )( 3 2 6 )

D = (

0 0 0

0 2 0

0 0 3

)

Extra

mercoledi 21 gennaio 2026 11:55

A = ( 1 4 20 1 00 4 3 )

H = ( 1 - lambda 4 20 1 - lambda 00 4 3 - lambda )

P(lambda) = (1 - lambda)²(3 - lambda) = 0

lambda = 1 m_A = 2

lambda = 3

H(1) = ( 0 4 20 0 00 4 2 ) -> 2y + z = 0 <=> z = -2y S: { beta, -2 alpha, alpha }

(0, -2, 1)(1, 0, 0)

H(3) = ( -2 4 20 -2 00 4 0 ) { -x + 2y + z = 0y = 0 } -> { x = zy = 0 }

S: { alpha, 0, alpha }

(1, 0, 1)

P = ( 0 1 1-2 0 01 0 1 )

D = ( 1 0 00 1 00 0 3 )

Extra 2

mercoledi 21 gennaio 2026    12:07

A =

( 3   -2   0 ) ( -2   3   0 ) ( 0   0   5 )

H =

( 3 - l   -2   0 ) ( -2   3 - l   0 ) ( 0   0   5 - l )

P(l) = (3 - l)²(5 - l) - 4(5 - l)

(5 - l)((3 - l)² - 4) = 0

l = 5

(3 - l)² = 4 <=> 3 - l = ± 2 <=> -l = ± 2 - 3 <=> -l = -1   )   l = 1 -l = -5

l = 1    m_a = 1 l = 5    m_a = 2

H(5) =

( -2   -2   0 ) ( -2   -2   0 ) ( 0   0   0 )

-x - y = 0 <=> x = -y     m_g = 2

S: { α, -α, β }

(1, -1, 0) (0, 0, 1)

H(1) =

( 2   -2   0 ) ( -2   2   0 ) ( 0   0   4 )

x - y = 0 -x + y = 0 z = 0

{ x = y z = 0 }

S: { α, α, 0 }

(1, 1, 0)

P =

( 1   1   0 ) ( -1   1   0 ) ( 0   0   1 )

D =

( 5   0   0 ) ( 0   1   0 ) ( 0   0   5 )