Diagonalizaci´ on de formas cuadr´ aticas con el m´ etodo matricial
Objetivos. Practicar el m´etodo matricial de diagonalizaci´on de formas cuadr´aticas.
Requisitos. Formas cuadr´aticas, cambio de base.
Copyright. Los ejemplos de esta secci´on pertenecen a Vadim D. Kryakvin.
1. Ejemplo.
q(x) = x21+ 2x22− 8x23+ 4x1x2− 2x1x3− 16x2x3.
Soluci´on. Con la matriz QE hacemos la operaciones por columnas y las operaciones por filas correspondientes, y con la matriz de cambio PE,F s´olo hacemos las operaciones por columnas.
1 2 −1 2 2 −8
−1 −8 −8
1 0 0
0 1 0
0 0 1
C2+= −2C1
C3+= C1
−−−−−−−→
1 0 0
0 −2 −6 0 −6 −9 1 −2 1
0 1 0
0 0 1
C3+= −3C2
−−−−−−−→
1 0 0
0 −2 0
0 0 9
1 −2 7 0 1 −3
0 0 1
Respuesta:
qF = diag(1, −2, 9), esto es, q(x) = y21− 2y22+ 9y32,
PE,F =
1 −2 7 0 1 −3
0 0 1
, esto es,
x1 = y1 − 2y2 + 7y3, x2 = y2 − 3y3,
x3 = y3.
Comprobaci´on:
2. Ejemplo.
q(x) = x21+ x22+ x23+ x24 + 2x1x2− 2x1x3+ 2x1x4 − 2x3x4. Soluci´on.
1 1 −1 1
1 1 0 0
−1 0 1 −1 1 0 −1 1
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
C2+= −C1
C3+= C1
C4+= −C1
−−−−−−→
1 0 0 0
0 0 1 −1
0 1 0 0
0 −1 0 0 1 −1 1 −1
0 1 0 0
0 0 1 0
0 0 0 1
C2+= C3
−−−−−→
1 0 0 0
0 2 1 −1
0 1 0 0
0 −1 0 0 1 0 1 −1
0 1 0 0
0 1 1 0
0 0 0 1
C3+= −12C2
C4+=12C2
−−−−−−−→
1 0 0 0
0 2 0 0
0 0 −1/2 1/2 0 0 1/2 −1/2
1 0 1 −1
0 1 −1/2 1/2 0 1 1/2 1/2
0 0 0 1
C4+= C3
−−−−−→
1 0 0 0
0 2 0 0
0 0 −1/2 0
0 0 0 0
1 0 1 0
0 1 −1/2 0 0 1 1/2 1
0 0 0 1
.
Comprobaci´on:
qF =
1 0 0 0
0 1 1 0
1 −1/2 1/2 0
0 0 1 1
1 1 −1 1
1 1 0 0
−1 0 1 −1 1 0 −1 1
1 0 1 0
0 1 −1/2 0 0 1 1/2 1
0 0 0 1
=
1 0 0 0
0 2 0 0
0 0 −1/2 0
0 0 0 0
= diag(1, 2, −1/2, 0).
q(x) = 6x1x3+ 4x1x4− 3x23− 2x3x4+ 2x1x2+ x22 + 2x2x3+ 2x2x4. Soluci´on.
0 1 3 2
1 1 1 1
3 1 −3 −1 2 1 −1 0
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
C1↔C2
−−−−→
1 1 1 1
1 0 3 2
1 3 −3 −1 1 2 −1 0
0 1 0 0
1 0 0 0
0 0 1 0
0 0 0 1
C2+= −C1
C3+= −C1
C4+= −C1
−−−−−−→
1 0 0 0
0 −1 2 1 0 2 −4 −2 0 1 −2 −1
0 1 0 0
1 −1 −1 −1
0 0 1 0
0 0 0 1
C3+= 2C2
C4+= C2
−−−−−−→
1 0 0 0
0 −1 0 0
0 0 0 0
0 0 0 0
0 1 2 1
1 −1 −3 −2
0 0 1 0
0 0 0 1
.
Comprobaci´on:
qF =
0 1 0 0 1 −1 0 0 2 −3 1 0 1 −2 0 1
0 1 3 2
1 1 1 1
3 1 −3 −1 2 1 −1 0
0 1 2 1
1 −1 −3 −2
0 0 1 0
0 0 0 1
=
1 0 0 0 0 −1 0 0 0 0 0 0 0 0 0 0
= diag(1, −1, 0, 0).
4. Ejemplo.
q(x) = 10x1x2− 4x2x3. Soluci´on.
0 5 0
5 0 −2 0 −2 0
1 0 0
0 1 0
0 0 1
C1+= C2
−−−−−→
10 5 −2 5 0 −2
−2 −2 0
1 0 0
1 1 0
0 0 1
C2+= −12C1
C3+=15C1
−−−−−−−→
10 0 0
0 −5/2 −1 0 −1 −2/5 1 −1/2 1/5 1 1/2 1/5
0 0 1
C3+= −15C2
−−−−−−−→
10 0 0
0 −5/2 0
0 0 0
1 −1/2 2/5 1 1/2 0
0 0 1
.
Comprobaci´on:
1 1 0
−1/2 1/2 0 2/5 0 1
0 5 0
5 0 −2 0 −2 0
1 −1/2 2/5 1 1/2 0
0 0 1
=
10 0 0 0 −5/2 0
0 0 0
.
1. q(x) = x21− 2x1x2− 6x1x3− 7x22+ 14x2x3+ 5x23; 2. q(x) = 2x21− 4x1x2+ 12x1x3+ 14x22− 48x2x3+ 45x23; 3. q(x) = −3x21− 6x1x2+ 18x1x3+ x22+ 30x2x3− 16x23; 4. q(x) = 2x21− 8x1x2− 8x1x3+ 16x22+ 24x2x3 + 11x23; 5. q(x) = −x21+ 2x1x2− 6x1x3− 19x22+ 30x2x3− 20x23; 6. q(x) = x21− 6x1x2− 2x1x3− 6x2x3− 3x23.
6. Ejercicios.
1. q(x) = x21+ 4x1x2− 2x1x3+ 2x1x4+ 4x22− 2x2x3+ 6x2x4+ x23− 4x3x4+ x24; 2. q(x) = x21− 4x1x2+ 2x1x3+ 4x1x4+ 4x22− 2x2x3− 8x2x4+ 3x24;
3. q(x) = x1x2 − x1x3+ x1x4+ x2x3− x3x4;
4. q(x) = 2x1x2− 2x1x3+ 2x1x4+ 2x2x3+ 2x2x4− 2x3x4; 5. q(x) = −x1x3 + x1x4− x2x3+ x2x4;
6. q(x) = x21+ 2x1x2+ 2x1x3− 4x1x4+ x22+ 4x2x3− 6x2x4 + 2x23− 6x3x4+ 5x24.
