Linear Algebra - Sebastian Pancratz

t = min{p, q} ψ 〈v1 +
vp+1, . . . , vt + vp+t, vp+q+1, . . . , vn〉 0 n − max{p, q}
ψ 0 ψ = 0 U ≤ V U ∩ X = {0} = U ∩ N
dim U ≤ n − p dim U ≤ n − q
ψ {v ∈ V : ψ(v, w) = 0 ∀w ∈ V } ker ψ =
〈vp+q+1, . . . , vn〉
ψ ker ψ = {0}
[ψ]B B n = p + q
Q V = R 3
Q(x1, x2, x3) = x 2 1 + x 2 2 + 2x 2 3 + 2x1x2 + 2x1x3 − 2x2x3.
Q
⎛
1 1
⎞
1
A = ⎝1
1 −1⎠
.
1 −1 2
Q(x1, x2, x3) = x 2 1 + x 2 2 + 2x 2 3 + 2x1x2 + 2x1x3 − 2x2x3
rank(Q) = 3 s(Q) = 2 − 1 = 1
= (x1 + x2 + x3) 2 + x 2 3 − 4x2x3
= (x1 + x2 + x3) 2 + (x3 − 2x2) 2 − (2x2) 2
P −1 ⎛
1 1
⎞
1
= ⎝0
−2 1⎠
P
0 2 0
T ⎛
1 0
⎞
0
AP = ⎝0
1 0 ⎠
0 0 −1
P P T AP
P = E1 · · · Ek
A → E T 1 AE1 → . . . → E T k · · · ET 1 AE1 · · · Ek = D
e1 Q(e1) = 0 e1 = (1, 0, 0) T
Q(e1) = 1 W = {v ∈ V : ψ(e, v) = 0} = {(a, b, c) T : a + b + c = 0}
e T 1 A = (1, 1, 1) e2 ∈ W Q(e2) = 0 e2 = (1, 0, −1) Q(e2) = 1
e3 ∈ W ψ(e2, e3) = 0 e3 = (a, b, c) T a + b + c = 0
2b − c = 0 e T 2 A = (0, 2, −1) e3 = 1
2 (−3, 1, 2)T Q(e3) = −1
s(Q)