Exercise 1
Given the vectors of R3
v1= (1, 0, 1), v2= (2, 1, −1), v3= (4, 2, −1) and the vectors of R2
w1= (1, 1), w2= (0, 2), w3= (−1, 0)
1 Check that {v1, v2, v3} is a basis of R3
2 Given the map f : R3→ R2defined by f (vi) = wi for i = 1, 2, 3, construct the matrix associated to f with respect to the bases B = {v1, v2v3}, E2= {ε1, ε2} (canonical basis)
3 Orthonormalise basis B
Exercise 1, answer
Since
1 0 1
2 1 −1 4 2 −1
= −1 + 2 + 4 − 4 = 1 6= 0
vectors v1, v2, v3 form a basis.
The matrix associated to f with respect to the bases B, E2is
1 0 −1
1 2 0
Exercise 1, answer (cont.)
To find an orthonormal basis starting from B, let:
v10 =v1= (1, 0, 1) v20 =v2−v2· v10
||v10||2v10= (2, 1, −1) −1
2(1, 0, 1) = 3 2, 1, −3
2
v30 =v3−v3· v10
||v10||2v10−v3· v20
||v20||2v20 =
=(4, 2, −1) −3
2(1, 0, 1) −
19 2 11 2
3 2, 1, −3
2
=
=
4 − 3
2 −57
22, 2 −19
11, −1 −3 2+57
22
=
−1 11, 3
11, 1 11
Exercise 1, answer (cont.)
Then an ortonormal basis is
1
||v10||v10,||v10 2||v20,||v10
3||v30 , where 1
||v10||v10 = √2
2 , 0,
√2 2
!
1
||v20||v20 = 1 q9
4+ 1 + 94
3 2, 1, −3
2
=
= 2
√22
3 2, 1, −3
2
=
3
√22, 2
√22, − 3
√22
1
||v30||v30 =√ 11
−1 11, 3
11, 1 11
=
− 1
√ 11, 3
√ 11, 1
√ 11
Exercise 2
Prove that the following matrix is positive definite
13 6 4 8
6 6 0 0
4 0 16 14
8 0 14 19
∈ Mat(R, 4, 4)
Exercise 2, answer
Using Sylvester’s criterion, it is enough to check that all leading principal minors are positive.
13 >0
13 6
6 6
=6(13 − 6) = 42 > 0
13 6 4
6 6 0
4 0 16
=24
13 6 4
1 1 0
1 0 4
= 24
−
6 4 0 4
+
13 4
1 4
=
=24(−24 + 52 − 4) > 0
Exercise 2, answer (cont.)
13 6 4 8
6 6 0 0
4 0 16 14
8 0 14 19
=12
13 6 4 8
1 1 0 0
2 0 8 7
8 0 14 19
=
=12
−
6 4 8
0 8 7
0 14 19
+
13 4 8
2 8 7
8 14 19
=
=12
−6
8 7
14 19
+ 13
8 7
14 19
− 4
2 7
8 19
+ 8
2 8
8 14
=
=12(7(152 − 98) − 4(38 − 56) + 8(28 − 64)) =
=12(7 · 54 + 4 · 18 − 8 · 36) =
=12(378 + 72 − 288) >
>0
Exercise 3
Compute the rank and the signature of the following quadratic forms.
(1)
F1(x , y , z) = y2+ 4z2+ 4xy + 6xz + 10yz (2)
F2(x , y , z) = x2− 3z2+ 4xy + 2xz − 2yz (3)
F3(x , y , z) = 4z2+ 4xy + 6xz − 2yz
Exercise 3, answer
(1) Quadratic form F1is associated to the matrix
0 2 3 2 1 5 3 5 4
. Its characteristic polynomial is
−λ 2 3
2 1 − λ 5
3 5 4 − λ
=
− λ(1 − λ)(4 − λ) + 30 + 30 + 9(λ − 1) + 4(λ − 4) + 25λ =
= −λ3+ 5λ2+ 34λ + 35
Notice that 0 is not an eigenvalue and, by Descartes’s rule, there is exactly one positive eigenvalue.
Thus the signature of F1is (1, 2), and the rank is 3.
Exercise 3, answer (cont.)
(2) Quadratic form F2is associated to the matrix
1 2 1
2 0 −1
1 −1 −3
. Its characteristic polynomial is
1 − λ 2 1
2 −λ −1
1 −1 −3 − λ
=
= λ(1 − λ)(3 + λ) − 2 − 2 + λ + 4(3 + λ) + λ − 1 =
= −λ3− 2λ2+ 9λ + 7
Notice that 0 is not an eigenvalue and, by Descartes’s rule, there is exactly one positive eigenvalue.
Thus the signature of F2is (1, 2), and the rank is 3.
Exercise 3, answer (cont.)
(3) Quadratic form F3is associated to the matrix
0 2 3
2 0 −1
3 −1 4
. Its characteristic polynomial is
−λ 2 3
2 −λ −1
3 −1 4 − λ
=
λ2(4 − λ) − 6 − 6 + 9λ + 4(λ − 4) + λ =
= −λ3+ 4λ2+ 14λ − 28
Notice that 0 is not an eigenvalue and, by Descartes’s rule, there are exactly two positive eigenvalues.
Thus the signature of F3is (2, 1) and the rank is 3.
