Source document: https://docs-lawep.s3.us-east-2.amazonaws.com/1697742502851.pdf (Georgia)
Reading [SB], Ch.

16.1-16.3, p. 375-393 1 Quadratic Forms A quadratic function f : R → R has the form f (x) = a · x2.

Generalization of this notion to two variables is the quadratic form Q(x1, x2) = a11x2 1 + a12x1x2 + a21x2x1 + a22x2 2.

Here each term has degree 2 (the sum of exponents is 2 for all summands).

A quadratic form of three variables looks as f (x1, x2, x3) = a11x2 1 + a12x1x2 + a13x1x3+ a21x2x1 + a22x2 2 + a23x2x3+ a31x1x3 + a32x3x2 + a33x2 3.
A general quadratic form of n variables is a real-valued function Q : Rn → R of the form Q(x1, x2, ..., xn) = a11x2 1 + a12x1x2 + ... + a1nx1xn+ 2 + ... + a2nx2xn+ a21x2x1 + a22x2 ... ... an1xnx1 + an2xnx2 + ... + annx2 n ... ...

In short Q(x1, x2, ..., xn) = n i,j aijxixj.

As we see a quadratic form is determined by the matrix    A =    .
a11 ... a1n .............. an1 ... ann 1.1 Matrix Representation of Quadratic Forms Let Q(x1, x2, ..., xn) = to see that n i,j aijxixj be a quadratic form with matrix A.

Easy Q(x1, ..., xn) = (x1, ..., xn) ·    a11 ... a1n .............. an1 ... ann     ·      .

x1 .. xn 1 Equivalently Q(x) = xT · A · x.

Example.
Example.

The quadratic form Q(x1, x2, x3) = 5x2 1 − 10x1x2 + x2 2 whose symmetric matrix is A = is the product of three matrices 5 −5 1 −5 (x1, x2, x3) · 5 −5 1 −5    ·    .

x1 x2 x3 1.1.1 Symmetrization of matrix The quadratic form Q(x1, x2, x3) = 5x2 example, by the following 2 × 2 matrices 1 − 10x1x2 + x2 2 can be represented, for 5 −2 1 −8 , 5 −3 1 −7 , 5 −5 1 −5 the last one is symmetric: aij = aji.
Theorem 1 Any quadratic form can be represented by symmetric matrix.

Indeed, if aij = aji we replace them by new a not change the corresponding quadratic form.

ij = a ji = aij +aji 2 , this does Generally, one can ﬁnd symmetrization A of a matrix A by A = A+AT .

2 1.2 Deﬁniteness of Quadratic Forms A quadratic form of one variable is just a quadratic function Q(x) = a · x2.

If a > 0 then Q(x) > 0 for each nonzero x.

If a < 0 then Q(x) < 0 for each nonzero x.


So the sign of the coeﬃcient a determines the sign of one variable quadratic form.

The notion of deﬁniteness described bellow generalizes this phenomenon for multivariable quadratic forms.

1.2.1 Generic Examples The quadratic form Q(x, y) = x2 + y2 is positive for all nonzero (that is (x, y) = (0, 0)) arguments (x, y).

Such forms are called positive deﬁnite.

The quadratic form Q(x, y) = −x2 − y2 is negative for all nonzero argu- ments (x, y).
Such forms are called negative deﬁnite.

The quadratic form Q(x, y) = (x − y)2 is nonnegative.

This means that Q(x, y) = (x − y)2 is either positive or zero for nonzero arguments.

Such forms are called positive semideﬁnite.

2 The quadratic form Q(x, y) = −(x − y)2 is nonpositive.

This means that Q(x, y) = (x − y)2 is either negative or zero for nonzero arguments.

Such forms are called negative semideﬁnite.


The quadratic form Q(x, y) = x2 − y2 is called indeﬁnite since it can take both positive and negative values, for example Q(3, 1) = 9 − 1 = 8 > 0, Q(1, 3) = 1 − 9 = −8 < 0.

1.2.2 Deﬁniteness Deﬁnition.


A quadratic form Q(x) = xT · A · x (equivalently a symmetric matrix A) is (a) positive deﬁnite if Q(x) > 0 for all x = 0 ∈ Rn; (b) positive semideﬁnite if Q(x) ≥ 0 for all x = 0 ∈ Rn; (c) negative deﬁnite if Q(x) < 0 for all x = 0 ∈ Rn; (d) negative semideﬁnite if Q(x) ≤ 0 for all x = 0 ∈ Rn; (e) indeﬁnite if Q(x) > 0 for some x and Q(x) < 0 for some other x.
1.2.3 Deﬁniteness and Optimality Determining the deﬁniteness of quadratic form Q is equivalent to determining wether x = 0 is max, min or neither.

Particularly: If Q is positive deﬁnite then x = 0 is global maximum; If Q is negative deﬁnite then x = 0 is global minimum.

1.2.4 Deﬁniteness of 2 Variable Quadratic Form Let Q(x1, x2) = ax2 1 + 2bx1x2 + cx2 2 = (x1, x2) · a b c b · x1 x2 be a 2 variable quadratic form.
Here A = a b c b a b c b Easy to see that determinant is the symmetric matrix of the quadratic form.

The = ac − b2 is called discriminant of Q. ax2 1 + 2bx1x2 + cx2 2 = a(x1 + b a x2)2 + ac − b2 a x2 2.

Let us use the notation D1 = a, D2 = ac − b2.

Actually D1 and D2 are leading principal minors of A.
Note that there exists one more principal (non leading) minor (of degree 1) D 1 = c. Then Q(x1, x2) = D1(x1 + b a x2)2 + D2 D1 x2 2.

From this expression we obtain: 3 1.

If D1 > 0 and D2 > 0 then the form is of x2 + y2 type, so it is positive deﬁnite; 2.

If D1 < 0 and D2 > 0 then the form is of −x2 − y2 type, so it is negative deﬁnite; 3.
If D1 > 0 and D2 < 0 then the form is of x2 − y2 type, so it is indeﬁnite; If D1 < 0 and D2 < 0 then the form is of −x2 + y2 type, so it is also indeﬁnite; Thus if D2 < 0 then the form is indeﬁnite.

Semideﬁniteness depends not only on leading principal minors D1, D2 but also on all principal minors, in this case on D 1 = c too.

4.

If D1 ≥ 0, D 1 ≥ 0 and D2 ≥ 0 then the form is positive semideﬁnite.
1 ≥ 0 here is absolutely necessary: consider the form Q(x1, x2) = −x2 Note that only D1 ≥ 0 and D2 ≥ 0 is not enough, the additional condition D 2 with a = 0, b = 0, c = −1, here D1 = a ≥ 0, D2 = ac − b2 ≥ 0, nevertheless the form is not positive semidieﬁnite.

5.

If D1 ≤ 0, D 1 ≤ 0 and D2 ≥ 0 then the form is negative semideﬁnite.
1 ≤ 0 again is absolutely necessary: consider the form Q(x1, x2) = x2 Note that only D1 ≤ 0 and D2 ≥ 0 is not enough,the additional condition D 2 with a = 0, b = 0, c = 1, here D1 = a ≤ 0, D2 = ac − b2 ≥ 0, nevertheless the form is not negative semidieﬁnite.

1.2.5 Deﬁniteness of 3 Variable Quadratic Form Let us start with the following Example.

Q(x1, x2, x3) = x2 matrix of this quadratic form is 1 + 2x2 2 − 7x2 3 − 4x1x2 + 8x1x3.

The symmetric    4 1 −2 2 0 0 −7 −2 4    .


The leading principal minors of this matrix are |D1| = 1 = 1, |D2| = 1 −2 2 −2 = −2, |D3| = 1 −2 4 0 2 0 −7 −2 4 = −18.
4 Now look: 1 − 4x1(x2 − 2x3) + 2x2 2 − 7x2 2 − 7x2 3 = 3 = 3 = x2 1 + 2x2 2 − 7x2 3 − 4x1x2 + 8x1x3 = Q(x1, x2, x3) = x2 1 − 4x1x2 + 8x1x3 + 2x2 x2 2 − 7x2 1 − 4x1(x2 − 2x3) + 4(x2 − 2x3) − 4(x2 − 2x3)] + 2x2 [x2 [x1 − 2x2 + 4x3]2 − 2x2 2 − 16x2x3 − 23x2 [x1 − 2x2 + 4x3]2 − 2(x2 2 − 8x2x3) − 23x2 2 − 8x2x3 + 16x2 [x1 − 2x2 + 4x3]2 − 2[x2 [x1 − 2x2 + 4x3]2 − 2[x2 − 4x3]2 − 16x2 [x1 − 2x2 + 4x3]2 − 2[x2 − 4x3]2 + 32x2 [x1 − 2x2 + 4x3]2 − 2[x2 − 4x3]2 + 9x2 l2 |D1|l2 3, 3 = 3 = 3 − 16x2 3) − 23x2 3 − 23x2 3 = 3] − 23x2 3 = 3 = l2 + D3 D2 1 + D2 D1 3 = where l1 = x1 −2x2 +4x3, x2 −4x3, l2 = x3.
l3 = That is (l1, l2, l3) are linear combinations of (x1, x2, x3).

More precisely where    l1 l2 l3     =   1 −2 0 0 4 1 −4 1 0     ·      x1 x2 x3    P =    1 −2 0 0 4 1 −4 1 0 is a nonsingular matrix (changing variables).

Now turn to general 3 variable quadratic form Q(x1, x2, x3) = (x1, x2, x3) · The following three determinants    a11 a12 a13 a21 a22 a23 a31 a32 a33     ·      .
x1 x2 x3 |D1| = a11 , |D2| = a11 a12 a21 a22 , |D3| = a11 a12 a13 a21 a22 a23 a31 a32 a33 are leading principal minors.
It is possible to show that, as in 2 variable case, if |D1| = 0, |D2| = 0, then Q(x1, x2, x3) = |D1|l2 1 + |D2| |D1| l2 2 + |D3| |D2| l2 3 5 where l1, l2, l3 are some linear combinations of x1, x2, x3 (this is called La- grange’s reduction).

This implies the following criteria: 1.

The form is positive deﬁnite iﬀ |D1| > 0, |D2| > 0, |D3| > 0, that is all principal minors are positive.

2.
2.

The form is negative deﬁnite iﬀ |D1| < 0, |D2| > 0, |D3| < 0, that is principal minors alternate in sign starting with negative one.

Example.

Determine the deﬁniteness of the form Q(x1, x2, x3) = 3x2 3x2 Solution.

The matrix of our form is 3 − 2x1x2 − 2x2x3.

1 +2x2 2 +    3 −1 −1 0 −1 0 2 −1 3    .


The leading principal minors are |D1| = 3 > 0, |D2| = 3 −1 2 −1 > 5, |D3| = 3 −1 −1 0 −1 0 2 −1 3 = 18 > 0, thus the form is positive deﬁnite.
1.2.6 Deﬁniteness of n Variable Quadratic Form Let Q(x1, ..., xn) = (x1, ..., xn) ·      a11 a21 ... an1 ... a1n ... a2n ... ... ... ann         · x1 ... xn    be an n variable quadratic form.
The following n determinants |D1| = a11 , |D2| = a11 a12 a21 a22 , ... , |Dn| = a11 a21 ... an1 ... a1n ... a2n ... ... ... ann are leading principal minors.
As in previous cases, it is possible to show that Q(x1, ... , x3) = |D1|l2 1 + |D2| |D1| l2 2 + ... + |Dn| |Dn−1| l2 n where (l1, l2, ... , ln) are linear combinations of (x1, x2, ... , xn), more precisely    l1 ... ln     =   p11 .

pn1 ... p1n .

.

... pnn     ·      x1 ... xn 6 where    P =    p11 .

pn1 ... p1n .

.

... pnn is a nonsingular matrix (changing variables).

Theorem 2 1.
Theorem 2 1.

A quadratic form is positive deﬁnite if and only if |D1| > 0, |D2| > 0, ... , |Dn| > 0, that is all principal minors are positive; 2.

A quadratic form is negative deﬁnite if and only if |D1| < 0, |D2| > 0, |D3| < 0, |D4| > 0, ... , that is principal minors alternate in sign starting with negative one.

3.

If some kth order leading principal minor is nonzero but does not ﬁt either of the above two sign patterns, then the form is indeﬁnite.
The situation with semideﬁniteness is more complicated, here are involved not only leading principal minors, but all principal minors.

Theorem 3 1.

A quadratic form is positive semideﬁnite if and only if all principal minors are ≥ 0; 2.

A quadratic form is negative semideﬁnite if and only if all principal minors of odd degree are ≤ 0, and all principal minors of even degree are ≥ 0.

1.3 Deﬁniteness and Eigenvalues As we know a symmetric n × n matrix has n real eigenvalues (maybe some multiple).
Theorem 4 Given a quadratic form Q(x) = xT Ax and let λ1, eigenvalues of A.

Then Q(x) is ... , λn be • positive deﬁnite iﬀ λi > 0, i = 1, ... , n; • negative deﬁnite iﬀ λi < 0, i = 1, ... , n; • positive semideﬁnite iﬀ λi ≥ 0, i = 1, ... , n; • negative semideﬁnite iﬀ λi ≤ 0, i = 1, ... , n; Proof.

Just xT Ax > 0 ⇒ ∀ λi > 0.

Let v be the normalized eigenvector of λi, that is Av = λiv.

Then 0 < vT Av = λivT v = λi.


7 1.4 Linear Constraints and Bordered Matrices 1.4.1 Two Variable Case The quadratic form Q(x1, x2) = x2 1 − x2 2 is indeﬁnite (why?).

But if we restrict Q to the subset (subspace) of R2 determined by the constraint x2 = 0 we obtain a one variable quadratic form q(x) = Q(x, 0) = x2 which is deﬁnitely positive deﬁnite.

Thus the restriction of Q on x1 axis is positive deﬁnite.

Similarly, the constraint x1 = 0 gives q(x) = Q(0, x) = −x2 which is negative deﬁnite.
Thus the restriction of Q on x2 axis is positive deﬁnite.

Now let us consider the constraint x1 − 2x2 = 0.

Solving x1 from this constraint we obtain x1 = 2x2.

Substituting in Q we obtain one variable quadratic form q(x) = Q(2x, x) = 4x2 − x2 = 3x2 which is positive deﬁnite.

Thus the restriction of Q on the line x1 + 2x2 = 0 is positive deﬁnite.
Let us repeat the last calculations for a general 2 variable quadratic form Q(x1, x2) = ax2 1 + 2bx1x2 + cx2 2 = (x1, x2) a b c b x1 x2 subject to the linear constraint Ax1 + Bx2 = 0.

Let us, as above, solve x1 from the constraint x1 = − B A x2 and substitute in Q: Q(x1, x2) = Q(− B A x2, x2) = a11(− B A x2)2 + 2a12(− B A x2)x2 + a22x2 2 = = aB2−2bAB+cA2 A2 x2 2.
So the deﬁniteness of Q(x1, x2) on the constraint set Ax1 + Bx2 = 0 depends on the sign of coeﬃcient aB2−2bAB+cA2 , more precisely on the nominator A2 aB2 − 2bAB + cA2, which is nothing else than It is easy to see that   −det   0 A B b A a c B b   .

8 This matrix    0 A B b A a c B b    is called bordered matrix of A.

Thus we have proved the 1 + 2a12x1x2 + 2 restricted on the constrained set Ax1 + Bx2 = 0 is positive (resp.
neg- Theorem 5 A two variable quadratic form Q(x1, x2) = a11x2 a22x2 ative) if and only if the determinant of the bordering matrix    det    0 A B b A a c B b is negative (resp.

positive).

1.4.2 n Variable m Constraint Case Analogous result holds in general case.

Let and x =    x1 ... xn     , A =      a11 ... an1 .. a1n ... ann Q(x1, ... , xn) = xT Ax be an n variable quadratic form.
Consider a constraint set    B11 ... Bm1 ... B1n ... Bmn       x1 ... xn     =      .

0 ... 0 In fact this set is the null space of B, is not it?

The bordered matrix for this situation looks as   H =             0 0 ... ... 0 ... 0 − − − ... Bm1 B11 ... ... Bmn B1n | B11 | | Bm1 ... B1n ... ... Bmn − − − a1n ... a11 ... ... ann an1 | | | = .
0 B BT A             This (m + n) × (m + n) matrix has m + n leading principal minors M1, M2, ... , Mm, Mm+1, ... , M2m−1, M2m, M2m+1, ... , Mm+n = H. 9 The ﬁrst m matrices M1, ... , Mm are zero matrices.

Next m − 1 matrices Mm+1, ... , M2m−1 have zero determinant.

The determinant of the next minor M2m is ±det B2 where B is the left m × m minor of B, it does not carry any information about A.
And only the determinants of last n−m matrices M2m+1, ... , Mm+n carry information about the matrix A, i.e.

about the quadratic form Q.

Exactly these minors are essential for constraint deﬁniteness.

Theorem 6 (i) If the determinant of H = Mm+n has the sign (−1)n and the signs of determinants of last m + n leading principal minors M2m+1, ... , Mm+n alternate in sign, then Q is negative deﬁnite on the constraint set Bx = 0, so x = 0 is a strict global max of Q on the constraint set Bx = 0.
(ii) If the determinants of all last m + n leading principal minors M2m+1, ... , Mm+n have the same sign (−1)m, then Q is positive deﬁnite on the constraint set Bx = 0, so x = 0 is a strict global min of Q on the constraint set Bx = 0.

(iii) If both conditions (i) and (ii) are violated by some nonzero minors from last m + n leading principal minors M2m+1, ... , Mm+n then Q is indeﬁnite on the constraint set Bx = 0, so x = 0 is neither max nor min of Q on the constraint set Bx = 0.
This table describes the above sign patterns: negative (−1)m+1 (−1)m positive Mm+m+1 Mm+m+2 (−1)m+2 (−1)m ... Mm+n−1 Mm+n (−1)n ... (−1)m ... (−1)n−1 (−1)m Example Determine the deﬁniteness of the following constrained quadratics 2 − x2 Q(x1, x2, x3) = x2 subject to x1 + x2 + x3 = 0.

1 + x2 3 + 4x1x3 − 2x1x2, 10 Solution.

Here n = 3, m = 1.

The bordered matrix here is         | 1 0 1 1 − − − − − 2 1 1 −1 1 1 0 0 −1 1 | | −1 2 |         .
The leading principal minors are |M1| = 0, |M2| = |M3| = 0 1 1 −1 1 1 1 −1 1 = −4 < 0, |M4| = = −1, 0 1 1 1 0 1 1 −1 2 1 1 1 1 1 −1 2 0 1 0 −1 = 16 > 0. in this case n = 3, m = 1, so the essential minors are M3 and M4.
For negative deﬁniteness the sign pattern must be (−1)n−1 = (−1)2 = ” + ”, (−1)n = (−1)3 = ” − ” And for positive deﬁniteness the sign pattern must be (−1)m = (−1)1 = ” − ”, (−1)m = (−1)1 = ” − ”, since we have −4 < 0, 16 > 0, which diﬀers from both patterns, our con- strained quadratic is indeﬁnite.

1.5 Change of variables* Let Q = xT Ax be a quadratic form of variable x ∈ Rn.

Let us step to new variable y ∈ Rn which is connected to x by x = P y where P is some nonsingular matrix .
Note that in this case xT = (P y)T = yT P T .

Then Q = xT Ax = yT P T AP y = yT (P T AP )y, so the matrix of new quadratic form of variable y is B = P T AP .

If A is symmetric, B = P T AP is symmetric too (prove it using the deﬁnition of symmetric matrix A = AT ).

Jacobi’s Theorem states that any symmetric matrix A can be trans- formed to a diagonal matrix Λ = P T AP by an orthogonal matrix P .

The elements of Λ are uniquely determined up to permutation.
If we allow P to be a nonsingular matrix, then A can be transformed to a diagonal matrix where each diagonal element is 1, -1 or 0.
11 Important remark: Let D1, D2, ... Dn be the leading principal minors of A, and let Λ be corresponding diagonal matrix with 1,0,-1 on the diagonal, then the Λ-s leading principal minors have the same sign pattern as Di-s. Silvester’s Law of inertia states that the number of 0-s n0, the number of 1-s n+ and the number of -1-s n− are invariant in the sense that any other diagonalization gives the same n0, n+ and n−.
The signature of a quadratic form is deﬁned as n+ − n−, so the signature is an invariant of quadratic form too.

A quadratic form Q is positive deﬁnite if n+ = n. A quadratic form Q is negative deﬁnite if n− = n. A quadratic form Q is positive semi deﬁnite if n− = 0.

A quadratic form Q is negative semi deﬁnite if n+ = 0.

A quadratic form Q is indeﬁnite if n+ > 0 and n− > 0.

1.6 Why Quadratic Forms?


* Recall Taylor formula f (x0 + h) = f (x0) + f (x0) · h + f (x0) 2 · h2 + f (x0) 3!

· h3 + ... and ask a question: is this function increasing at x = x0?

That is if h > 0 is f (x0 + h) − f (x0) = f (x0) · h + f (x0) 2 · h2 + f (x0) 3!

· h3 + ... positive?


Notice, that for small enough h the quadratic term f (x0) · h2, the cubical term f (x0) · h3 etc are smaller that the linear term f (x0) · h, so the positivity of f (x0 + h) − f (x0) depends on positivity of the coeﬃcient f (x0) of that linear form f (x0) · h, that is on the positivity of the derivative at x0.

3!

2 Well, from this easily follows, that a local extremum we can expect at a critical point f (x0) = 0.
Now ask a question: how to recognize, is a critical point x0 min or max?

Address again to Taylor formula which now, since f (x0) = 0, looks as f (x0 + h) = f (x0) + f (x0) 2 · h2 + f (x0) 3!

· h3 + ... .

Consider the diﬀerence f (x0 + h) − f (x0) = f (x0) 2 · h2 + f (x0) 3!

· h3 + ... .

12 The point x0 is min if this diﬀerence f (x0 + h) − f (x0) is positive for all small enough values of h. f (4)(x0) 4!
For small enough h the cubical term f (x0) · h4 etc are smaller that the quadratic term f (x0) · h3, the term of degree 4 · h2, so the positivity of the diﬀerence f (x0 + h) − f (x0) depends on positivity of the coeﬃcient f (x0) · h2, that is on the positivity of the second 2 of that quadratic form f (x0) 3!

2 2 derivative at x0.


So, at h = 0 this form has minimum (it is positive deﬁnite) if its coeﬃcient f (x0) is positive (our good old second order condition).

Now consider a function of two variables F (x1, x2).

Again there exists Taylor for two variables F (x1 + h1, x2 + h2) = F (x1, x2) + ∂F ∂x1 1 (x1, x2) · h2 2 1 (x1, x2) · h3 3 (x1, x2) · h1 + ∂F ∂x2 1 + ∂2F ∂x1∂x2 .
1 + ... ∂2F ∂x2 1 ∂3F ∂x3 1 (x1, x2) · h1h2 + 1 2 (x1, x2) · h2+ ∂2F ∂x2 2 (x1, x2) · h2 2+ As in one-variable case, an optimum (min or max) can be expected at a (x1, x2) · h2 vanishes.

critical point, where the linear form ∂F ∂x1 And the minimality or maximality depends on the quadratic form (x1, x2) · h1 + ∂F ∂x2 Q(h1, h2) = 1 2 ∂2F ∂x2 1 (x1, x2) · h2 1 + ∂2F ∂x1∂x2 (x1, x2) · h1h2 + 1 2 ∂2F ∂x2 2 (x1, x2) · h2 2.
Namely, if this form is positive deﬁnite, that is if Q(h1, h2) > 0 for all (h1, h2) = (0, 0), then (x1, x2) is a point of minimum, and if the form is negative deﬁnite, that is Q(h1, h2) < 0 for all (h1, h2) = (0, 0), then (x1, x2) is a point of maximum.

13 Exercises 1.

Write the following quadratic forms in matrix form (a) Q(x1, x2) = x2 (b) Q(x1, x2) = 5x2 (c) Q(x1, x2, x3) = x2 1 − 2x1x2 + x2 2.

1 − 10x1x2 + x2 2.

2 + 3x2 3 + 4x1x2 − 6x1x3 + 8x2x3.

1 + 2x2 2.


By direct matrix multiplication express each of the following matrix products as a quadratic form (a) x1 x2 · 4 2 2 3 x1 x2 , (b) u v · 5 2 4 0 · .

u v · 3.
u v · 3.

Write the quadratic form corresponding to the matrix A and then ﬁnd a symmetric matrix which determines the same quadratic form for (a) A = 2 −1 1 −2 , (b) A =    1 2 0 3 4 5 0 7 6    (c) A =           .

1 0 3 0 0 2 0 5 5 0 4 0 0 1 0 6 4.

Write an example of 2 variable quadratic form Q(x1, x2) which is (a) positive deﬁnite.

(b) negative deﬁnite.

(c) positive semideﬁnite.

(d) negative semideﬁnite.

(e) indeﬁnite.

5.


Write an example of 3 variable quadratic form Q(x1, x2, x3) which is (a) positive deﬁnite.

(b) negative deﬁnite.

(c) positive semideﬁnite.

(d) negative semideﬁnite.

(e) indeﬁnite.

6.


Determine deﬁniteness of the following symmetric matrices (quadratic forms) (a)    (e) , (b)     (f )   2 −1 1 −1 1 2 0 2 4 5 0 5 6 , (c) −3 4 4 −5 , −3 4 4 −6  −1 1 1 −1 0 0 0 0 −2    (g)     1 0 3 0 0 2 0 5 3 0 4 0 0 5 0 6 (d)      .

 , 2 4 4 8 7.
, 2 4 4 8 7.

Formulate criteria for positive and negative deﬁniteness of the quadratic form given by a diagonal matrix 14         a1 0 ... 0 0 0 a2 ... 0 0 0 ... 0 ... ... ... ... an−1 ... 0         .

0 0 ... 0 an 8.

Determine the deﬁniteness of the following constrained quadratics 1 + x2 2, subject to x1 + x2 = 0.

2, subject to x1 + x2 = 0.


3 + 4x1x3 − 2x1x2, subject to 1 + 2x1x2 − x2 1 + 2x1x2 − x2 2 − x2 (a) Q(x1, x2) = x2 (b) Q(x1, x2) = 4x2 (c) Q(x1, x2, x3) = x2 x1 + x2 + x3 = 0.

(d) Q(x1, x2, x3) = x2 x1 + x2 + x3 = 0.

(e) Q(x1, x2, x3) = x2 x1 + x2 − x3 = 0.

(f ) Q(x1, x2, x3, x4) = x2 1 − x2 subject to x1 + x2 − x3 + x4 = 0. x1 − 9x2 + x4 = 0.

3 + 4x1x2 − 6x2x3, subject to 3 + 4x1x3 − 2x1x2, subject to 4 + 4x2x3 − 2x1x4, 1 − x2 3 + x2 1 + x2 2 + x2 2 + x2 Homework 5, 6(g), 7, 8(e), 8(f).
Essay (optional, please type) 1.

Lagrangian reduction.

If all leading minors |Dk| of the symmetric matrix of a quadratic form are nonzero then Q(x1, ... , x3) = |D1|l2 1 + |D2| |D1| l2 2 + ... + |Dn| |Dn−1| l2 n where l1, ... , ln are some linear combinations of x1, ... , xn We have shown it for n = 2 in general case and for some particular example for n = 3 (see above).

(a) Give a general proof for n = 3.

(b) Deduce the criterion for deﬁniteness from this reduction.

2.
2.

Linear restriction of a quadratic form.

We have actually proved the criterion for deﬁniteness of restricted quadratic form in terms of bordered matrix for n = 2 and m = 1 (see above).

Do the same for n = 3 and m = 1.

1 + 2a12x1x2 + a22x2 3.

Deﬁniteness of restricted form.

Show that if a quadratic form Q(x1, x2) = a11x2 2 is positive (negative) deﬁnite then its restriction on a linear constraint set Ax1 + Bx2 = 0 is also positive (nega- tive) deﬁnite.
Hint: Calculate the determinant of bordered matrix.

Or just substitute in Q the value x2 = − A B x1.

But if a form is indeﬁnite then the restricted form can be whatever you wish (give examples).

15 Short Summary Quadratic Forms Q(x1, x2, ..., xn) = n i,j aijxixj = (x1, ..., xn)·    a11 ... a1n .............. an1 ... ann    ·   x1 .. xn    = xT ·A·x A is symmetric.

If not, take its symmetrization A = A+AT 2 .
Deﬁniteness of Q(x): (a) positive deﬁnite if Q(x) > 0 for all x = 0 ∈ Rn; (b) positive semideﬁnite if Q(x) ≥ 0 for all x = 0 ∈ Rn; (c) negative deﬁnite if Q(x) < 0 for all x = 0 ∈ Rn; (d) negative semideﬁnite if Q(x) ≤ 0 for all x = 0 ∈ Rn; (e) indeﬁnite if Q(x) > 0 for some x and Q(x) < 0 for some other x. Deﬁniteness and Optimality If Q is positive deﬁnite then x = 0 is global maximum; If Q is negative deﬁnite then x = 0 is global minimum.
Leading principal minors |D1| = a11 , |D2| = a11 a12 a21 a22 , ... , |Dn| = |A|.

A quadratic form Q(x) is: Positive deﬁnite iﬀ |D1| > 0, Negative deﬁnite iﬀ |D1| < 0, Indeﬁnite iﬀ some nonzero Dk violates above sign patterns.

... |D3| < 0, |D2| > 0, |D2| > 0, , |Dn| > 0.

|D4| > 0, ... .

Positive semideﬁnite iﬀ all principal minors Mk ≥ 0.
Negative semideﬁnite iﬀ all M2k+1 ≤ 0 and M2k ≥ 0.

Deﬁniteness of Q(x) = xT · A · x on the constrained set B · x = 0: negative (−1)m+1 (−1)m positive Mm+m+1 Mm+m+2 (−1)m+2 (−1)m ... Mm+n−1 Mm+n (−1)n ... (−1)m ... (−1)n−1 (−1)m where M2m+1, ..., Mm+n are last n − m minors of the bordered matrix              0 0 ... ... 0 0 ... − − − ... Bm1 B11 ... ... Bmn B1n              .

| B11 | | Bm1 ... B1n ... ... Bmn − − − a1n ... a11 ... ... ann an1 | | | 16