Nonlinear Control Sy.. - Free

2.4. MATRICES 43
(i) A is positive definite if and only if all of its (real) eigenvalues are positive.
(ii) A is positive semi definite if and only if \, > 0, Vi = 1, 2, n.
(iii) A is negative definite if and only if all of its eigenvalues are negative.
(iv) A is negative semi definite if and only if \, < 0, Vi = 1, 2,- n.
(v) Indefinite if and only if it has positive and negative eigenvalues.
The following theorem will be useful in later sections.
Theorem 2.5 (Rayleigh Inequality) Consider a nonsingular symmetric matrix Q E Rnxn,
and let Aman and A,,,ay be respectively the minimum and maximum eigenvalues of Q. Under
these conditions, for any x E IItn,
amin(Q)IIXI12 < xTQx < am.(Q)11_112. (2.11)
Proof: The matrix Q, which is symmetric, is diagonalizable and it must have a full set of
linearly independent eigenvectors. Moreover, we can always assume that the eigenvectors
u1i U2, , un associated with the eigenvalues ,11, a2, . , ,1n are orthonormal. Given that
the set of eigenvectors Jul, u2i , un} form a basis in Rn, every vector x can be written as
a linear combination of the elements of this set. Consider an arbitrary vector x. We can
assume that I xji = 1 (if this is not the case, divide by 11x11). We can write
for some scalars x1i x2, , xn. Thus
which implies that
since
x = x1u1 +x2U2 +"' +xnun
xT Qx = XT Q [XI U1 +... + xnun]
= xT [Alxlu1 + ... + Anxnun]
= AixT x1u1 + ....+.nxT xnun
= Ai x112+...+Anlxn12
Amin(Q) < xTQx < A.(Q)
n
1x112=1.
(2.12)
a=1
Finally, it is worth nothing that (2.12) is a special case of (2.11) when the norm of x is 1.