Chapter 1 LINEAR COMPLEMENTARITY PROBLEM, ITS ...

22 Chapter 1. Linear Complementarity Problem, Its Geometry, and Applications
D is already symmetric, and all its diagonal elements are positive. The rst step of
the algorithm requires performing the operation: (row 3){2(row 1) on D. This leads
to
8 1
0
D1 =
>: 0
0
0
2
4
0
2
4
0
5
9 0
0
5>
3
:
Since the third diagonal element in D1 is not strictly positive, D is not PD.
Algorithm for Testing Positive Semide niteness
Let F = (fij) be the given square matrix. Obtain D = F + F T . If any diagonal
element of D is 0, all the entries in the row and column of the zero diagonal entry
must be zero. Otherwise D (and hence F ) is not PSD and we terminate. Also, if any
diagonal entries in D are negative, D cannot be PSD and we terminate. If termination
has not occurred, reduce the matrix D by striking o the rows and columns of zero
diagonal entries.
Start o by performing the row operations as in (ii) above, that is, transform D
into D1. If any diagonal element in D1 is negative, D is not PSD. Let E1 be the
submatrix of D1 obtained by striking o the rst row and column of D1. Also, if a
diagonal element in E1 is zero, all entries in its row and column in E1 must be zero.
Otherwise D is not PSD. Terminate. Continue if termination does not occur.
In general, after r steps we willhave a matrix Dr as in (iii) above. Let Er be the
square submatrix of Dr obtained by striking o the rst r rows and columns of Dr.
If any diagonal element in Er is negative, D cannot be PSD. If any diagonal element
of Er is zero, all the entries in its row and column in Er must be zero otherwise D is
not PSD. Terminate. If termination does not occur, continue.
Let dss be the rst nonzero (and, hence, positive) diagonal elementinEr. Subtract
suitable multiples of row s in Dr from rows i, i>s, so that all the entries in column
s and rows i, i > s in Dr, are transformed into 0. This transforms Dr into Ds and
we repeat the same operations with Ds. If termination does not occur until Dn;1 is
obtained and, if the diagonal entries in Dn;1 are nonnegative, D and hence F are
PSD.
In the process of obtaining Dn;1, if all the diagonal elements in all the matrices
obtained during the algorithm are strictly positive, D and hence F is not only PSD
but actually PD.
Example 1.5
Is the matrix
2 0
6 2
6 F = 6 3
4 4
;2
3
3
0
;3
3
3
0
;4
0
0
8
3 5
0 7 07
5 4
;5 0 0 4 2
PSD? D = F + F T =
2
6
4
3
0
0
0
0
0
6
6
0
0
6
6
0
0
0
0
16
0
0 7 07
5 8
0 0 0 8 4
: