12.4.2-RdctnMtrx

12.4.2 Reductions Among Matrix Problems

NOTATIONS:

MQ is the problem of multiplying arbitrary square matrices.
MT is the problem of multiplying upper triangular square matrices.
MS is the problem of multiplying symmetric matrices.
IT is the problem of inverting non-singular upper triangular matrices.

These problems are all ("under reasonable assumptions") linearly equivalent.

(The problem of inverting an *arbitrary* nonsingular matrix is also essentially linearly equivalent to these problems, but the proof is not given in our text.)

We know from the discussion in section 7.6 that MQ can be done with a O(n^2.81) or even O(n^2.376) algorithm. Because of the linear equivalences stated above, we know there are also O(n^2.376) algorithms for all these other problems. The equivalence also tells us that if we find an algorithm with a smaller big-Θ for any of the problems we can use it to create better algorithms for the other problems too.

THEOREM 12.4.5 [MT] <=^L [MQ] and [MS] <=^L [MQ]

PROOF: [MT] <=^L [MQ] and [MS] <=^L [MQ] are obvious: any algorithm for multiplying *arbitrary* square matrices can be used to multiply two upper triangular square matrices or to multiply two symmetric matrices.

THEOREM: 12.4.6 [MQ] <=^L [MT], assuming MT is smooth.

PROOF: Suppose that triangular square matrices can be multiplied by an algorithm that is O(t(n)), where t(n) is smooth. If A and B are two arbitrary NXN matrices that we wish to multiply, we can construct the two (3N)X(3N) matrices depicted on the left below and form the product shown with our algorithm for triangular matrices.

0  A  0      0  0  0      0  0  AB  
0  0  0  X   0  0  B  =   0  0  0
0  0  0      0  0  0      0  0  0

The work involved will be O(t(3n)), including the "overhead", which is O(n²). Since t is smooth, t(3n) is O(t(n)), and it follows that the transformation indicated provides us with an O(t(n)) algorithm for multiplying square matrices.

THEOREM 12.4.7 [MQ] <=^L [MS], assuming MS is smooth.

PROOF: Similar to proof of Theorem 12.4.6. Use:

0    A       0  B^t        AB    0
         X            =    
A^t   0       B  0         0    A^tB^t

(Here, A^t and B^t are the transposes of A and B, respectively.)

THEOREM 12.4.8 [MQ] <=^L [IT], assuming IT is smooth.

PROOF: Suppose that A and B are two arbitrary NXN matrices that we wish to multiply. Consider the (3N)X(3N) matrices depicted below.

I   A   0       I  -A   AB       I  0  0
0   I   B   X   0   I   -B   =   0  I  0
0   0   I       0   0    I       0  0  I

The equation above makes it clear that the second matrix above is the inverse of the first. The first matrix above is upper triangular and the second matrix above contains AB.

It should be clear now that if we have an O(t(n)) algorithm that can invert non-singular upper triangular matrices, and if t(n) is smooth, then we can also construct AB from A and B with an O(t(n)) algorithm.

LEMMA: Assuming IT2 is smooth, [IT] <=^L [IT2]

PROOF: (IT2 is the problem of inverting a non-singular upper triangular matrix whose size is a power of 2). If we want to invert an arbitrary NXN non-singular upper triangular matrix A, we can find the smallest 2^j such that N <= 2^j, and then form the 2^j by 2^j matrix B:

A  0
0  I

Then, assuming we have an algorithm for IT2 that is O(t(n)) where t(n) is smooth, we can use the algorithm to produce the inverse of B, which is:

 A^-1  0
 0    I

This gives us the inverse of A. Because 2^j is less than 2*N, it's easy to use the smoothness of t(n) to show that the amount of work done is O(t(n)). We invert B with at most t(2^j) work. This is t(2*2^j-1). Since t is smooth, t(2*2^j-1) <= c*t(2^j-1) for some constant c, and sufficiently large n. Since t is eventually non- decreasing (part of being smooth), t(2^j-1) <= t(N). Thus the work to invert B is no more than c*t(N). (The "overhead" is O(N²) and is dominated by t.)

LEMMA 12.4.10 [IT2] <=^L [MQ], assuming MQ is strongly quadratic.

PROOF: The proof is omitted here. See Brassard and Bratley, pp. 436-437. The second half of the proof is a bit involved, but the first half is straightforward and conveys the basic idea of the proof well.

THEOREM 12.4.11 [IT] <=^L [MQ]< assuming MQ is strongly quadratic.

PROOF: This is "almost immediate" from the preceding two results. There is a technical impediment to the proof. The authors give hints and suggest we work out the details. We'll decline this time.