Download The Minimax Theorem

A PROOF OF THE MINIMAX THEOREM
ANTHONY MENDES
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A probability vector is a vector x = x1 , · · · , xn
∈ Rm such that xi ≥ 0 and x1 + · · · + xm = 1. Let
1 denote the vector of all 1’s and 0 denote is the zero vector (the dimensions of these vectors will be clear
from context). Then x is a probability vector if
x≥0
x> 1 = 1> x = 1
and
where the relation x ≥ y for vectors means that every entry in x is greater than or equal to the corresponding
entry in y.
Definition 1. Let A be an m × n matrix. Let v be the maximum number for which there is a probability
vector x such that
x> A ≥ v1> .
The number v is the row value and a vector x which satisfies this inequality is an optimal row strategy.
Similarly, let w be the minimum number for which there is a probability vector y such that
Ay ≤ w1.
The number w is the column value and a vector y which satisfies this inequality is an optimal column
strategy.
Theorem 1. There is a row value, column value, optimal row strategy, and optimal column strategy for
every matrix A.
Proof. Let C = { Ay : y is a probability vector}. Since the matrix multiplication Ay is a convex combination
of the columns of A, the set C is the convex hull formed by the columns of A. Among other things, this
implies C is a compact subset of Rm .
The function f : C → R which selects the maximum component of a vector c ∈ C is a continuous
function on a compact set and therefore attains its minimum. This minimum is the column value and a
probability vector y for which f ( Ay) = w is an optimal column strategy. The existence proof for the row
value and optimal row strategy follows similarly.
Lemma 1. Let v be the row value and w be the column value of a matrix A. Then v ≤ w.
Proof. If x is an optimal row strategy and y an optimal column strategy, then
v = v1> y ≤ x> Ay ≤ x> w1 = w.
Lemma 2. Let v be the row value and w be the column value of an m × n matrix A. Let 1 denote the m × n
matrix with every entry equal to the number 1. Then the row value for ( A + k1) is v + k and the column
value of ( A + k1) is equal to w + k.
Proof. Since x> ( A + k1) = x> A + k1> , the entries in x> A and x> ( A + k1) only differ by the constant k.
The statement about the rows follows because maximizing the minimum entry in one of these vectors is the
same as maximizing the minimum entry in another. The statement about the columns follows similarly. After proving the next theorem in 1928, John von Neumann was quoted as saying “As far as I can see,
there could be no theory of games without that theorem. I thought there was nothing worth publishing
until the Minimax Theorem was proved.”
Theorem 2 (The Minimax Theorem). The row value v and the column value w are equal for every m × n
matrix A.
1
2
ANTHONY MENDES
Proof. Without loss of generality, assume v = 0. Indeed, if v 6= 0, Lemma 2 allows us to consider the matrix
A − v1 instead of A. Our goal is to show that w = 0. Lemma 1 tells us 0 ≤ w, so we will assume 0 < w and
look for a contradiction.
Let C = { Ay : y is a probability vector}. Just as in the proof of Theorem 1, C is a convex compact subset
of Rm . For any c ∈ C, let Pc be the m × m diagonal matrix with diagonal entry equal to 1 if the ith coordinate
of c is positive and 0 if not. Then Pc c is the vector c with all negative entries replaced with 0’s. Simple
observations about this matrix include Pc > = Pc and k Pc xk2 ≤ kxk2 for all c ∈ C and x ∈ Rm . Furthermore,
each c ∈ C must have at least one positive coordinate because otherwise we would have c ≤ 0, implying
w ≤ 0. This means that Pc c 6= 0 for all c ∈ C.
Let f : C → R be the function defined by f (c) = k Pc ck2 . This is a continuous function on a compact set
>
and therefore attains is minimum. Let a = a1 , . . . , am ∈ C minimize f (a). This means that for all c ∈ C,
0 < k Pa ak2 ≤ k Pc ck2 .
(1)
>
For any c = c1 , . . . , cm ∈ C, select λ ∈ (0, 1) close enough to 1 such that
1. If ai > 0, then λai + (1 − λ)ci > 0 for all i,
2. If ai < 0, then
λai + (1 − λ)ci < 0 for all i, and
3. 0 < 1 − λ2 k Pa ak2 − (1 − λ)2 kck2 .
It is easier to see that such a choice for λ is possible when this last condition is written as
(1 − λ) kck2 < (1 + λ) k Pa ak2 ;
by taking λ close to 1, we can make the left hand side of the equation a number close to 0.
The first two conditions on λ ensures that the ith diagonal entry of Pa and Pλa+(1−λ)c can only differ if
ai = 0. Therefore Pa a = Pλa+(1−λ)c a and Pa Pλa+(1−λ)c = Pa .
Using equation (1),
2
k Pa ak2 ≤ Pλa+(1−λ)c (λa + (1 − λ)c)
> = λPa a + (1 − λ) Pλa+(1−λ)c c
λPa a + (1 − λ) Pλa+(1−λ)c c
2
= λ2 k Pa ak2 + 2λ(1 − λ)a> Pa > Pλa+(1−λ)c c + (1 − λ)2 Pλa+(1−λ)c c
≤ λ2 k Pa ak2 + 2λ(1 − λ)a> Pa c + (1 − λ)2 kck2
Rewriting this, we find
1 − λ2 k Pa ak2 − (1 − λ)2 kck2 ≤ 2λ(1 − λ)a> Pa c.
Comparing this with our last condition when choosing λ, we find a> Pa c > 0.
Define x ∈ Rm to be the vector
Pa a
x= >
.
1 Pa a
By construction, x is a probability vector and x> c > 0 for any c ∈ C. This means that x> Ay > 0 for all
probability vectors y. Therefore we have found a probability vector x for which x> A > 0. This, however,
tells us that the row value v is greater than 0. We have found our contradiction.
M ATHEMATICS D EPARTMENT, C ALIFORNIA P OLYTECHNIC S TATE U NIVERSITY, S AN L UIS O BISPO , C ALIFORNIA 93407.
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