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For ordered bases $\mathcal{B}=(v_1,\dots ,v_n)$ of $V$ and $\mathcal{C}=(w_1,\dots ,w_m)$ of $W$, the matrix representation of a linear map $\gamma :V\to W$ is the matrix whose $j$th column consists of the $\mathcal{C}$-coordinates of $\gamma (v_j)$. Thus

The operators $\alpha ,\beta :V\to V$ are conjugate linear operators when there is an isomorphism $s:V\to V$ such thatIn one basis their matrices therefore satisfy $[\beta ]=[s{]}^{-1}[\alpha ][s]$, so they are similar matrices.

For invertible $\beta$, the mapis linear, and its inverse is $A\mapsto \beta A{\beta }^{-1}$; hence it is a linear isomorphism of $\mathrm{End}(V)$. If ${\beta }^{\prime }=s^{-1}\beta s$, define ${\Phi }_s(A)=s^{-1}As$. A direct substitution givesso ${\varphi }_{{\beta }^{\prime }}$ and ${\varphi }_{\beta }$ are conjugate. This is the conjugation operator on an endomorphism space.

It remains to compute the Jordan normal form over $\mathbb{C}$. By the preceding conjugacy, we may put $\beta$ in Jordan form.

Ifthen the matrix units $E_{ij}$ are eigenvectors becauseThusThis includes the scalar case $\lambda =\mu$, when ${\varphi }_{\beta }$ is the identity.

Otherwise $\beta$ has one size-two Jordan block. Multiplying $\beta$ by a nonzero scalar does not change ${\varphi }_{\beta }$, and conjugating within its Jordan class allows us to use $J=I+N$ with $N=E_{12}$. Put $D={\varphi }_J-I$. Direct multiplication givesHence $D^3=0$, $\mathrm{rank}D=2$, and $\mathrm{rank}{D}^2=1$. The nilpotent Jordan blocks of $D$ therefore have sizes three and one. Adding the identity givesThese two cases are the Jordan normal form of conjugation on two-by-two matrices.

Fix $x\in X$. By assumption there is a neighbourhood $U$ of $x$ on which $f_n\to f$ with uniform convergence. Each restriction $f_n{|}_U$ is continuous, and a uniform limit of continuous functions is continuous. Thus $f{|}_U$ is continuous, in particular at $x$. Since $x$ was arbitrary, $f$ is continuous on $X$.

Now let $K\subseteq X$ be compact and let $\epsilon >0$. For each $x\in K$, choose a neighbourhood $U_x$ on which convergence is uniform. The sets $U_x$ cover $K$, so compactness gives a finite subcoverFor each $j$, choose $N_j$ such thatTaking $N={\max }_j{N}_j$, every $y\in K$ belongs to one of these finitely many neighbourhoods, and thereforeThus $f_n\to f$ uniformly on every compact subset. This proves the local uniform convergence on compact subsets principle.

A bilinear form $B$ on $V$ is nondegenerate whenIn finite dimensions this is equivalent to nondegeneracy in the first argument.

Define the linear mapNondegeneracy makes ${\beta }_1$ injective. Since $V$ and its dual space have the same finite dimension, ${\beta }_1$ is an isomorphism. Similarly define ${\beta }_2(w)=B_2(-,w)$ and setThen $\alpha$ is linear and

If $w\in \ker \alpha$, this identity gives $B_2(v,w)=0$ for every $v$. Conversely, if $B_2(v,w)=0$ for every $v$, then $B_1(v,\alpha w)=0$ for every $v$, so nondegeneracy gives $\alpha w=0$. HenceThis is the representation of a bilinear form relative to a nondegenerate bilinear form.

Let $p\in {\mathbb{R}}^m$ be Player I's row-strategy distribution and let $e$ denote an all-ones vector of the required dimension. Against column $j$, the expected payoff is $(p^{T}A{)}_j$. Thus Player I's matrix-game optimization problem isEquivalently, Player I maximizes ${\min }_j(p^{T}A{)}_j$ over the probability simplex.

Let $q^{\ast }$ be an optimal mixed strategy for Player II and let the game value be $v^{\ast }$. A sufficient condition for a probability vector $p$ to be optimal for Player I isIndeed, this makes $p$ guarantee at least $v^{\ast }$ against every pure column and hence every mixed strategy. On the other hand, the optimality of $q^{\ast }$ prevents any row strategy from obtaining more than $v^{\ast }$ against $q^{\ast }$. Therefore $p$ is optimal. This is the mixed-strategy optimality certificate for a matrix game.

Now suppose that $A$ is invertible and symmetric and that $A^{-1}e\ge 0$. PutThe vector $A^{-1}e$ is nonzero and nonnegative, so $c>0$ and $p,q$ are probability vectors. Since $A$ is symmetric,Thus $p$ guarantees $1/c$ and $q$ holds the payoff to $1/c$. By the minimax theorem,This proves the symmetric inverse formula for a matrix-game equilibrium.

For the card game, the payoff matrix to Player I isIt is symmetric and invertible, and direct solution of $Ax=e$ givesThe preceding result therefore gives the same optimal strategy for both players:Thus each player chooses cards $1,2,3$ with probabilities $41/50,2/25,1/10$, respectively, and the value to Player I isThis is the three-card threshold-sum zero-sum game.