Let A ∈ M₂, D = diag {α₁, α₂}. If D is an invertible diagonal unitary matrix then |α_i| = 1 for i = 1, 2. If P = I, then

If P=P12=[0110], then

Suppose i = j = 1 and A = B = E₁₁, then

That is φ(E₁₁)² is a diagonal matrix with diagonal entries 0 and 1. Since φ (E₁₁)² has distinct eigenvalues and commutes with φ (E₁₁), by proposition (2.1), φ (E₁₁) is a polynomial in φ (E₁₁)². It follows φ (E₁₁) is a diagonal matrix with diagonal entries 0 and 1 or −1 as φ(E11)2=E112 or E222. Then φ(E₁₁) = c₁₁E₁₁ or φ(E₁₁) = c₁₁E₂₂, where c₁₁ = ±1.

Without loss of generality we may assume that c₁₁ = 1 otherwise replace φ by −φ. So we assume φ(E₁₁) = E₁₁. Similarly φ(E₂₂) = c₂₂E₁₁ or c₂₂E₂₂, where c₂₂ = ±1. Suppose φ(E₂₂) = c₂₂E₁₁. If A = E₁₁, B = E₂₂

But,

As c₂₂ = ±1, this is a contradiction. Hence φ(E₂₂) = c₂₂E₂₂, where c₂₂ = ±1.

Suppose i ≠ j and φ(E₂₁) = [b_ij ] ∈ M₂. Then

But,

Therefore,

Taking A = E₁₁, B = E₂₁, we have

Note that G(E₂₁) is a disk with center 0 and radius 1. Hence

is a disk with centre 0 and radius 1. It follows that |2b₁₁| < 1 and |b₂₁| ≤ 1. Two cases arise (i) |b₂₁| = 1 or (ii) |b₂₁| < 1, in which case b₁₁ = 0 and |b₁₂| = 1.

Case (i): If |b₂₁| = 1, take A = E₂₂, B = E₂₁. Then

which is a disk with centre 0 and radius 1. But

As |b₂₁| = 1 and c₂₂ = ±1, b₂₂ = 0. Otherwise this disk will not be identical with the above disk.

Now from the set of equations (1) we have b₁₁ = 0 and b₁₂ = 0 and hence φ(E₂₁) = b₂₁E₂₁, where |b₂₁| = 1

Case (ii): If |b₂₁| < 1, in which case b₁₁ = 0 and |b₁₂| = 1, from the set of equations (1) we have b₂₁ = 0 and b₂₂ = 0 and hence φ(E₂₁) = b₁₂E₁₂, where |b₁₂| = 1.

This implies φ(E₂₁) = c₂₁PE₂₁P^t, where c₂₁ = b₁₂ and P = P₁₂.

Similarly, φ(E₁₂) = c₁₂PE₁₂P^t.

Thus φ(E_ij) = c_ijPE_ijP^t. If P ≠ I₂, then we replace the map φ by P^tφP and get φ(E_ij) = c_ijE_ij.

Claim :c_ijc_ji = 1

If i = j, then c_ijc_ji = 1. If i ≠ j, take A = E₁₂ and B = E₂₁, then

But,

Therefore, c₁₂c₂₁ = 1.

For A, B ∈ M₂ let φ(A) = c(PD)A(PD)⁻¹, φ(B) = c(PD)B(PD)⁻¹, where c = ±1, P is permutation matrix and D is a unitary diagonal matrix. Then

Suppose φ : M₂ → M₂ satisfies

Let A = [a_ij] in M₂. First we claim φ(A) = [c_ija_ij], with |c_ij | = 1. Proof of this claim is divided into four steps. Let φ(A) = [d_ij ]

• Step 1: To show that d_ij = c_ija_ij for i ≠ j with |c_ij | = 1.

• Step 2:φ(I) = I, where I is the identify matrix of order two.

• Step 3: φ [100-1]=[100-1].

• Step 4: Using the results of steps 1, 2 and 3 we prove d_ij = a_ij for i = j.

In view of proposition 3.3, we have φ(E_ij) = c_ijE_ij, where |c_ij | = 1, c₂₂ = ±1 and c₁₁ = 1.

Step 1: d_ij = c_ija_ij for i ≠ j.

Now, take B = E₂₁, then

Since both the sides are single disks comparing the centers we get,

Similarly, d₂₁ = c₂₁a₂₁. Hence d_ij = c_ija_ij for i ≠ j.

Step 2: φ(I) = I

Let φ(I) = [b_ij ]. Taking A = E₁₁, B = I, we have

Now, take A = E₂₂, B = I. Then

Therefore φ(I)=[100c22].

Taking A = E₁₂, B = I, we get

But, G(2E₁₂) is a disk with center 0 and radius 2. This gives

Thus φ(I) = I.

Step 3 : φ [100-1]=[100-1]

Take A = I and B=[100-1]. Then

If φ(B)=[-1001], taking A = E₁₁, we get

Which is not possible and our supposition is wrong. Therefore

Step 4: d₁₁ = a₁₁and d₂₂ = a₂₂. For A = [a_ij] and B=[100-1], we have

If a₁₁ = −a₂₂ then the right hand side is one degenerated disk i.e. single point and therefore

If a₁₁ ≠ −a₂₂ then we get two degenerated disks on the right hand side and there are two cases

• Case (i)a₁₁ = d₁₁ and a₂₂ = d₂₂, then there is nothing to prove.

• Case (ii)a₁₁ = −d₂₂ and a₂₂ = −d₁₁.

Now, we claim this case does not arise. We have either |a₁₂| = |a₂₁| or |a₁₂| > |a₂₁| or |a₁₂| < |a₂₁|. Taking B = E₁₁, we have

If they represent two disks then comparison of centers gives a₁₁ = d₁₁ and then from case (i) a₂₂ = d₂₂. Suppose they represent a single disk.

If |a₁₂| = |a₂₁| then |d₁₂| = |d₂₁|.

By comparison of centers, we get a₁₁ = d₁₁ = 0 and then from case (i) a₂₂ = d₂₂.

If |a₁₂| > |a₂₁|, then from step (i) |d₁₂| > |d₂₁|.

By comparison of centers, we get a₁₁ = d₁₁ and hence a₂₂ = d₂₂ from case (i).

If |a₁₂| < |a₂₁|, then |d₁₂| < |d₂₁|.

Taking B = E₂₂ we have

As |a₁₂| < |a₂₁|, |c₂₂d₁₂| < |c₂₂d₂₁|, then by comparison of centers we have a₂₂ = d₂₂ and hence a₁₁ = d₁₁ from case (i).

Therefore case (ii) does not arise.

Thus d_ij = c_ija_ij with |c_ij | = 1, c_ijc_ji = 1 and c_ij = 1 for i = j. Thus

Let D=[100c21], then D-1=[100c12]. Hence

Recall that in proposition 3.3 we obtained φ(E_ij) = c_ijPE_ijP^t. When P ≠ I we replaced the map φ by P^tφP and assumed φ(E_ij) = c_ijE_ij and the rest of the proof was carried out under this assumption. Hence

and this completes the proof.