A Note on k-Potence Preservers on Matrix Spaces over Complex Field

Theorem 1. Suppose ϕ : S_n → M_n satisfy that A − λB is a k-potent matrix in S_n implying that ϕ(A) − λϕ(B) is a k-potent matrix in M_n, where λ ∈ ℂ. Then there exist invertible P ∈ M_n and ϵ ∈ ℂ with ϵ^k = ϵ such that ϕ(X) = ϵP⁻¹XP for every X ∈ S_n.

Corollary 2. Suppose ϕ : S_n → S_n satisfy that A − λB is a k-potent matrix in S_n implying that ϕ(A) − λϕ(B) is a k-potent matrix in S_n, where λ ∈ ℂ. Then there exist invertible P ∈ M_n and ϵ ∈ ℂ with ϵ^k = ϵ such that ϕ(X) = ϵP⁻¹XP for every X ∈ S_n, where PP^t = aI_n for some nonzero a ∈ ℂ.

Lemma 3 (see [2].)Suppose X, Y ∈ Γ_n, and X + ϵY ∈ Γ_n  for every ϵ ∈ Λ; then X and Y are orthogonal.

Lemma 4 ([7, Lemma 1]). Suppose A₁, A₂, …, A_n are n  ×  n mutually orthogonal nonzero k-potent matrices; then there exists P ∈ GL_n such that P⁻¹A_iP = c_iE_ii with $$ for every i ∈ 〈n〉.

Lemma 5. Suppose Z ∈ M_n−1, p, q, g, h ∈ ℂ^(n−1)×1 with gh^t ≠ 0, δ ∈ ℂ, for arbitrary nonzero α ∈ ℂ with h^tg + α² ≠ 0 and τ = (α⁻¹h^tg + α) ⁻¹, $$$$. Then Z = 0, δ = 0, and there exist λ₁, λ₂ ∈ ℂ with (λ₁ + 1)(λ₂ + 1) = 1 such that p = λ₁g and q = λ₂h.

Proof. By the assumption of α and τ, $$ is idempotent. Denote this matrix by X, and then we can get the following equation:

Since the matrices on both sides of X satisfy the following equation:

then the following matrix is k-potent by the assumption of lemma:

We denote by A the following matrix:

then the following equation is obvious: Unfolding it, we get $$; that is, τ^kA^k + τ^k−1(⋯) _k−1 + ⋯+τ(⋯) ₁ − τA = 0, where (⋯)  _i is the coefficient matrix of τⁱ for every i ∈ 〈k − 1〉.

Let $$, then we calculate it and get the following equations:

It is easy to get $$ and the following equation:

Note that the highest degree of α in τ⁻²A is 2; then the highest degree of α in τ^−3k+i(⋯) _k−i is less or equal to 3k − i for every i with 2 ≤ i ≤ k − 1, and the highest degree of α in $$ is 3k − 1, where Z is the coefficient matrix of α^3k−1 in Z₁ and δ is the coefficient of α^3k−1 in δ₁.

By the assumption of α, we have Z = 0 and δ = 0. Then the following equations are true:

and τ^−3k+1Z₁ = τ^−3k+1[−α⁻¹gq^t + α⁻¹gq^tτα⁻¹gh^t − pτh^t] = τ^−3k+2[−τ⁻¹α⁻¹gq^t + α⁻²gq^tgh^t − ph^t], where the highest degree of α is 3k − 2 and −gq^t − ph^t is the coefficient matrix of α^3k−2.

Now, we calculate the upper left part of τ^−3k+2(⋯) ₂.

When k = 2, τ^−3k+2(⋯) ₂ = τ⁻⁴A², of which the upper left part is τ⁻⁴[pq^t(I_n−1 − τα⁻¹gh^t) − q^tpα⁻¹gτh^t] = τ⁻⁴[pq^t − τα⁻¹pq^tgh^t − τα⁻¹q^tpgh^t]. Then in the upper left part of $$, the highest degree of α is 4, and the coefficient matrix is pq^t + gq^t + ph^t.

When k > 2, if $$ appears in the left (or right) end of an additive item of τ^−3k+2(⋯) ₂, then the upper left part of this item is 0. So, the upper left part of τ^−3k+2(⋯) ₂ is equal to the upper left part of $$; that is, the upper left part is $$, and the highest degree of α is 3k − 2 with pq^t as the coefficient matrix of α^3k−2.

By the assumption of α, we have pq^t + gq^t + ph^t = 0.

By gh^t ≠ 0, we have g ≠ 0, h ≠ 0, and p = 0 if and only if q = 0. When p ≠ 0, we can get p = λ₁g by p(q^t + h^t) + gq^t = 0, and q = λ₂h by (p + g)q^t + ph^t = 0, where λ₁ and λ₂ satisfy λ₁λ₂gh^t + λ₂gh^t + λ₁gh^t = 0; that is, λ₁λ₂ + λ₂ + λ₁ = 0 by gh^t ≠ 0, which is equivalent to (λ₁ + 1)(λ₂ + 1) = 1. When p = q = 0, λ₁ = λ₂ = 0.

Remark 6. Replacing gh^t ≠ 0 with gh^t = 0 in Lemma 5, we have g = 0 implies p = 0 or q + h = 0, and h = 0 implies q = 0 or p + g = 0. These cases will not appear in the proof of Theorem 1, but are necessary for the weak preservers from M_n to M_n.

Lemma 7. Suppose $$  for arbitrary a, b ∈ ℂ with a ≠ b, where λ : ℂ → ℂ is a map satisfying λ(x) ≠ 0 for every x ∈ ℂ. Then there exists nonzero λ₀ ∈ ℂ such that λ(x) = λ₀ for every x ∈ ℂ.

Proof. Since the trace of A is equal to 1, then (λ(a) − λ(b))(λ⁻¹(a) − λ⁻¹(b))/(a − b) ² = 0, or −1, especially, when equal to −1, k − 1 = 6p with p ∈ Z⁺. Denote λ(a)/λ(b) by y, and a − b by c; then we have (2 − y − y⁻¹)/c² = 0 or −1.

• (1)

  If (2 − y − y⁻¹)/c² = 0, then y = 1, that is, λ(a) = λ(b);

• (2)

  if (2 − y − y⁻¹)/c² = −1, then $$. When c = 1, $$; when c = 2, $$. But $$ implies λ(b + 2)/λ(b) = (λ(b + 2)/λ(b + 1))(λ(b + 1)/λ(b)) = 1, or $$. It is a contradiction! So it is impossible that (2 − y − y⁻¹)/x² = −1.

Hence, there exists nonzero λ₀ ∈ ℂ such that λ(x) = λ₀ for every x ∈ ℂ.

Lemma 8 (see [4], Lemma 4.)Suppose ϕ ∈ Φ_n, A and B are n × n orthogonal k-potent matrices; then ϕ(A) and ϕ(B) are orthogonal.

Lemma 9 (see [4], Lemma 5.)Suppose ϕ ∈ Φ_n; then ϕ are homogeneous; that is, ϕ(λX) = λϕ(X) for every X ∈ S_n and every λ ∈ ℂ.

Corollary 10. Suppose ϕ ∈ Φ_n, A + B, C ∈ SΓ_n, and for every ϵ ∈ Λ, A + B + ϵC ∈ SΓ_n, ϕ(B + ϵC) = ϕ(B) + ϕ(ϵC). Then ϕ(A) + ϕ(B) and ϕ(C) are orthogonal.

Proof. By the assumption and Lemma 9, we have ϕ(A) + ϕ(B) ∈ Γ_n, ϕ(C) ∈ Γ_n, ϕ(A)  +  ϕ(B  +  ϵC) = ϕ(A) + ϕ(B) + ϵϕ(C) ∈ Γ_n. By Lemma 3, ϕ(A) + ϕ(B) and ϕ(C) are orthogonal.

Corollary 11. Suppose ϕ ∈ Φ_n and ϕ(D_n) = D_n for arbitrary diagonal matrix D_n ∈ M_n. Then for every i, j ∈ 〈n〉 with i ≠ j, $$, where λ_ij ∈ ℂ is only decided by i and j.

Proof. Let A = (1/2)(E_ij + E_ji + D_n), B = (1/2)(E_ii + E_jj − D_n), and C = ∑_l≠i,j E_ll; then A, B and C satisfy the assumption of Corollary 10, and ϕ(A) + ϕ(B) and ϕ(C) are orthogonal; that is, ϕ((E_ij + E_ji + D_n)) = α_iiE_ii + β_ijE_ij + γ_jiE_ji + δ_jjE_jj + D_n for some α_ii, β_ij, γ_ji, and δ_jj ∈ ℂ.

Since (η⁻¹ + η) ⁻¹[(E_ij + E_ji + D_n)−(D_n − η⁻¹E_ii − ηE_jj)] = (η⁻¹ + η) ⁻¹(η⁻¹E_ii + E_ij + E_ji + ηE_jj) ∈ SΓ_n for arbitrary nonzero η ∈ ℂ with 1 + η² ≠ 0, after applying ϕ, we have (η⁻¹ + η) ⁻¹[α_iiE_ii + β_ijE_ij + γ_jiE_ji + δ_jjE_jj + η⁻¹E_ii + ηE_jj] = (η⁻¹ + η) ⁻¹[α_iiE_ii + (β_ij − 1)E_ij + (γ_ji − 1)E_ji + δ_jjE_jj]+(η⁻¹ + η) ⁻¹(η⁻¹E_ii + E_ij + E_ji + ηE_jj) ∈ Γ_n. By Lemma 5, α_ii = δ_jj = 0, β_ijγ_ji = 1.

Let $$, where x_l ∈ ℂ for every l ∈ 〈n〉; then β_ij is the function of i, j, and x_l and denote by β_ij(D_n) the value of β_ij on x₁, …, x_n, i, and j.

Fix i, j, and D_n and add a free variable x to x_l for some l ∈ 〈n〉; then β_ij(D_n + xE_ll) becomes into a map of x. Since (1/(a − b))(E_ij + E_ji + D_n + aE_jj)−(1/(a − b))(E_ij + E_ji + D_n + bE_jj) ∈ SΓ_n for arbitrary a and b ∈ ℂ with a − b ≠ 0, then by $$ and $$, we can derive that $$Γ_n. By Lemma 7, β_ij(D_n + aE_jj) = β_ij(D_n + bE_jj) for fixed i, j, and D_n; that is, β_ij(D_n + xE_jj) = β_ij(D_n) for arbitrary x ∈ ℂ. Similarly, we can prove β_ij(D_n + xE_ii) = β_ij(D_n) for arbitrary x ∈ ℂ.

In fact, we have proved that β_ij(D_n + xE_ii) = β_ij(D_n) and β_ij(D_n + yE_jj) = β_ij(D_n) for arbitrary x, y ∈ ℂ and arbitrary D_n; then β_ij(D_n + xE_ii + yE_jj) = β_ij(D_n + xE_ii)( = β_ij(D_n + yE_jj)) = β_ij(D_n) follows.

Since β_ij(D_n + xE_jj + yE_ll) = β_ij(D_n + yE_ll) for fixed i, j, and l with l ≠ i, j, and arbitrary x, y ∈ ℂ, then (1/(a − b))(E_ij + E_ji + D_n + (a − b)E_jj + aE_ll)−(1/(a − b))(E_ij + E_ji + D_n + bE_ll) ∈ SΓ_n implies $$∈Γ_n. By Lemma 7, we can get β_ij(D_n + aE_ll) = β_ij(D_n + bE_ll) for arbitrary a and b ∈ ℂ with a − b ≠ 0; that is, β_ij(D_n + xE_ll) = β_ij(D_n) for arbitrary x ∈ ℂ.

Until now, we have proved that $$ for arbitrary D_n; that is, β_ij is only decided by i and j.

Remark 12. The proof of Corollary 11 presents the basic procedure of proof of Theorem 1. In order to decide the image of matrix A, we use Corollary 10 and the images of B and C, which usually are diagonal matrices or some matrices with images already decided.

If ϕ is a weak preserver from M_n to M_n, then Corollary 11 is also true. Let A = E_ij + D_n, B = −(E_ij + E_ji + D_n) + E_ii, and C = ∑_l≠i,j E_ll; then we can prove ϕ(A) = a_iiE_ii + a_ijE_ij + a_jiE_ji + a_jjE_jj + D_n similarly as proving ϕ((E_ij + E_ji + D_n)) = α_iiE_ii + β_ijE_ij + γ_jiE_ji + δ_jjE_jj + D_n, and $$. Since α⁻¹A + α⁻¹(−(E_ij + E_ji + D_n) + αE_ii) = −α⁻¹E_ji + E_ii ∈ Γ_n for arbitrary nonzero α, then the following matrix is k-potent:

Remark 6 tells us that a_ii = a_jj = 0,  a_ij − λ_ij = 0, or $$; that is, ϕ(A) = λ_ijE_ij + D_n, or $$, $$. Similarly, we can prove $$, ϕ(E_ji + D_n) = λ_ijE_ij + D_n, or $$. Since D_n is arbitrary, we set D_n = 0 for convenience.

If $$; then (1/3)E_ij + (1/3)(E_ij + E_ji + 2E_ii + E_jj) = (1/3)(2E_ij + E_ji + 2E_ii + E_jj) ∈ Γ_n implies $$; that is, −2/9 ∈ Δ, which is a contradiction. Hence, we proved that it is impossible $$ or $$.

If ϕ(E_ij) = λ_ijE_ij and ϕ(E_ji) = λ_ijE_ij, then (1/2)(E_ii + E_ij + E_ji + E_jj) ∈ Γ_n implies (1/2)(ϕ(E_ij) + ϕ(E_ii + E_ji + E_jj)) ∈ Γ_n; that is, (1/2)(E_ii + 2λ_ijE_ij + E_jj) ∈ Γ_n, which is a contradiction. Hence, we proved that ϕ(E_ij) = λ_ijE_ij and $$, or $$ and ϕ(E_ji) = λ_ijE_ij.

Proposition 13. Suppose i, j ∈ 〈n〉 with i ≠ j; then ϕ(E_ii) = 0 if and only if ϕ(E_jj) = 0.

Proof. Suppose ϕ(E_ii) = 0 and ϕ(E_jj) ≠ 0 for some i, j ∈ 〈n〉 with i ≠ j. At first, we prove that ϕ(aE_ii + E_jj) = ϕ(E_jj) for arbitrary a ∈ ℂ. Since the equation is already true when a = 0, then we assume a ≠ 0 in the following proof.

Let A = a⁻¹(aE_ii + E_jj), B = −a⁻¹E_jj, and C = E_jj; then it is easy to verify A, B, and C satisfying the assumption of Corollary 10. So ϕ(a⁻¹(aE_ii + E_jj)) + ϕ(−a⁻¹E_jj) and ϕ(E_jj) are orthogonal. Moreover, we can derive ϕ(aE_ii + E_jj) ∈ Γ_n from (aE_ii + E_jj) − aE_ii ∈ SΓ_n and ϕ(E_ii) = 0. Let a⁻¹(ϕ(aE_ii + E_jj) − ϕ(E_jj)) = D, then D and ϕ(E_jj) are orthogonal k-potent matrices. While ϕ(aE_ii + E_jj) ∈ Γ_n implies aD + ϕ(E_jj) ∈ Γ_n; then aD ∈ Γ_n. There are two cases on a.

• (1)

  If a ∉ Λ, then D = 0; that is, ϕ(aE_ii + E_jj) = ϕ(E_jj);

• (2)

  if a ∈ Λ, we can derive that (1/3)ϕ(aE_ii + E_jj)−(1/3)ϕ[(a − 3)E_ii + E_jj] ∈ Γ_n from (1/3)(aE_ii + E_jj)−(1/3)[(a − 3)E_ii + E_jj] ∈ SΓ_n. Note that a − 3 ∉ Λ, so it is true that ϕ[(a − 3)E_ii + E_jj] = ϕ(E_jj); that is, (1/3)ϕ(aE_ii + E_jj)−(1/3)ϕ(E_jj) = (a/3)D ∈ Γ_n. Finally, we can derive D = 0 from a/3 ∉ Λ and D ∈ Γ_n. At the same time, ϕ(aE_ii + E_jj) = ϕ(E_jj).

Anyway, ϕ(aE_ii + E_jj) = ϕ(E_jj) for arbitrary a ∈ ℂ.

Since (b⁻¹ + b) ⁻¹(b⁻¹E_ii + E_ij + E_ji + bE_jj) ∈ SΓ_n for every nonzero b ∈ ℂ with 1 + b² ≠ 0, then (b⁻¹ + b) ⁻¹[ϕ(E_ij + E_ji) + ϕ(b⁻¹E_ii + bE_jj)] ∈ Γ_n, and (b⁻¹ + b) ⁻¹[ϕ(E_ij + E_ji) + bϕ(E_jj)] ∈ Γ_n by ϕ(b⁻¹E_ii + bE_jj) = bϕ(E_jj). While the equation (b⁻¹ + b) ^−k[ϕ(E_ij + E_ji) + bϕ(E_jj)] ^k = (b⁻¹ + b) ⁻¹[ϕ(E_ij + E_ji) + bϕ(E_jj)] is equivalent to b^k−1[ϕ(E_ij + E_ji) + bϕ(E_jj)] ^k = (1 + b²) ^k−1[ϕ(E_ij + E_ji) + bϕ(E_jj)]. Note that ϕ(E_ij + E_ji) is the constant term of the equation; then ϕ(E_ij + E_ji) = 0 by the infinite property of b, and (b⁻¹ + b) ⁻¹bϕ(E_jj) ∈ Γ_n follows. Then we can derive ϕ(E_jj) = 0 which is a contradiction to the assumption.

Proposition 14. Suppose ϕ(E_ii) = 0 for every i ∈ 〈n〉; then ϕ(X) = 0 for arbitrary X ∈ S_n.

Proof. The proof will be completed by induction on the following equation for arbitrary X ∈ S_n with X[i, i] = x_i for every i ∈ 〈n〉:

where 1 ≤ m ≤ n − 1.

When m = 1, (10) is equivalent to $$ for arbitrary $$.

At first, by the assumption, it is already true that ϕ(E_ii) = 0 for every i ∈ 〈n〉.

Suppose $$ for every s ∈ 〈n − 1〉 with 1 ≤ i₁ < ⋯<i_s ≤ n; then by the homogeneity of ϕ, we just need to prove the following equation for i_s+1 with i_s < i_s+1 ≤ n:

There are two cases on $$.

• (1)

  If B_s ∉ SΓ_n, then there exists l ∈ 〈s〉 such that $$, and the following statements are true:

  $$ ()

Note that ϕ(B_s) = 0 and $$ by the assumption; then the following statements are true:

Since $$, then $$, and $$ follows.

• (2)

  If B_s ∈ SΓ_n, then we have the following statements:

  $$ ()

Since $$; then $$ by case 1, and $$ follows. While $$, hence we get $$.

Anyway, we prove $$; then by the induction, (10) is true for m = 1.

Suppose (10) is true for m ∈ 〈n − 1〉, then we prove the case on m + 1.

Let X_m = X_{[1,…,m;  1,…,m]}, g = X_{[1,…,m;m+1]}, $$; then we have g^t = X_{[m+1;  1,…,m]} and the following equation:

We will prove the following equation which is equivalent to (10) on m + 1:

For arbitrary nonzero α ∈ ℂ with g^tg + α² ≠ 0, the following n × n matrix B is idempotent:

where τ = (α⁻¹g^tg + α) ⁻¹.

Note that $$ and A_n−m−1 satisfy the following equation:

After applying ϕ on the above matrices, we have τϕ(X_m+1 ⊕ A_n−m−1) ∈ Γ_n by the inductive assumption. Then ϕ(X_m+1 ⊕ A_n−m−1) = 0 because of the assumption of α; that is, (10) holds for m + 1.

Finally, we prove that ϕ(X) = 0 for every X ∈ S_n by the induction.

Remark 15. If ϕ is a weak k-potence preserver from M_n to M_n; then Propositions 13 and 14 (replacing g^t with h^t for arbitrary X ∈ M_n in the proof of Proposition 14) hold since Corollary 10 is true under this assumption.

Proposition 16. Suppose ϕ(E_ii) ≠ 0 for every i ∈ 〈n〉, then there exist P ∈ GL_n and c ∈ Λ such that ϕ(X) = cP⁻¹XP for every X ∈ S_n.

Proof. The proof will be completed in the following 4 steps.

Step 1. ϕ(E_ii) = c_iE_ii, where c_i ∈ Λ for every i ∈ 〈n〉:

Since ϕ(E_ii) is nonzero k-potent, then we can derive from Lemma 4 that there exists P₁ ∈ GL_n such that $$ for every i ∈ 〈n〉, where c_i ∈ Λ. It is obvious that the following map φ ∈ Φ_n and φ(E_ii) = c_iE_ii for every i ∈ 〈n〉.

Without loss of generality, we can assume ϕ(E_ii) = c_iE_ii.

Step 2. $$, for arbitrary diagonal matrix $$.

The proof of this step can be seen in Step 3, Section  3 in [5].

Step 3. c_i = c ∈ Λ for every i ∈ 〈n〉.

Let A = (1/2)(E_ij + E_ji), B = (1/2)(E_ii + E_jj), and C = ∑_{l∈<n>∖{i,j}} E_ll, we can derive the following equation from Step 2 and Corollary 10:

where α₀, β₀, γ₀, δ₀ ∈ ℂ, i, j ∈ 〈n〉 with i ≠ j.

Note that pE_ii + q(E_ij + E_ji)+(1 − p)E_jj ∈ SΓ_n for p, q ∈ ℂ with q² = p(1 − p). In fact, 0 and 1 are all the eigenvalues of this matrix. Applying ϕ on the matrix q(E_ij + E_ji)+[pE_ii + (1 − p)E_jj], we have H(p) = q(α₀E_ii + β₀E_ij + γ₀E_ji + δ₀E_jj) + pc_iE_ii + (1 − p)c_jE_jj = (pc_i + qα₀)E_ii + qβ₀E_ij + qγ₀E_ji + ((1 − p)c_j + qδ₀)E_jj ∈ Γ_n.

Since k is fixed, then Δ is the finite set which contains all of eigenvalues of H(p), and there exists w ∈ {c + d∣c, d ∈ Δ} such that the trace of H(p) is w for infinite choices of p; that is, there exist (p₁, p₂) with p₁ ≠ p₂ such that the traces of H(p₁) and H(p₂) are all equal to w; then we have the following equation:

which is equivalent to where $$, for s = 1, 2.

Naturally, there are infinite choices of p₂ for fixed p₁ such that the above equation is true. If (q₁ − q₂)/(p₂ − p₁) is equal to some a ∈ ℂ, where p₂ ≠ p₁, p₁ and q₁ are fixed, then we can derive from the following equation:

that there are infinite choices of p₂ for constant (q₁ − q₂)/(p₂ − p₁) if and only if a² + 1 = 2aq₁ + 2a²p₁ + 1 = (q₁ + ap₁) ² = 0. While a² + 1 = (q₁ + ap₁) ² = 0 and $$ imply p₁ = q₁ = 0, which is a contradiction to 2aq₁ + 2a²p₁ + 1 = 0, hence (q₁ − q₂)/(p₂ − p₁) varies with p₂.

Since α₀ + δ₀ and c_i − c_j are all fixed numbers for fixed ϕ, then α₀ + δ₀ ≠ 0 implies that there are at least two different values of c_i − c_j = (q₁ − q₂)/(p₂ − p₁)(α₀ + δ₀) for fixed p₁ and infinite choices of p₂; it is a contradiction. So α₀ + δ₀ = 0 and c_i = c_j follows. Hence c_i = c ∈ Λ for every i ∈ 〈n〉.

Step 4. ϕ(X) = X for every X ∈ S_n.

After the discussion in Steps 1, 2, and 3, we already have the following equation:

where c ∈ Λ, a_i ∈ ℂ for every i ∈ 〈n〉. Since the map c⁻¹ϕ ∈ Φ_n, then we can assume $$ without loss of generality.

The proof in this step will be completed by induction on the following equation for arbitrary X ∈ S_n with X[i, i] = x_i for every i ∈ 〈n〉:

where 1 ≤ i₁ < ⋯<i_m ≤ n with 2 ≤ m ≤ n − 1.

When m = 2, (25) is equivalent to ϕ(E_ij + E_ji + D_n) = E_ij + E_ji + D_n for arbitrary diagonal matrix D_n ∈ S_n and i, j ∈ 〈n〉 with i < j, since ϕ is homogeneous. The proof will be completed in the following (1) and (2).

• (1)

  ϕ(E_ii+1 + E_i+1i + D_n) = E_ii+1 + E_i+1i + D_n for every i ∈ 〈n − 1〉.

We already derive from Corollary 11 that $$ for every i ∈ 〈n − 1〉, where λ_i ∈ ℂ is only decided by i.

Suppose the map ρ : S_n → M_n satisfies the following equation for every X ∈ S_n,

then ρ ∈ Φ_n, and for arbitrary diagonal matrix D_n and every i ∈ 〈n − 1〉, ρ(D_n) = D_n and ρ(E_ii+1 + E_i+1i + D_n) = E_ii+1 + E_i+1i + D_n.

Without loss of generality, we can assume ϕ(E_ii+1 + E_i+1i + D_n) = E_ii+1 + E_i+1i + D_n for every i ∈ 〈n − 1〉 and arbitrary D_n.

• (2)

  Suppose ϕ(E_ij + E_ji + D_n) = E_ij + E_ji + D_n for every i, j with 1 ≤ j − i < s < n − 1; then ϕ(E_ij + E_ji + D_n) = E_ij + E_ji + D_n for every i, j with j − i = s.

At first, we have to prove that ϕ(x_ii+1(E_ii+1 + E_i+1i) + x_i+1i+m(E_i+1i+m + E_i+mi+1) + D_n) = x_ii+1(E_ii+1 + E_i+1i) + x_i+1i+m(E_i+1i+m + E_i+mi+1)  +D_n for arbitrary nonzero x_ii+1 and x_i+1i+m ∈ ℂ.

By the assumption, we already have the following equations:

Let X₁ = x_ii+1(E_ii+1 + E_i+1i) + x_i+1i+m(E_i+1i+m + E_i+mi+1) + D_n, X₂ = x_ii+1(E_ii+1 + E_i+1i) + D_n, and X₃ = x_i+1i+m(E_i+1i+m + E_i+mi+1) + D_n. Then the following statements are true

where x_i+1i+m(E_i+1i+m + E_i+mi+1) + a_i+1E_i+1i+1 + a_i+mE_i+mi+m and x_ii+1(E_ii+1 + E_i+1i) + b_iE_ii + b_i+1E_i+1i+1 are k-potent.

Let A = X₁, B = −(X₂ − a_i+1E_i+1i+1 − a_i+mE_i+mi+m), and $$, then A, B, and C satisfy the assumption of Corollary 10. Hence we get ϕ(A) + ϕ(B) and ϕ(C) are orthogonal; that is,

Similarly, we can derive the following equation from Corollary 10:

Comparing the above two equations, we have z_i = y_i+m = 0, z_i+1 = y_i+1, z_ii+1 = z_i+1i = x_ii+1, and y_i+1i+m = y_i+mi+1 = x_i+1i+m, that is, ϕ(X₁) = X₁ + y_i+1E_i+1i+1.

We will prove z_i+1 = y_i+1 = 0. For arbitrary nonzero α with $$, let $$, and $$; then τX₁ + τX₄ ∈ SΓ_n implies τϕ(X₁) + τϕ(X₄) ∈ Γ_n; that is, the following matrix is k-potent since ϕ(X₄) = X₄ by the assumption

by Lemma 5, y_i+1 = 0. Hence we prove ϕ(X₁) = X₁.

Now we prove ϕ(E_ii+m + E_i+mi + D_n) = E_ii+m + E_i+mi + D_n.

By Corollary 11, we already have $$.

For arbitrary nonzero α with 2 + α² ≠ 0, (2α⁻¹ + α) ⁻¹(E_ii+m + E_i+mi + D_n)−(2α⁻¹ + α) ⁻¹(−α⁻¹(E_ii + E_ii+1 + E_i+1i + E_i+1i+1) − E_i+1i+m − E_i+mi+1 − αE_i+mi+m + D_n) = (2α⁻¹ + α) ⁻¹(α⁻¹(E_ii + E_ii+1 + E_i+1i + E_i+1i+1)+(E_ii+m + E_i+1i+m)+(E_i+mi + E_i+mi+1) + αE_i+mi+m) is idempotent.

After applying ϕ on the above matrices, we have (2α⁻¹ + α) ⁻¹ϕ(E_ii+m + E_i+mi + D_n)−(2α⁻¹ + α) ⁻¹ϕ(−α⁻¹(E_ii + E_ii+1 + E_i+1i + E_i+1i+1) − E_i+1i+m − E_i+mi+1 − αE_i+mi+m + D_n) = $$.

Then λ_ii+m = 1 by Lemma 5.

By the induction, we prove ϕ(E_ij + E_ji + D_n) = E_ij + E_ji + D_n for every i, j with 1 ≤ i < j ≤ n.

• (3)

  Suppose (25) is true for every s with 2 ≤ s < m ≤ n; then we prove it holds on m.

For arbitrary X ∈ S_n with X[i, i] = x_i for every i ∈ 〈n〉, let A, B, U, V, $$, and τ satisfy the following equations:

Then $$ is idempotent for arbitrary nonzero α with $$. Applying ϕ on it, we have $$. Let $$; then by $$ for every ϵ ∈ Λ, we have $$.

Note that $$ and ϕ(C) = C by the assumption; then τϕ(A) + τϕ(C) and $$ are orthogonal by Corollary 10; that is, $$ for some Y ∈ M_n.

On the other hand, $$ implies $$∈Γ_n by τϕ(A) + τϕ(C) ∈ Γ_n. By Lemma 5, we can derive the following equations:

that is, $$.

Let B₁ and B₂ satisfy the following equations:

then we can prove

Comparing the above three sets of equations, we can get ϕ(A) = A, which is equivalent to (25) on m.

By the induction, we prove that ϕ(X) = X for arbitrary X ∈ S_n.

Remark 17. If ϕ is a weak k-potence preserver from M_n to M_n, then the proof in Steps 1, 2, and 3 of Proposition 16 holds, and we prove ϕ(X) = X or ϕ(X) = X^t in Step 4. We omit the detailed proof since the case on X^t is totally the same after changing relevant notations.