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

Abstract

Let ℂ be the field of all complex numbers, 𝑀_𝑛 the space of all 𝑛×𝑛 matrices over ℂ, and 𝑆_𝑛 the subspace of 𝑀_𝑛 consisting of all symmetric matrices. The map 𝜙∶𝑆_𝑛→𝑀_𝑛 satisfies that 𝐴−𝜆𝐵 is 𝑘-potent in 𝑆_𝑛 implying that 𝜙(𝐴)−𝜆𝜙(𝐵) is 𝑘-potent in 𝑀_𝑛, where 𝜆∈ℂ, then there exist an invertible matrix 𝑃∈𝑀_𝑛 and 𝜖∈ℂ with 𝜖^𝑘=𝜖 such that 𝜙(𝑋)=𝜖𝑃⁻¹(𝑋)𝑃 for every 𝑋∈𝑆_𝑛. Moreover, the inductive method used in this paper can be used to characterise similar maps from 𝑀_𝑛 to 𝑀_𝑛.

1. Introduction

Let ℂ be the field of all complex numbers, 𝑀_𝑛 the space of all 𝑛×𝑛 matrices over ℂ, 𝑇_𝑛 the subspace of 𝑀_𝑛 consisting of all triangular matrices, and 𝑆_𝑛 the subspace of 𝑀_𝑛 consisting of all symmetric matrices. For fixed integer 𝑘≥2, 𝐴∈𝑀_𝑛 is called a 𝑘-potent matrix if 𝐴^𝑘=𝐴; especially, 𝐴 is an idempotent matrix when 𝑘=2. The map 𝜙∶𝑆_𝑛→𝑀_𝑛 satisfies 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, is a kind of the so-called weak preservers. While replacing “implying that” with “if and only if,” 𝜙 is called strong preserver. Obviously, a strong preserver must be a weak preserver, while a weak preserver may not be a strong preserver.

The preserver problem in this paper is from LPPs but without linear assumption (more details about LPP in [1–3]). You and Wang characterized the strong 𝑘-potence preservers from 𝑀_𝑛 to 𝑀_𝑛 in [4]; then Song and Cao extended the result to weak preservers from 𝑀_𝑛 to 𝑀_𝑛 in [5]. In [6], Wang and You characterized the strong 𝑘-potence preservers from 𝑇_𝑛 to 𝑀_𝑛. In this paper, the authors characterized the weak 𝑘-potence preservers from 𝑆_𝑛 to 𝑀_𝑛 and proved the following theorem.

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.

Furthermore, we can derive the following corollary from Theorem 1.

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.

In fact, the proof of Theorem 1 through some adjustments is suitable for the weak 𝑘-potence preserver from 𝑀_𝑛 to 𝑀_𝑛, and more details can be seen in remarks.

2. Notations and Lemmas

Γ_𝑛 denotes the set of all 𝑘-potent matrices in 𝑀_𝑛, while 𝑆Γ_𝑛=Γ_𝑛∩𝑆_𝑛. Λ denotes the set of all complex number 𝜖 satisfying 𝜖^𝑘−1=1, Δ=Λ∪{0}. 𝐸_𝑖𝑗 denotes matrices in 𝑀_𝑛 with 1 in (𝑖,𝑗) and 0 elsewhere, and 𝐼_𝑛 denotes the unit matrix in 𝑀_𝑛. ⟨𝑛⟩ denotes the set of integer 𝑠 satisfy 1≤𝑠≤𝑛. 𝐺𝐿_𝑛 denotes the general linear group consisting of all invertible matrices in 𝑀_𝑛. 𝐷_𝑛 denotes an arbitrary diagonal matrix in 𝑀_𝑛. For 𝐴, 𝐵∈𝑀_𝑛, 𝐴 and 𝐵 are orthogonal if 𝐴𝐵=𝐵𝐴=0. ℂ^𝑛×1 denotes the space of all 𝑛×1 matrices over ℂ. Φ_𝑛 denotes the set of all maps 𝜙∶𝑆_𝑛→𝑀_𝑛 satisfying 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.

For an arbitrary matrix 𝑋∈𝑀_𝑛, we denote by 𝑋[𝑖,𝑗] the term in (𝑖,𝑗) position of 𝑋, by 𝑋_{[𝑖1,…,𝑖𝑠;𝑗1,…,𝑗𝑡]} the 𝑠×𝑡 matrix with the term in its (𝑝,𝑞) position equal to 𝑋[𝑖_𝑝,𝑗_𝑞], where 𝑖₁<⋯<𝑖_𝑠 and 𝑗₁<⋯<𝑗_𝑡. Moreover, we denote by 𝑋_{{𝑖1,…,𝑖𝑠;𝑗1,…,𝑗𝑡}} the 𝑛×𝑛 matrix with the term in its (𝑖_𝑝,𝑗_𝑞) position equal to 𝑋[𝑖_𝑝,𝑗_𝑞] and terms elsewhere equal to 0. We especially simplify it with 𝑋_{{𝑖1,…,𝑖𝑠}} when 𝑠=𝑡, and 𝑖_𝑙=𝑗_𝑙 for every 𝑙∈⟨𝑠⟩. Naturally, 𝑋_{𝑖}=𝑋[𝑖,𝑖]𝐸_𝑖𝑖 for every 𝑖∈⟨𝑛⟩.

Without fixing 𝑋, 𝑋_{{𝑖1,…,𝑖𝑠;𝑗1,…,𝑗𝑡}} also denotes a matrix in 𝑀_𝑛 with 0 in its (𝑝,𝑞) position, where 𝑝∉{𝑖₁,…,𝑖_𝑠}, 𝑞∉{𝑗₁,…,𝑗_𝑡}, and 1≤𝑖₁<⋯<𝑖_𝑠≤𝑛, 1≤𝑗₁<⋯<𝑗_𝑡≤𝑛.

At first, we need the following Lemmas 3, 4, 5, and 7, which are about 𝑘-potent matrices and orthogonal matrices.

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 𝑐_𝑖^𝑘−1=1 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, 𝜏_𝑞^𝑍𝑝_𝑡𝛿+𝜏𝛼⁻¹𝑔ℎ^𝑡𝑔ℎ^𝑡𝛼∈Γ_𝑛. 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, 𝜏𝛼⁻¹𝑔ℎ^𝑡𝑔ℎ^𝑡𝛼 is idempotent. Denote this matrix by 𝑋, and then we can get the following equation: 𝐼_𝑛−1−𝛼⁻¹𝑔𝜏ℎ^𝑡1−𝜏𝛼⁻¹ℎ^𝑡𝑔𝑋𝐼_𝑛−1−𝜏𝛼⁻¹𝑔ℎ^𝑡𝛼⁻¹𝑔−𝜏ℎ^𝑡1=.0001 (1)
Since the matrices on both sides of 𝑋 satisfy the following equation: 𝐼_𝑛−1−𝛼⁻¹𝑔𝜏ℎ^𝑡1−𝜏𝛼⁻¹ℎ^𝑡𝑔𝐼_𝑛−1−𝜏𝛼⁻¹𝑔ℎ^𝑡𝛼⁻¹𝑔−𝜏ℎ^𝑡1=𝐼_𝑛 (2) then the following matrix is 𝑘-potent by the assumption of lemma: 𝐼_𝑛−1−𝛼⁻¹𝑔𝜏ℎ^𝑡1−𝜏𝛼⁻¹ℎ^𝑡𝑔𝜏𝑞𝑍𝑝_𝑡𝛿×𝐼+𝑋_𝑛−1−𝜏𝛼⁻¹𝑔ℎ^𝑡𝛼⁻¹𝑔−𝜏ℎ^𝑡1. (3)
We denote by 𝐴 the following matrix: 𝐼_𝑛−1−𝛼⁻¹𝑔𝜏ℎ^𝑡1−𝜏𝛼⁻¹ℎ^𝑡𝑔𝑞𝑍𝑝_𝑡𝛿𝐼_𝑛−1−𝜏𝛼⁻¹𝑔ℎ^𝑡𝛼⁻¹𝑔−𝜏ℎ^𝑡1; (4) then the following equation is obvious: 𝜏𝐴+0001^𝑘=𝜏𝐴+.0001 (5) Unfolding it, we get 𝜏^𝑘𝐴^𝑘+𝜏^𝑘−1(⋯)_𝑘−1+⋯+𝜏(⋯)₁+⁰⁰₀₁=𝜏𝐴+⁰⁰₀₁; that is, 𝜏^𝑘𝐴^𝑘+𝜏^𝑘−1(⋯)_𝑘−1+⋯+𝜏(⋯)₁−𝜏𝐴=0, where (⋯)_𝑖 is the coefficient matrix of 𝜏^𝑖 for every 𝑖∈⟨𝑘−1⟩.
Let 𝐴=^𝑍1𝑝1_{𝑞𝑡1𝛿1}, then we calculate it and get the following equations: 𝑍₁=𝑍−𝛼⁻¹𝑔𝑞^𝑡𝐼_𝑛−1−𝜏𝛼⁻¹𝑔ℎ^𝑡−𝑝−𝛿𝛼⁻¹𝑔𝜏ℎ^𝑡,𝑝₁=𝑍−𝛼⁻¹𝑔𝑞^𝑡𝛼⁻¹𝑔+𝑝−𝛿𝛼⁻¹𝑞𝑔,_𝑡1=𝜏ℎ^𝑡𝑍+𝑞^𝑡−𝜏𝛼⁻¹𝑞^𝑡ℎ^𝑡𝑔𝐼_𝑛−1−𝜏𝛼⁻¹𝑔ℎ^𝑡−𝜏ℎ^𝑡𝑝+𝛿−𝛿𝜏𝛼⁻¹ℎ^𝑡𝑔𝜏ℎ^𝑡,𝛿₁=𝜏ℎ^𝑡𝑍+𝑞^𝑡−𝜏𝛼⁻¹𝑞^𝑡ℎ^𝑡𝑔𝛼⁻¹𝑔+𝜏ℎ^𝑡𝑝+𝛿−𝛿𝜏𝛼⁻¹ℎ^𝑡𝑔. (6)
It is easy to get 𝜏(⋯)₁=𝜏^0𝑝1_{𝑞𝑡1𝑘𝛿1} and the following equation: 𝜏^𝑘𝐴^𝑘+𝜏^𝑘−1(⋯)_𝑘−1+⋯+𝜏²(⋯)₂+𝜏−𝑍₁00(𝑘−1)𝛿₁=0. (7) 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 𝜏^{−3𝑘+1−𝑍10}_{0(𝑘−1)𝛿1} 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: 𝑍₁=−𝛼⁻¹𝑔𝑞^𝑡𝐼_𝑛−1−𝜏𝛼⁻¹𝑔ℎ^𝑡−𝑝𝜏ℎ^𝑡,𝑝₁=−𝛼⁻¹𝑔𝑞^𝑡𝛼⁻¹𝑞𝑔+𝑝,_𝑡1=𝑞^𝑡−𝜏𝛼⁻¹𝑞^𝑡ℎ^𝑡𝑔𝐼_𝑛−1−𝜏𝛼⁻¹𝑔ℎ^𝑡−𝜏ℎ^𝑡𝑝𝜏ℎ^𝑡,𝛿₁=𝑞^𝑡−𝜏𝛼⁻¹𝑞^𝑡ℎ^𝑡𝑔𝛼⁻¹𝑔+𝜏ℎ^𝑡𝑝 (8) 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 𝜏⁻⁴𝐴²+𝜏^{−5−𝑍10}_{0(𝑘−1)𝛿1}, the highest degree of 𝛼 is 4, and the coefficient matrix is 𝑝𝑞^𝑡+𝑔𝑞^𝑡+𝑝ℎ^𝑡.
When 𝑘>2, if 0001 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 𝜏^−3𝑘+2𝐴⁰⁰₀₁^𝑘−2𝐴; that is, the upper left part is 𝜏^−3𝑘+2𝑝₁𝑞^𝑡₁=𝜏^−3𝑘+4[𝜏⁻¹𝛼𝑝𝑞^𝑡−(𝑞^𝑡𝑔+ℎ^𝑡𝑝)𝑝ℎ^𝑡−𝜏⁻¹𝛼⁻¹𝑔𝑞^𝑡𝑔𝑞^𝑡+𝛼⁻²𝑞^𝑡𝑔(𝑞^𝑡𝑔+ℎ^𝑡𝑝)𝑔ℎ^𝑡], 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 𝐴=^{0(𝜆(𝑎)−𝜆(𝑏))/(𝑎−𝑏)}_{(𝜆(𝑎)−1−𝜆(𝑏)−1)/(𝑎−𝑏)1}∈Γ₂  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 𝑦=(2+𝑐²±√4𝑐²+𝑐⁴)/2. When 𝑐=1, 𝜆(𝑏+1)/𝜆(𝑏)=(3±√5)/2; when 𝑐=2, 𝜆(𝑏+2)/𝜆(𝑏)=(6±√32)/2=3±2√2. But 𝜆(𝑏+1)/𝜆(𝑏)=(3±√5)/2 implies 𝜆(𝑏+2)/𝜆(𝑏)=(𝜆(𝑏+2)/𝜆(𝑏+1))(𝜆(𝑏+1)/𝜆(𝑏))=1, or (7±3√5)/2. 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.

We can prove the following Lemmas 8 and 9 similar as Lemmas 4 and  5 in [4].

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, 𝜙(𝐸_𝑖𝑗+𝐸_𝑗𝑖+𝐷_𝑛)=𝜆_𝑖𝑗𝐸_𝑖𝑗+𝜆⁻¹_𝑖𝑗𝐸_𝑗𝑖+𝐷_𝑛, 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 𝐷_𝑛=∑^𝑛_𝑙=1𝑥_𝑙𝐸_𝑙𝑙, 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 𝜙(𝐸_𝑖𝑗+𝐸_𝑗𝑖+𝐷_𝑛+𝑎𝐸_𝑗𝑗)=𝛽_𝑖𝑗(𝐷_𝑛+𝑎𝐸_𝑗𝑗)𝐸_𝑖𝑗+𝛽⁻¹_𝑖𝑗(𝐷_𝑛+𝑎𝐸_𝑗𝑗)𝐸_𝑗𝑖+𝐷_𝑛+𝑎𝐸_𝑗𝑗 and 𝜙(𝐸_𝑖𝑗+𝐸_𝑗𝑖+𝐷_𝑛+𝑏𝐸_𝑗𝑗)=𝛽_𝑖𝑗(𝐷_𝑛+𝑏𝐸_𝑗𝑗)𝐸_𝑖𝑗+𝛽⁻¹_𝑖𝑗(𝐷_𝑛+𝑏𝐸_𝑗𝑗)𝐸_𝑗𝑖+𝐷_𝑛+𝑏𝐸_𝑗𝑗, we can derive that ((𝛽_𝑖𝑗(𝐷_𝑛+𝑎𝐸_𝑗𝑗)−𝛽_𝑖𝑗(𝐷_𝑛+𝑏𝐸_𝑗𝑗))/(𝑎−𝑏))𝐸_𝑖𝑗+((𝛽⁻¹_𝑖𝑗(𝐷_𝑛+𝑎𝐸_𝑗𝑗)−𝛽⁻¹_𝑖𝑗(𝐷_𝑛+𝑏𝐸_𝑗𝑗))/(𝑎−𝑏))𝐸_𝑗𝑖+𝐸_𝑗𝑗∈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 ((𝛽_𝑖𝑗(𝐷_𝑛+𝑎𝐸_𝑙𝑙)−𝛽_𝑖𝑗(𝐷_𝑛+𝑏𝐸_𝑙𝑙))/(𝑎−𝑏))𝐸_𝑖𝑗+((𝛽⁻¹_𝑖𝑗(𝐷_𝑛+𝑎𝐸_𝑙𝑙)−𝛽⁻¹_𝑖𝑗(𝐷_𝑛+𝑏𝐸_𝑙𝑙))/(𝑎−𝑏))𝐸_𝑗𝑖+𝐸_𝑗𝑗+𝐸_𝑙𝑙∈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 𝛽_𝑖𝑗(𝐷_𝑛)=𝛽_𝑖𝑗(∑^𝑛_𝑙=1𝑥_𝑙𝐸_𝑙𝑙)=𝛽_𝑖𝑗(∑^𝑛−1_𝑙=1𝑥_𝑙𝐸_𝑙𝑙)=⋯=𝛽_𝑖𝑗(𝑥₁𝐸₁₁)=𝛽_𝑖𝑗(0) 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 (𝑎_𝑖𝑖+1)𝐸_𝑖𝑖+(𝑎_𝑖𝑗−𝜆_𝑖𝑗)𝐸_𝑖𝑗+(𝑎_𝑗𝑖−𝜆⁻¹_𝑖𝑗)𝐸_𝑗𝑖+𝑎_𝑗𝑗𝐸_𝑗𝑗∈Γ_𝑛. 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: 𝛼⁻¹𝑎_𝑖𝑖𝑎_𝑖𝑗−𝜆_𝑖𝑗𝑎_𝑗𝑖−𝜆⁻¹_𝑗𝑖𝑎_𝑗𝑗+𝛼⁻¹.𝛼000 (9) Remark 6 tells us that 𝑎_𝑖𝑖=𝑎_𝑗𝑗=0,  𝑎_𝑖𝑗−𝜆_𝑖𝑗=0, or 𝑎_𝑗𝑖−𝜆⁻¹_𝑖𝑗=0; that is, 𝜙(𝐴)=𝜆_𝑖𝑗𝐸_𝑖𝑗+𝐷_𝑛, or 𝜙(𝐴)=𝜆⁻¹_𝑖𝑗𝐸_𝑗𝑖+𝐷_𝑛, 𝜙(𝐴)=𝜆_𝑖𝑗𝐸_𝑖𝑗+𝜆⁻¹_𝑖𝑗𝐸_𝑗𝑖+𝐷_𝑛. Similarly, we can prove 𝜙(𝐸_𝑗𝑖+𝐷_𝑛)=𝜆⁻¹_𝑖𝑗𝐸_𝑗𝑖+𝐷_𝑛, 𝜙(𝐸_𝑗𝑖+𝐷_𝑛)=𝜆_𝑖𝑗𝐸_𝑖𝑗+𝐷_𝑛, or 𝜙(𝐸_𝑗𝑖+𝐷_𝑛)=𝜆_𝑖𝑗𝐸_𝑖𝑗+𝜆⁻¹_𝑖𝑗𝐸_𝑗𝑖+𝐷_𝑛. Since 𝐷_𝑛 is arbitrary, we set 𝐷_𝑛=0 for convenience.
If 𝜙(𝐸_𝑖𝑗)=𝜆_𝑖𝑗𝐸_𝑖𝑗+𝜆⁻¹_𝑖𝑗𝐸_𝑗𝑖; 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 (1/3)(𝜆_𝑖𝑗𝐸_𝑖𝑗+𝜆⁻¹_𝑖𝑗𝐸_𝑗𝑖)+(1/3)(𝜆_𝑖𝑗𝐸_𝑖𝑗+𝜆⁻¹_𝑖𝑗𝐸_𝑗𝑖+2𝐸_𝑖𝑖+𝐸_𝑗𝑗)=(1/3)(2𝜆_𝑖𝑗𝐸_𝑖𝑗+2𝜆⁻¹_𝑖𝑗𝐸_𝑗𝑖+2𝐸_𝑖𝑖+𝐸_𝑗𝑗)∈Γ_𝑛; that is, −2/9∈Δ, which is a contradiction. Hence, we proved that it is impossible 𝜙(𝐸_𝑖𝑗)=𝜆_𝑖𝑗𝐸_𝑖𝑗+𝜆⁻¹_𝑖𝑗𝐸_𝑗𝑖 or 𝜙(𝐸_𝑗𝑖)=𝜆_𝑖𝑗𝐸_𝑖𝑗+𝜆⁻¹_𝑖𝑗𝐸_𝑗𝑖.
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 𝜙(𝐸_𝑗𝑖)=𝜆⁻¹_𝑖𝑗𝐸_𝑗𝑖, or 𝜙(𝐸_𝑖𝑗)=𝜆⁻¹_𝑖𝑗𝐸_𝑗𝑖 and 𝜙(𝐸_𝑗𝑖)=𝜆_𝑖𝑗𝐸_𝑖𝑗.

3. Proof of Theorem 1

Suppose 𝜙∈Φ_𝑛, then we can derive Theorem 1 from Propositions 13, 14, and 16.

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⟩: 𝜙𝑋_{1,…,𝑚}+^𝑛_𝑖=𝑚+1𝑥_𝑖𝐸_𝑖𝑖=0, (10) where 1≤𝑚≤𝑛−1.
When 𝑚=1, (10) is equivalent to 𝜙(∑^𝑛_𝑖=1𝑎_𝑖𝐸_𝑖𝑖)=0 for arbitrary 𝐷_𝑛=∑^𝑛_𝑖=1𝑎_𝑖𝐸_𝑖𝑖∈𝑆_𝑛.
At first, by the assumption, it is already true that 𝜙(𝐸_𝑖𝑖)=0 for every 𝑖∈⟨𝑛⟩.
Suppose 𝜙(∑^𝑠_𝑗=1𝑎_𝑖𝑗𝐸_{𝑖𝑗𝑖𝑗})=0 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: 𝜙^𝑠_𝑗=1𝑎_𝑖𝑗𝐸_{𝑖𝑗𝑖𝑗}+𝐸_{𝑖𝑠+1𝑖𝑠+1}=0. (11) There are two cases on 𝐵_𝑠=∑^𝑠_𝑗=1𝑎_𝑖𝑗𝐸_{𝑖𝑗𝑖𝑗}. (1) If 𝐵_𝑠∉𝑆Γ_𝑛, then there exists 𝑙∈⟨𝑠⟩ such that 𝑎_𝑖𝑙∉Δ, and the following statements are true: 𝐵_𝑠+𝐸_{𝑖𝑠+1𝑖𝑠+1}−𝐵_𝑠=𝐸_{𝑖𝑠+1𝑖𝑠+1}∈𝑆Γ_𝑛,𝑎_𝑖⁻¹_𝑙𝐵_𝑠+𝐸_{𝑖𝑠+1𝑖𝑠+1}−𝑎_𝑖⁻¹_𝑙𝐵_𝑠+𝐸_{𝑖𝑠+1𝑖𝑠+1}−𝑎_𝑖𝑙𝐸_{𝑖𝑙𝑖𝑙}=𝐸_{𝑖𝑙𝑖𝑙}∈𝑆Γ_𝑛. (12)
Note that 𝜙(𝐵_𝑠)=0 and 𝜙(𝐵_𝑠+𝐸_{𝑖𝑠+1𝑖𝑠+1}−𝑎_𝑖𝑙𝐸_{𝑖𝑙𝑖𝑙})=0 by the assumption; then the following statements are true: 𝜙𝐵_𝑠+𝐸_{𝑖𝑠+1𝑖𝑠+1}∈Γ_𝑛,𝑎_𝑖⁻¹_𝑙𝜙𝐵_𝑠+𝐸_{𝑖𝑠+1𝑖𝑠+1}∈Γ_𝑛. (13)
Since 𝑎_𝑖𝑙∉Δ, then 𝑎_𝑖⁻¹_𝑙∉Δ, and 𝜙(𝐵_𝑠+𝐸_{𝑖𝑠+1𝑖𝑠+1})=0 follows.(2) If 𝐵_𝑠∈𝑆Γ_𝑛, then we have the following statements: 𝐵_𝑠+𝐸_{𝑖𝑠+1𝑖𝑠+1}∈𝑆Γ_𝑛,13𝐵_𝑠+𝐸_{𝑖𝑠+1𝑖𝑠+1}−13−3𝐸_𝑖1𝑖1+𝐵_𝑠+𝐸_{𝑖𝑠+1𝑖𝑠+1}=𝐸_𝑖1𝑖1∈𝑆Γ_𝑛. (14)
Since 𝑎_𝑖1−3∉Δ; then 𝜙(−3𝐸_𝑖1𝑖1+𝐵_𝑠+𝐸_{𝑖𝑠+1𝑖𝑠+1})=0 by case 1, and (1/3)𝜙(𝐵_𝑠+𝐸_{𝑖𝑠+1𝑖𝑠+1})∈Γ_𝑛 follows. While 𝜙(𝐵_𝑠+𝐸_{𝑖𝑠+1𝑖𝑠+1})∈Γ_𝑛, hence we get 𝜙(𝐵_𝑠+𝐸_{𝑖𝑠+1𝑖𝑠+1})=0.
Anyway, we prove 𝜙(∑^𝑠_𝑗=1𝑎_𝑖𝑗𝐸_{𝑖𝑗𝑖𝑗}+𝐸_{𝑖𝑠+1𝑖𝑠+1})=0; 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]}, 𝐴_𝑛−𝑚=∑^𝑛−𝑚_𝑖=1𝑥_𝑖+𝑚𝐸_𝑖𝑖∈𝑀_𝑛−𝑚; then we have 𝑔^𝑡=𝑋_{[𝑚+1;1,…,𝑚]} and the following equation: 𝜙𝑋_𝑚00𝐴_𝑛−𝑚=0. (15) We will prove the following equation which is equivalent to (10) on 𝑚+1: 𝜙𝑋_𝑚𝑔𝑔0_𝑡𝑥_𝑚+1000𝐴_{𝑛−𝑚−1}=0 (16)
For arbitrary nonzero 𝛼∈ℂ with 𝑔^𝑡𝑔+𝛼²≠0, the following 𝑛×𝑛 matrix 𝐵 is idempotent: 𝛼𝐵=𝜏⁻¹𝑔𝑔^𝑡𝑔𝑔0_𝑡𝛼0,000 (17) where 𝜏=(𝛼⁻¹𝑔^𝑡𝑔+𝛼)⁻¹.
Note that 𝑋_𝑚+1=^𝑋𝑚𝑔_{𝑔𝑡𝑥𝑚+1} and 𝐴_{𝑛−𝑚−1} satisfy the following equation: 𝜏𝑋_𝑚𝑔𝑔0_𝑡𝑥_𝑚+1000𝐴_{𝑛−𝑚−1}𝑋−𝜏^𝑚−𝛼⁻¹𝑔𝑔^𝑡000𝑥_𝑚+1−𝛼000𝐴_{𝑛−𝑚−1}=𝐵. (18)
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 𝑃₁⁻¹𝜙(𝐸_𝑖𝑖)𝑃₁=𝑐_𝑖𝐸_𝑖𝑖 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⟩. 𝜑(𝑋)=𝑃₁⁻¹𝜙(𝑋)𝑃₁. (19) Without loss of generality, we can assume 𝜙(𝐸_𝑖𝑖)=𝑐_𝑖𝐸_𝑖𝑖.
Step 2. 𝜙(∑^𝑛_𝑖=1𝑎_𝑖𝐸_𝑖𝑖)=∑^𝑛_𝑖=1𝑎_𝑖𝜙(𝐸_𝑖𝑖), for arbitrary diagonal matrix ∑^𝑛_𝑖=1𝑎_𝑖𝐸_𝑖𝑖.
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: 𝜙𝐸_𝑖𝑗+𝐸_𝑗𝑖=𝛼₀𝐸_𝑖𝑖+𝛽₀𝐸_𝑖𝑗+𝛾₀𝐸_𝑗𝑖+𝛿₀𝐸_𝑗𝑗, (20) 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: 𝑝₁𝑐_𝑖+𝑞₁𝛼₀+1−𝑝₁𝑐_𝑗+𝑞₁𝛿₀=𝑝₂𝑐_𝑖+𝑞₂𝛼₀+1−𝑝₂𝑐_𝑗+𝑞₂𝛿₀ (21) which is equivalent to 𝑞₁−𝑞₂𝛼₀+𝛿₀=𝑝₂−𝑝₁𝑐_𝑖−𝑐_𝑗, (22) where 𝑞²_𝑠=𝑝_𝑠(1−𝑝_𝑠), 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: 𝑎²+1𝑝²₂−2𝑎𝑞₁+2𝑎²𝑝₁+1𝑝₂+𝑞₁+𝑎𝑝₁²=0 (23) 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 𝑞²₁=𝑝₁(1−𝑝₁) 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: 𝜙^𝑛_𝑖=1𝑎_𝑖𝐸_𝑖𝑖=𝑐^𝑛_𝑖=1𝑎_𝑖𝐸_𝑖𝑖, (24) where 𝑐∈Λ, 𝑎_𝑖∈ℂ for every 𝑖∈⟨𝑛⟩. Since the map 𝑐⁻¹𝜙∈Φ_𝑛, then we can assume 𝜙(∑^𝑛_𝑖=1𝑎_𝑖𝐸_𝑖𝑖)=∑^𝑛_𝑖=1𝑎_𝑖𝐸_𝑖𝑖 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⟩: 𝜙𝑋_{{𝑖1,…,𝑖𝑚}}+_{𝑗∈<𝑛>⧵{𝑖1,…,𝑖𝑚}}𝑥_𝑗𝐸_𝑗𝑗=𝑋_{{𝑖1,…,𝑖𝑚}}+_{𝑖𝑗∈<𝑛>⧵1,…,𝑖𝑚}𝑥_𝑗𝐸_𝑗𝑗, (25) 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 𝜙(𝐸_𝑖𝑖+1+𝐸_𝑖+1𝑖+𝐷_𝑛)=𝜆_𝑖𝐸_𝑖𝑖+1+𝜆_𝑖⁻¹𝐸_𝑖+1𝑖+𝐷_𝑛 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, 𝜌(𝑋)=diag1,𝜆₁,𝜆₁𝜆₂,…,^𝑛−1_𝑖=1𝜆_𝑖𝜙(𝑋)×diag1,𝜆₁⁻¹,𝜆₁⁻¹𝜆₂⁻¹,…,^𝑛−1_𝑖=1𝜆_𝑖⁻¹; (26) 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: 𝜙𝑥_𝑖𝑖+1𝐸_𝑖𝑖+1+𝐸_𝑖+1𝑖+𝐷_𝑛=𝑥_𝑖𝑖+1𝐸_𝑖𝑖+1+𝐸_𝑖+1𝑖+𝐷_𝑛,𝜙𝑥_{𝑖+1𝑖+𝑚}𝐸_{𝑖+1𝑖+𝑚}+𝐸_{𝑖+𝑚𝑖+1}+𝐷_𝑛=𝑥_{𝑖+1𝑖+𝑚}𝐸_{𝑖+1𝑖+𝑚}+𝐸_{𝑖+𝑚𝑖+1}+𝐷_𝑛. (27)
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 𝑋₁−𝑋₂−𝑎_𝑖+1𝐸_𝑖+1𝑖+1−𝑎_𝑖+𝑚𝐸_{𝑖+𝑚𝑖+𝑚}∈𝑆Γ_𝑛,𝑋₁−𝑋₂−𝑎_𝑖+1𝐸_𝑖+1𝑖+1−𝑎_𝑖+𝑚𝐸_{𝑖+𝑚𝑖+𝑚}+𝜖^𝑛_{𝑙≠𝑖+1,𝑖+𝑚}𝐸_𝑙𝑙∈𝑆Γ_𝑛,𝑋₁−𝑋₃−𝑏_𝑖𝐸_𝑖𝑖−𝑏_𝑖+1𝐸_𝑖+1𝑖+1∈𝑆Γ_𝑛,𝑋₁−𝑋₃−𝑏_𝑖𝐸_𝑖𝑖−𝑏_𝑖+1𝐸_𝑖+1𝑖+1+𝜖^𝑛_{𝑙≠𝑖,𝑖+1}𝐸_𝑙𝑙∈𝑆Γ_𝑛, (28) 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 𝐶=∑^𝑛_{𝑙≠𝑖+1,𝑖+𝑚}𝐸_𝑙𝑙, 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, 𝜙𝑋₁=𝑋₂+𝑦_𝑖+1𝐸_𝑖+1𝑖+1+𝑦_𝑖+𝑚𝐸_{𝑖+𝑚𝑖+𝑚}+𝑦_{𝑖+1𝑖+𝑚}𝐸_{𝑖+1𝑖+𝑚}+𝑦_{𝑖+𝑚𝑖+1}𝐸_{𝑖+𝑚𝑖+1}. (29)
Similarly, we can derive the following equation from Corollary 10: 𝜙𝑋₁=𝑋₃+𝑧_𝑖𝐸_𝑖𝑖+𝑧_𝑖+1𝐸_𝑖+1𝑖+1+𝑧_𝑖𝑖+1𝐸_𝑖𝑖+1+𝑧_𝑖+1𝑖𝐸_𝑖+1𝑖. (30)
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 𝑥²_{𝑖+1𝑖+𝑚}+𝛼²≠0, let 𝜏=(𝛼⁻¹𝑥²_{𝑖+1𝑖+𝑚}+𝛼)⁻¹, and 𝑋₄=−𝑋₂+𝛼⁻¹𝑥²_{𝑖+1𝑖+𝑚}𝐸_𝑖+1𝑖+1+𝛼𝐸_{𝑖+𝑚𝑖+𝑚}; 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 𝜏𝑦_𝑖+10𝛼00+𝜏⁻¹𝑥²_{𝑖+1𝑖+𝑚}𝑥_{𝑖+1𝑖+𝑚}𝑥_{𝑖+1𝑖+𝑚}𝛼 (31) 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 𝜙(𝐸_{𝑖𝑖+𝑚}+𝐸_{𝑖+𝑚𝑖}+𝐷_𝑛)=𝜆_{𝑖𝑖+𝑚}𝐸_{𝑖𝑖+𝑚}+𝜆⁻¹_{𝑖𝑖+𝑚}𝐸_{𝑖+𝑚𝑖}+𝐷_𝑛.
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)=(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})+(2𝛼⁻¹+𝛼)⁻¹((𝜆_{𝑖𝑖+𝑚}−1)𝐸_{𝑖𝑖+𝑚}+(𝜆⁻¹_{𝑖𝑖+𝑚}−1)𝐸_{𝑖+𝑚𝑖})∈Γ_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, 𝑦_𝑖𝑚, and 𝜏 satisfy the following equations: 𝐴=𝑋_{{𝑖1,…,𝑖𝑚}}+_{⟨𝑛⟩𝑗∈⧵{𝑖1,…,𝑖𝑚}}𝑥_𝑗𝐸_𝑗𝑗,𝐵=𝑋_{{𝑖1,…,𝑖𝑚−1}}+_{⟨𝑛⟩𝑗∈⧵{𝑖1,…,𝑖𝑚−1}}𝑥_𝑗𝐸_𝑗𝑗,𝑈=𝑋_{{𝑖1,…,𝑖𝑚−1;𝑖𝑚}},𝑉=𝑋_{{𝑖𝑚;𝑖1,…,𝑖𝑚−1}},𝑦_𝑖𝑚=𝑋_{{𝑖𝑚;𝑖1,…,𝑖𝑚−1}}𝑋_{{𝑖1,…,𝑖𝑚−1;𝑖𝑚}}𝑖_𝑚,𝑖_𝑚,𝜏=𝛼⁻¹𝑦_𝑖𝑚+𝛼⁻¹. (32) Then 𝜏𝐴+𝜏(−𝐵+𝛼⁻¹𝑈𝑉+𝛼𝐸_{𝑖𝑚𝑖𝑚}) is idempotent for arbitrary nonzero 𝛼 with 𝑦_𝑖𝑚+𝛼²≠0. Applying 𝜙 on it, we have 𝜏𝜙(𝐴)+𝜏𝜙(−𝐵+𝛼⁻¹𝑈𝑉+𝛼𝐸_{𝑖𝑚𝑖𝑚})∈Γ_𝑛. Let 𝐶=−𝐵+𝛼⁻¹𝑈𝑉+𝛼𝐸_{𝑖𝑚𝑖𝑚}; then by 𝜏𝐴+𝜏𝐶+𝜖∑_{𝑗∈⟨𝑛⟩⧵{𝑖1,…,𝑖𝑚}}𝐸_𝑗𝑗∈𝑆Γ_𝑛 for every 𝜖∈Λ, we have 𝜏𝜙(𝐴)+𝜙(𝜏𝐶+𝜖∑_{𝑗∈⟨𝑛⟩⧵{𝑖1,…,𝑖𝑚}}𝐸_𝑗𝑗)∈Γ_𝑛.
Note that 𝜙(𝜏𝐶+𝜖∑_{𝑗∈⟨𝑛⟩⧵{𝑖1,…,𝑖𝑚}}𝐸_𝑗𝑗)=𝜏𝐶+𝜖∑_{𝑗∈⟨𝑛⟩⧵{𝑖1,…,𝑖𝑚}}𝐸_𝑗𝑗 and 𝜙(𝐶)=𝐶 by the assumption; then 𝜏𝜙(𝐴)+𝜏𝜙(𝐶) and ∑_{𝑗∈⟨𝑛⟩⧵{𝑖1,…,𝑖𝑚}}𝐸_𝑗𝑗 are orthogonal by Corollary 10; that is, 𝜙(𝐴)=𝑌_{{𝑖1,…,𝑖𝑚}}+∑_{𝑗∈⟨𝑛⟩⧵{𝑖1,…,𝑖𝑚}}𝑥_𝑗𝐸_𝑗𝑗 for some 𝑌∈𝑀_𝑛.
On the other hand, 𝐶=−(𝑋_{{𝑖1,…,𝑖𝑚−1}}+∑_{𝑗∈⟨𝑛⟩⧵{𝑖1,…,𝑖𝑚−1}}𝑥_𝑗𝐸_𝑗𝑗)+𝛼⁻¹𝑈𝑉+𝛼𝐸_{𝑖𝑚𝑖𝑚}=−(𝑋_{{𝑖1,…,𝑖𝑚}}+∑_{𝑗∈⟨𝑛⟩⧵{𝑖1,…,𝑖𝑚}}𝑥_𝑗𝐸_𝑗𝑗)+𝛼⁻¹𝑈𝑉+𝛼𝐸_{𝑖𝑚𝑖𝑚}+𝑈+𝑉 implies 𝜏(𝑌_{{𝑖1,…,𝑖𝑚}}−𝑋_{{𝑖1,…,𝑖𝑚}}+𝛼⁻¹𝑈𝑉+𝛼𝐸_{𝑖𝑚𝑖𝑚}+𝑈+𝑉)=𝜏(𝑌_{{𝑖1,…,𝑖𝑚}}−𝑋_{{𝑖1,…,𝑖𝑚}})+𝜏(𝛼⁻¹𝑈𝑉+𝛼𝐸_{𝑖𝑚𝑖𝑚}+𝑈+𝑉)∈Γ_𝑛 by 𝜏𝜙(𝐴)+𝜏𝜙(𝐶)∈Γ_𝑛. By Lemma 5, we can derive the following equations: 𝑌_{{𝑖1,…,𝑖𝑚−1}}=𝑋_{{𝑖1,…,𝑖𝑚−1}},𝑌𝑖_𝑚,𝑖_𝑚=𝑋𝑖_𝑚,𝑖_𝑚,𝑌_{{𝑖1,…,𝑖𝑚−1},{𝑖𝑚}}𝑌=𝜂𝑈,_{{𝑖𝑚},{𝑖1,…,𝑖𝑚−1}}=𝜂⁻¹𝑉 (33) that is, 𝜙(𝐴)=𝑋_{{𝑖1,…,𝑖𝑚−1}}+𝜂𝑈+𝜂⁻¹𝑉+∑_{𝑗∈⟨𝑛⟩⧵{𝑖1,…,𝑖𝑚−1}}𝑥_𝑗𝐸_𝑗𝑗.
Let 𝐵₁ and 𝐵₂ satisfy the following equations: 𝐵₁=𝑋_{{𝑖2,…,𝑖𝑚}}+_{⟨𝑛⟩𝑗∈⧵{𝑖2,…,𝑖𝑚}}𝑥_𝑗𝐸_𝑗𝑗,𝐵₂=𝑋_{{𝑖1,…,𝑖𝑚−2,𝑖𝑚}}+_{⟨𝑛⟩𝑗∈⧵{𝑖1,…,𝑖𝑚−2,𝑖𝑚}}𝑥_𝑗𝐸_𝑗𝑗; (34) then we can prove 𝑌_{{𝑖2,…,𝑖𝑚}}=𝑋_{{𝑖2,…,𝑖𝑚}},𝑌𝑖₁,𝑖₁=𝑋𝑖₁,𝑖₁,𝑌_{{𝑖2,…,𝑖𝑚},{𝑖1}}=𝛽𝑋_{{𝑖2,…,𝑖𝑚},{𝑖1}},𝑌_{{𝑖1},{𝑖2,…,𝑖𝑚}}=𝛽⁻¹𝑋_{{𝑖1},{𝑖2,…,𝑖𝑚}},𝑌_{{𝑖1,…,𝑖𝑚−2,𝑖𝑚}}=𝑋_{{𝑖1,…,𝑖𝑚−2,𝑖𝑚}},𝑌𝑖_𝑚−1,𝑖_𝑚−1=𝑋𝑖_𝑚−1,𝑖_𝑚−1,𝑌_{{𝑖1,…,𝑖𝑚−2,𝑖𝑚},{𝑖𝑚−1}}=𝛾𝑋_{{𝑖1,…,𝑖𝑚−2,𝑖𝑚},{𝑖𝑚−1}},𝑌_{{𝑖𝑚−1},{𝑖1,…,𝑖𝑚−2,𝑖𝑚}}=𝛾⁻¹𝑋_{{𝑖𝑚−1},{𝑖1,…,𝑖𝑚−2,𝑖𝑚}}. (35)
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.