Let $\mathbb{C}$ be the field of complex numbers and $\mathbb{C^{\ast}}=\mathbb{C}-0$. For a positive integer $n$, let $\mathcal{M}_{n}$ be the set of all $n \times n $ complex matrices over $\mathbb{C}$. The symbols rank $(A)$, $A^{\ast}$, $\mathcal{R}(A)$, and $\mathcal{N}(A)$ stand for the rank, conjugate transpose, the range space, and the null space of $A\in \mathcal{M}_{n}$, respectively. Recall that a matrix $A\in \mathcal{M}_{n}$ is tripotent if $A^{3} = A $.

The nonsingularity of linear combinations of idempotent matrices and k-potent matrices was studied in, for example [1-4]. The nonsingularities of the combinations $c_{1}P+c_{2}Q-c_{3}PQ$ and $c_{1}P+c_{2}Q-c_{3}PQ-c_{4}QP-c_{5}PQP$ of two idempotent matrices $P, Q$ were investigated in [5] and [6], respectively. The considerations of this paper are inspired by Benítez et al. [7]. They established necessary and sufficient conditions for the nonsingularity of combinations $T=c_{1}T_{1}+c_{2}T_{2}+c_{3}T_{3}-c_{4}(T_{1}T_{2}+T_{3}T_{1}+T_{2}T_{3})$ of three trioptent matrices and gave some formulae for the inverse of $T=c_{1}T_{1}+c_{2}T_{2}+c_{3}T_{3}-c_{4}(T_{1}T_{2}+T_{3}T_{1}+T_{2}T_{3})$ under some conditions.

In this paper we consider a combination of the form

$ \begin{equation}\label{eq:1} T=c_{1}T_{1}+c_{2}T_{2}+c_{3}T_{3}+c_{4}T_{4}, \end{equation} $

(1.1)

where $c_{1}, c_{2}, c_{3}, c_{4}\in\mathbb{C^{\ast}}$ and $ T_{1}, T_{2}, T_{3}, T_{4}\in \mathcal{M}_{n}$ are three tripotent matrices. The purpose of this paper is mainly twofold: first, to establish necessary and sufficient conditions for the nonsingularity of combinations of form (1.1); second, to give some formulae for the inverse of them.

Now, let us give the following additional concepts and properties. A given matrix $A\in \mathcal{M}_{n}$ is said to be group invertible if there exists a matrix $X\in \mathcal{M}_{n}$ such that

$ \begin{equation}AXA = A; XAX = X; AX = XA \end{equation} $

(1.2)

hold. If such an X $\in \mathcal{M}_{n}$ exists, then it is unique, customarily denoted by $A^{\sharp}$ (see e.g. [8]). A matrix $A\in \mathcal{M}_{n}$ is group invertible if and only if there exist nonsingular $S\in \mathcal{M}_{n}$, $C\in \mathcal{M}_{r}$ such that $A = S (C\oplus 0) S^{-1}$, $r$ being the rank of $A$ (see [9], Exercise 5.10.12). In this situation, one has $A^{\sharp} = S (C^{-1}\oplus 0) S^{-1}$. This latter representation implies that any diagonalizable matrix is group invertible. Moreover, it is well known that A $\in \mathcal{M}_{n}$ is nonsingular if and only if $\mathcal{N}(A)={0}$. Furthermore, if $A\in \mathcal{M}_{n}$ and $k$ is a natural number greater than 1, then $A$ satisfies $ A^{k}= A $ if and only if $A$ is diagonalizable and the spectrum of $A$ is contained in $\sqrt[k-1]{1}\cup{0}$ (see e.g. [10]).

Special types of matrices, such as idempotents, tripotents, etc., are very useful in many contexts and they have been extensively studied in the literature. For example, quadratic forms with idempotent matrices are used extensively in statistical theory. So it is worth to stress and spread these kinds of results. Evidently, if $T$ is a tripotent matrix, then $T$ is group invertible and $T^{\sharp}= T$. Many of the results given in this work will be given in terms of group invertible matrices.

If $A\in \mathcal{M}_{n}$ satisfying $A^{2}=I_{n}$, We call $A$ an involutory matrix. On the inverse of linear combinations of involutory matrices, we have the following results.

Lemma 2.1  Let $A, B, C, D\in \mathcal{M}_{n}$ be involutory matrices and they are mutually commuting, $a, b, c, d\in\mathbb{C^{\ast}}$ and

$ \begin{eqnarray*}&& (a+b+c+d)(a-b+c+d)(a+b-c+d)(a+b+c-d)\\ && (a-b-c+d)(a-b+c-d)(a+b-c-d)(a-b-c-d)\neq 0, \end{eqnarray*} $

then

$ \begin{eqnarray}\label{eq:A}&& (aA+bB+cC+dD)^{-1}\nonumber\\ &=&\frac{1}{m}(x_{1}A+x_{2}B+x_{3}C+x_{4}D+z_{1}ABC+z_{2}ABD+z_{3}ACD+z_{4}BCD), \end{eqnarray} $

(2.1)

where

$ \begin{eqnarray*} m&=&ax_{1}+bx_{2}+cx_{3}+dx_{4}=(a+b+c+d)(a-b+c+d)\\ &&(a+b-c+d)\\ &&(a+b+c-d)(a-b-c+d)(a-b+c-d)(a+b-c-d)\\ &&(a-b-c-d), \\ z_{1}&=&2bc(a^{5}-2a^{3}b^{2}+ab^{4}-2a^{3}c^{2}-2ab^{2}c^{2}+ac^{4}+2a^{3}d^{2}\\ &&+2ab^{2}d^{2}+2ac^{2}d^{2}-3ad^{4}), \\ z_{2}&=&2bd(a^{5}-2a^{3}b^{2}+ab^{4}+2a^{3}c^{2}+2ab^{2}c^{2}-3ac^{4}-2a^{3}d^{2}\\ &&-2ab^{2}d^{2}+2ac^{2}d^{2}+ad^{4}), \\ z_{3}&=&2cd(a^{5}+2a^{3}b^{2}-3ab^{4}-2a^{3}c^{2}+2ab^{2}c^{2}+ac^{4}-2a^{3}d^{2}\\ &&+2ab^{2}d^{2}-2ac^{2}d^{2}+ad^{4}), \\ z_{4}&=&2bcd(-3a^{4}+2a^{2}b^{2}+b^{4}+2a^{2}c^{2}-2b^{2}c^{2}+c^{4}+2a^{2}d^{2}\\ &&-2b^{2}d^{2}-2c^{2}d^{2}+d^{4}), \\ x_{1}&=&-(-a^{7}+3a^{5}b^{2}-3a^{3}b^{4}+ab^{6}+3a^{5}c^{2}-2a^{3}b^{2}c^{2}\\ &&-ab^{4}c^{2}-3a^{3}c^{4}-ab^{2}c^{4}+ac^{6}+3a^{5}d^{2}\\ && -2a^{3}b^{2}d^{2}-ab^{4}d^{2}-2a^{3}c^{2}d^{2}+10ab^{2}c^{2}d^{2}\\ &&-ac^{4}d^{2}-3a^{3}d^{4}-ab^{2}d^{4}-ac^{2}d^{4}+ad^{6}), \\ x_{2}&=&-b(a^{6}-3a^{4}b^{2}+3a^{2}b^{4}-b^{6}-a^{4}c^{2}-2a^{2}b^{2}c^{2}\\ &&+3ab^{4}c^{2}-a^{2}c^{4}-3b^{2}c^{4}+c^{6}-a^{4}d^{2}\\ &&-2a^{2}b^{2}d^{2}+3b^{4}d^{2}+10a^{2}c^{2}d^{2}-2b^{2}c^{4}d^{2}\\ &&-c^{4}d^{2}-a^{2}d^{4}-3b^{2}d^{4}-c^{2}d^{4}+d^{6}), \\ x_{3}&=&-c(a^{6}-a^{4}b^{2}-a^{2}b^{4}+b^{6}-3a^{4}c^{2}-2a^{2}b^{2}c^{2}\\ &&-3b^{4}c^{2}+3a^{2}c^{4}+3b^{2}c^{4}-c^{6}-a^{4}d^{2}\\ && +10a^{2}b^{2}d^{2}-b^{4}d^{2}-2a^{2}c^{2}d^{2}-2b^{2}c^{2}d^{2}\\ &&+3c^{4}d^{2}-a^{2}d^{4}-b^{2}d^{4}-3c^{2}d^{4}+d^{6}), \\ x_{4}&=&d(-a^{6}+a^{4}b^{2}+a^{2}b^{4}-b^{6}+a^{4}c^{2}-10a^{2}b^{2}c^{2}\\ &&+b^{4}c^{2}+a^{2}c^{4}+b^{2}c^{4}-c^{6}+3a^{4}d^{2}+2a^{2}b^{2}d^{2}\\ && +3b^{4}d^{2}+2a^{2}c^{2}d^{2}+2b^{2}c^{2}d^{2}+3c^{4}d^{2}\\ &&-3a^{2}d^{4}-3b^{2}d^{4}-3c^{2}d^{4}+d^{6}).\end{eqnarray*} $

Corollary 2.1  Let $A, B, C\in \mathcal{M}_{n}$ be involutiory matrices and they are mutually commuting, $a, b, c\in\mathbb{C^{\ast}}$ and $(a+b+c)(a-b+c)(a+b-c)(a-b-c)\neq 0$, then

$ \begin{equation}\label{eq:B} (aA+bB+cC)^{-1}=\frac{1}{m}(xA+yB+zC+wABC), \end{equation} $

(2.2)

where

$ \begin{eqnarray*}&& m=ax_{1}+bx_{2}+cx_{3}+dx_{4}=(a+b+c)(a-b+c)(a+b-c)(a-b-c), \\ && x=a^{3}-ab^{2}-ac^{2}, y=b^{3}-bc^{2}-ba^{2}, z=c^{3}-ca^{2}-cb^{2}, w=2abc.\end{eqnarray*} $

Proof  In Lemma 2.1, put $D=0$ and $d=0$, we will obtain Corollary 2.1.

About group inverses of linear combinations of three tripotent matrices, we give the following Lemmas.

Lemma 2.2(see [7], Theorem 2.2)  Let $T_{1}, T_{2}, T_{3}\in \mathcal{M}_{n}\setminus \{0\}$ be three mutually commuting tripotent matrices and $c_{1}, c_{2}, c_{3}\in\mathbb{C^{\ast}}$ such that $(c_{i}+c_{j})(c_{i}-c_{j})\neq 0~ (i, j=1, 2, 3$ and $i\neq j)$ and

$ (c_{1}+c_{2}+c_{3})(c_{1}-c_{2}+c_{3})(c_{1}+c_{2}-c_{3})(c_{1}-c_{2}-c_{3})\neq 0, $

then

$ \mathcal{R}(T^{2}_{1}+T^{2}_{2}+T^{2}_{3})= \mathcal{R}(c_{1}T_{1}+c_{2}T_{2}+c_{3}T_{3}), \mathcal{N}(T^{2}_{1}+T^{2}_{2}+T^{2}_{3})\\ =\mathcal{N}(c_{1}T_{1}+c_{2}T_{2}+c_{3}T_{3}), $

$c_{1}T_{1}+c_{2}T_{2}+c_{3}T_{3}$ is group invertible, and

$ \begin{eqnarray} (c_{1}T_{1}+c_{2}T_{2}+c_{3}T_{3})^{\sharp}&=& q(T_{1}, T_{2}, T_{3})T^{2}_{1}T^{2}_{2}T^{2}_{3}+p_{c_{1}, c_{2}}(T_{1}, T_{2}) T^{2}_{1}T^{2}_{2}(I_{n}-T^{2}_{3})\nonumber\\ && \label{C} +p_{c_{1}, c_{3}}(T_{1}, T_{3})T^{2}_{1}(I_{n}-T^{2}_{2})+p_{c_{2}, c_{3}}(T_{2}, T_{3})(I_{n}-T^{2}_{1}), \end{eqnarray} $

(2.3)

where $p_{a, b}:\mathbb{C}^{2}\rightarrow \mathbb{C}$ and $q:\mathbb{C}^{3}\rightarrow \mathbb{C}$ are the following complex polynomials,

$ \begin{eqnarray}&& p_{a, b}(z, w)=\\ &&\frac{b^{2}}{a(a^{2}-b^{2})}zw^{2}+\frac{a^{2}}{b(b^{2}-a^{2})}z^{2}w+\frac{1}{a}z+\frac{1}{b}w~~(a, b\in\mathbb{C^{\ast}}, a^{2}\neq b^{2}), \nonumber\\ &&\label{D} q(z, w, u)=\\ &&\frac{(c^{3}_{1}-c_{1}c_{2}^{2}-c_{1}c_{3}^{2})z+(c^{3}_{2}-c_{2}c_{3}^{2}-c_{2}c_{1}^{2})w +(c^{3}_{3}-c_{3}c_{1}^{2}-c_{3}c_{2}^{2})u +2c_{1}c_{2}c_{3}zwu}{(c_{1}+c_{2}+c_{3})(c_{1}-c_{2}+c_{3})(c_{1}+c_{2}-c_{3})(c_{1}-c_{2}-c_{3})}.\nonumber\\ \end{eqnarray} $

(2.4)

In particular, if $T^{2}_{1}+T^{2}_{2}+T^{2}_{3}$ is nonsingular, then $c_{1}T_{1}+c_{2}T_{2}+c_{3}T_{3}$ is nonsingular and $(c_{1}T_{1}+c_{2}T_{2}+c_{3}T_{3})^{-1}$ is given by (2.3).

Lemma 2.3(see [7], Theorem 2.3)  Let $T_{1}, T_{2}, T_{3}\in \mathcal{M}_{n}$ be three mutually commuting tripotent matrices, then $T_{1}+T_{2}+T_{3}$ is nonsingular if and only if

$ I_{n}+T_{1}T_{2}+T_{2}T_{3}+T_{3}T_{1}+T_{1}T_{2}T_{3} $

and $T^{2}_{1}+T^{2}_{2}+T^{2}_{3}$ are nonsingular.

Now we give the nonsingularity and group inverses of linear combinations of four tripotent matrices. And we denote

$ \begin{eqnarray*}&& (aA+bB+cC+dD)^{-1}\\ &=&\frac{1}{m}(x_{1}A+x_{2}B+x_{3}C+x_{4}D+z_{1}ABC+z_{2}ABD+z_{3}ACD+z_{4}BCD)\\ &=&h_{a, b, c, d}(A, B, C, D)\end{eqnarray*} $

in (2.1), $(c_{1}T_{1}+c_{2}T_{2}+c_{3}T_{3})^{\sharp}=k_{c_{1}, c_{2}, c_{3}}(T_{1}, T_{2}, T_{3})$ in (2.3), $p_{a, b}(z, w)$ and $q(z, w, u)$ are the same in (2.4).

Theorem 2.1  Let $T_{1}, T_{2}, T_{3}, T_{4}\in \mathcal{M}_{n}\setminus \{0\}$ be four mutually commuting tripotent matrices and $c_{1}, c_{2}, c_{3}, c_{4}\in\mathbb{C^{\ast}}$ such that

$ \begin{eqnarray*}&& (c_{i}+c_{j})(c_{i}-c_{j})\neq 0~~ (i, j=1, 2, 3, 4~~{\rm and}~~i\neq j), \\ && (c_{i}+c_{j}+c_{k})(c_{i}-c_{j}+c_{k})(c_{i}+c_{j}-c_{k})(c_{i}-c_{j}-c_{k})\neq 0~~ \\ &&(i, j=1, 2, 3, 4 ~~{\rm and}~~i < j < k), \\ && (c_{1}+c_{2}+c_{3}+c_{4})(c_{1}-c_{2}+c_{3}+c_{4})(c_{1}+c_{2}-c_{3}+c_{4})\\ &&(c_{1}+c_{2}+c_{3}-c_{4})\\ && (c_{1}-c_{2}-c_{3}+c_{4}) (c_{1}-c_{2}+c_{3}-c_{4})(c_{1}+c_{2}-c_{3}-c_{4})\\ &&(c_{1}-c_{2}-c_{3}-c_{4})\neq 0, \end{eqnarray*} $

then

$ \begin{eqnarray*}&& \mathcal{R}(T^{2}_{1}+T^{2}_{2}+T^{2}_{3}+T^{2}_{4})= \mathcal{R}(c_{1}T_{1}+c_{2}T_{2}+c_{3}T_{3}+c_{4}T_{4}), \\ && \mathcal{N}(T^{2}_{1}+T^{2}_{2}+T^{2}_{3}+T^{2}_{4})=\mathcal{N}(c_{1}T_{1}+c_{2}T_{2}+c_{3}T_{3}+c_{4}T_{4}), \end{eqnarray*} $

$c_{1}T_{1}+c_{2}T_{2}+c_{3}T_{3}+c_{4}T_{4}$ is group invertible, and

$ \begin{eqnarray} && (c_{1}T_{1}+c_{2}T_{2}+c_{3}T_{3}+c_{4}T_{4})^{\sharp} = h_{c_{1}, c_{2}, c_{3}, c_{4}}(T_{1}, T_{2}, T_{3}, T_{4})T^{2}_{1}T^{2}_{2}T^{2}_{3}T^{2}_{4}\nonumber\\ && +q(T_{1}, T_{2}, T_{3})T^{2}_{1}T^{2}_{2}T^{2}_{3}(I_{n}-T^{2}_{4})+k_{c_{1}, c_{2}, c_{4}}(T_{1}, T_{2}, T_{4})T^{2}_{1}T^{2}_{2}(I_{n} -T^{2}_{3})\nonumber \\ && +k_{c_{1}, c_{3}, c_{4}}(T_{1}, T_{3}, T_{4})T^{2}_{1}(I_{n}-T^{2}_{2})+k_{c_{2}, c_{3}, c_{4}}(T_{2}, T_{3}, T_{4})(I_{n}-T^{2}_{1}). \label{E} \end{eqnarray} $

(2.5)

Proof  By (see [9], Exercise 5.10.12), there exist nonsingular matrices $S_{1}\in \mathcal{M}_{n}$ and $X_{1} \in \mathcal{M}_{n-t}$ such that $T_{1} = S_{1}(X_{1} \oplus 0)S_{1}^{-1}$. The tripotency of $T_{1}$ and the nonsingularity of $X_{1}$ lead to $X_{1}^{2}= I_{n-t}$ as $T_{1}T_{j} = T_{j}T_{1}, j=2, 3, 4.$ We can write matrices $T_{2}, T_{3}$ and $T_{4}$ as follows

$ \begin{eqnarray}\label{T_{2, 3, 4}}T_{2}=S_{1}\left(\begin{array}{cc} X_{2}&0\\0&E_{2}\end{array}\right)S_{1}^{-1}, \quad T_{3}=S_{1}\left(\begin{array}{cc} X_{3}&0\\0&E_{3}\end{array}\right)S_{1}^{-1}, \quad T_{4}=S_{1}\left(\begin{array}{cc} X_{4}&0\\0&E_{4}\end{array}\right)S_{1}^{-1} \end{eqnarray} $

(2.6)

with $E_{2}, E_{3}, E_{4}\in \mathcal{M}_{t}$, and

$ \begin{equation}\label{E_{2, 3, 4}} X_{i}X_{j}=X_{j}X_{i} \quad \quad (i, j=1, 2, 3, 4). \end{equation} $

(2.7)

Let us notice that matrices $X_{2}, X_{3}, X_{4}, E_{2}, E_{3}, E_{4}$ are tripotent because $T_{2}, T_{3} $ and $T_{4}$ are tripotent. By applying again (see [9], Exercise 5.10.12), there exist nonsingular matrices $S_{2} \in \mathcal{M}_{n-t}$ and $Y_{2} \in \mathcal{M}_{n-t-s}$ such that $X_{2} = S_{2}(Y_{2} \oplus 0)S_{2}^{-1}$, where $Y_{2}^{2}=I_{n-t-s}$. From (2.7) we can write

$ \begin{eqnarray}\label{X_{1, 3, 4}}X_{1}=S_{2}\left(\begin{array}{cc} Y_{1}&0\\0&D_{1}\end{array}\right)S_{2}^{-1}, \quad X_{3}=S_{2}\left(\begin{array}{cc} Y_{3}&0\\0&D_{3}\end{array}\right)S_{2}^{-1}, \quad X_{4}=S_{2}\left(\begin{array}{cc} Y_{4}&0\\0&D_{4}\end{array}\right)S_{2}^{-1} .\end{eqnarray} $

(2.8)

Observe that $Y_{1}^{2}=I_{n-t-s}, D_{1}^{2}=I_{s}, Y_{i}^{3}=Y_{i}$, $D_{i}^{3}=D_{i}, (i, j=3, 4)$ and

$ \begin{equation}\label{Y_} Y_{i}Y_{j}=Y_{j}Y_{i}~~ D_{i}D_{j}=D_{j}D_{i}~~ (i, j=1, 2, 3, 4). \end{equation} $

(2.9)

By applying again (see [9], Exercise 5.10.12), there exist nonsingular matrices $S_{3} \in \mathcal{M}_{n-t-s}$ and $Z_{3} \in \mathcal{M}_{n-t-s-r}$ such that $Y_{3} = S_{3}(Z_{3} \oplus 0)S_{3}^{-1}$, where $Z_{3}^{2}=I_{n-t-s-r}$. From (2.8) we can write

$ \begin{eqnarray}\label{Y_{1, 2, 4}}Y_{1}=S_{3}\left(\begin{array}{cc}Z_{1}&0\\0&C_{1}\end{array}\right)S_{3}^{-1}, \quad Y_{2}=S_{3}\left(\begin{array}{cc}Z_{2}&0\\0&C_{2}\end{array}\right)S_{3}^{-1}, \quad Y_{4}=S_{3}\left(\begin{array}{cc}Z_{4}&0\\0&C_{4}\end{array}\right)S_{3}^{-1} .\end{eqnarray} $

(2.10)

Observe that $Z_{1}^{2}=Z_{2}^{2}=I_{n-t-s-r}, C_{1}^{2}=C_{2}^{2}=I_{r}$, $Z_{4}^{3}=Z_{4}, C_{4}^{3}=C_{4}$and $ Z_{i}Z_{j}=Z_{j}Z_{i}$, $C_{i}C_{j}=C_{j}C_{i}~~ (i, j=1, 2, 3, 4). $

Finally, utilize again (see [9], Exercise 5.10.12) to matrix $Z_{4}$ to obtain nonsingular matrices $S_{4} \in \mathcal{M}_{n-t-s-r}$ and $A_{4} \in \mathcal{M}_{u}$ such that $Z_{4} = S_{4}(A_{4} \oplus 0)S_{4}^{-1}$ with $A_{4}^{2}=I_{u}$. By carrying out the same routine as before, we can write

$ \begin{eqnarray}\label{Z_{1, 2, 3}}Z_{1}=S_{4}\left(\begin{array}{cc}A_{1}&0\\0&B_{1}\end{array}\right)S_{4}^{-1}, Z_{2}=S_{4}\left(\begin{array}{cc}A_{2}&0\\0&B_{2}\end{array}\right)S_{4}^{-1}, Z_{3}=S_{4}\left(\begin{array}{cc}A_{3}&0\\0&B_{3}\end{array}\right)S_{4}^{-1}.\end{eqnarray} $

(2.11)

where $A_{i}^{2}=I_{n-t-s-r-u}$ , $B_{i}^{2}=I_{u}~~(i=1, 2, 3)$and $ A_{i}A_{j}=A_{j}A_{i}$, $B_{i}B_{j}=B_{j}B_{i} ~~ (i, j=1, 2, 3, 4).$

Let us define $m = n-t-r-u$. By setting $S = S_{1}(S_{2} \oplus I_{t})(S_{3} \oplus I_{s}\oplus I_{t})$ $(S_{4}\oplus I_{r}\oplus I_{s}\oplus I_{t})$, one easily has

$ \begin{eqnarray*}&& T_{1} = S(A_{1} \oplus B_{1} \oplus C_{1} \oplus D_{1}\oplus 0)S^{-1}, T_{2} = S(A_{2} \oplus B_{2} \oplus C_{2} \oplus 0 \oplus E_{2})S^{-1}, \\ && T_{3} = S(A_{3} \oplus B_{3} \oplus 0 \oplus D_{3}\oplus E_{3})S^{-1}, ~ T_{4} = S(A_{4} \oplus 0 \oplus C_{4} \oplus D_{4} \oplus E_{4})S^{-1}, \end{eqnarray*} $

and the matrices $A_{i}^{2}=I_{m}~~(i=1, 2, 3, 4)$; $B_{i}^{2}=I_{u}~~(i=1, 2, 3)$; $C_{i}^{2}=I_{r}~~ (i=1, 2), C_{4}^{3}=C_{4}$; $D_{1}^{2}=I_{s}, D_{i}^{3}=D_{i}(i=3, 4)$; $E_{i}^{3}=E_{i}~~(i=2, 3, 4)$. In addition, the families $\{A_{i}\}_{i=1, 2, 3, 4}$, $\{B_{i}\}_{i=1, 2, 3} $, $ \{C_{i}\}_{i=1, 2, 4}$, $\{D_{i}\}_{i=1, 3, 4}$, $\{E_{i}\}_{i=2, 3, 4}$ are commutative.

Observe that

$ \begin{equation}\label{T^{2}} T^{2}_{1} + T_{2}^{2} + T^{2}_{3}+ T^{2}_{4} = S[4I_{m}\oplus 3I_{u}\oplus(C_{1}^{2}+C_{2}^{2}+C_{4}^{2})\\ \oplus(D_{1}^{2}+D_{3}^{2}+D_{4}^{2})\oplus(E_{2}^{2}+E_{3}^{2}+E_{4}^{2})]S^{-1} \end{equation} $

(2.12)

and

$ \begin{eqnarray}&&c_{1}T_{1} + c_{2}T_{2} + c_{3}T_{3}+ c_{4}T_{4} = S[(c_{1}A_{1} + c_{2}A_{2} + c_{3}A_{3}+ c_{4}A_{4})\\ &&\oplus (c_{1}B_{1} + c_{2}B_{2} + c_{3}B_{3})\nonumber\\ &&\oplus \label{cT} (c_{1}C_{1} + c_{2}C_{2}+ c_{4}C_{4})\oplus(c_{1}D_{1} + c_{3}D_{3} + c_{4}D_{4})\\ &&\oplus(c_{2}E_{2} + c_{3}E_{3} + c_{4}E_{4})]S^{-1}. \end{eqnarray} $

(2.13)

By Lemma 2.1, Corollary 2.1 and Lemma 2.2, we have that $c_{1}A_{1} + c_{2}A_{2} + c_{3}A_{3}+ c_{4}A_{4}$ and $c_{1}B_{1} + c_{2}B_{2} + c_{3}B_{3}$ are nonsingular and

$ \begin{eqnarray*}&& (c_{1}A_{1} + c_{2}A_{2} + c_{3}A_{3}+ c_{4}A_{4})^{-1}=h_{c_{1}, c_{2}, c_{3}, c_{4}}(A_{1}, A_{2}, A_{3}, A_{4}), \\ && (c_{1}B_{1} + c_{2}B_{2} + c_{3}B_{3})^{-1}=q(B_{1}, B_{2} , B_{3}).\end{eqnarray*} $

Since $c_{1}A_{1} + c_{2}A_{2} + c_{3}A_{3}+ c_{4}A_{4}$ and $c_{1}B_{1} + c_{2}B_{2} + c_{3}B_{3}$ are nonsingular, then

$ \mathcal{N}(c_{1}A_{1} + c_{2}A_{2} + c_{3}A_{3}+ c_{4}A_{4})=\mathcal{N}(4I_{m}), \mathcal{N}(c_{1}B_{1} + c_{2}B_{2} + c_{3}B_{3})=\mathcal{N}(3I_{u}). $

Lemma 2.2 leads to

$ \begin{eqnarray*}&& \mathcal{N}(c_{1}C_{1} + c_{2}C_{2} + c_{4}C_{4})=\mathcal{N}(C_{1}^{2}+C_{2}^{2} + C_{4}^{2}), \\ && \mathcal{N}(c_{1}D_{1} + c_{3}D_{3} + c_{4}D_{4})=\mathcal{N}(D_{1}^{2}+D_{3}^{2} + D_{4}^{2}), \\ && \mathcal{N}(c_{2}E_{2} + c_{3}E_{3} + c_{4}E_{4})=\mathcal{N}(E_{2}^{2}+E_{3}^{2} + E_{4}^{2}), \end{eqnarray*} $

and analogous identities for the range space. By considering (2.12) and (2.13), and Lemma 2.2 we get that the null space (range space) of $c_{1}T_{1} + c_{2}T_{2} + c_{3}T_{3}+ c_{4}T_{4}$ equals to the null space (range space) $T^{2}_{1}+T^{2}_{2}+T^{2}_{3}+T^{2}_{4}$.

By Lemma 2.2 we have the group invertibility of $c_{1}C_{1} + c_{2}C_{2}+ c_{4}C_{4}$, $c_{1}D_{1} + c_{3}D_{3} + c_{4}D_{4}$, and $c_{2}E_{2} + c_{3}E_{3} + c_{4}E_{4}$, we get

$ \begin{eqnarray*}&& (c_{1}C_{1} + c_{2}C_{2}+ c_{4}C_{4})^{\sharp}=k_{c_{1}, c_{2}, c_{4}, }(C_{1}, C_{2}, C_{4}), \\ && (c_{1}D_{1} + c_{3}D_{3} + c_{4}D_{4})^{\sharp}=k_{c_{1}, c_{3}, c_{4}}(D_{1}, D_{3}, D_{4}), \\ && (c_{2}E_{2} + c_{3}E_{3} + c_{4}E_{4})^{\sharp}=k_{c_{2}, c_{3}, c_{4}}(E_{2}, E_{3}, E_{4}).\end{eqnarray*} $

The second part of Lemma 2.3 leads to the group invertibility of $ c_{1}T_{1} + c_{2}T_{2} + c_{3}T_{3}+ c_{4}T_{4} $ and

$ \begin{eqnarray}&& (c_{1}T_{1} + c_{2}T_{2} + c_{3}T_{3}+ c_{4}T_{4} )^{\sharp}=S(h_{c_{1}, c_{2}, c_{3}, c_{4}}(A_{1}, A_{2}, A_{3}, A_{4})\oplus q(B_{1}, B_{2}, B_{3}) \nonumber\\ && \label{cThk}\oplus k_{c_{1}, c_{2}, c_{4}, }(C_{1}, C_{2}, C_{4}) \oplus k_{c_{1}, c_{3}, c_{4}}(D_{1}, D_{3}, D_{4}) \oplus k_{c_{2}, c_{3}, c_{4}}(E_{2}, E_{3}, E_{4}))S^{-1}. \end{eqnarray} $

(2.14)

Now, observe that

$ \begin{eqnarray*}&& S(I_{m}\oplus 0 \oplus 0 \oplus 0 \oplus 0)S^{-1}=T^{2}_{1}T^{2}_{2}T^{2}_{3}T^{2}_{4}, \\ && S(0 \oplus I_{u} \oplus 0 \oplus 0 \oplus 0)S^{-1}=T^{2}_{1}T^{2}_{2}T^{2}_{3}(I_{n}-T^{2}_{4}), \\ && S(0 \oplus 0 \oplus I_{r} \oplus 0 \oplus 0)S^{-1}=T^{2}_{1}T^{2}_{2}(I_{n}-T^{2}_{3}), \\ && S(0 \oplus 0 \oplus 0 \oplus I_{s} \oplus 0)S^{-1}=T^{2}_{1}(I_{n}-T^{2}_{2}), \\ && S(0 \oplus 0 \oplus 0 \oplus 0 \oplus I_{t})S^{-1}=I_{n}-T^{2}_{1}. \end{eqnarray*} $

Thus we have

$ \begin{eqnarray}&& S(h_{c_{1}, c_{2}, c_{3}, c_{4}}(A_{1}, A_{2}, A_{3}, A_{4})\oplus q(B_{1}, B_{2}, B_{3}) \oplus 0 \oplus 0 \oplus 0)S^{-1}\nonumber\\ \label{hq}&=& h_{c_{1}, c_{2}, c_{3}, c_{4}}(T_{1}, T_{2}, T_{3}, T_{4})T^{2}_{1}T^{2}_{2}T^{2}_{3}T^{2}_{4}+q(T_{1}, T_{2}, T_{3})T^{2}_{1}T^{2}_{2}T^{2}_{3}(I_{n}-T^{2}_{4}), \end{eqnarray} $

(2.15)

$\begin{eqnarray} \label{k1}&& S(0 \oplus 0 \oplus k_{c_{1}, c_{2}, c_{4}, }(C_{1}, C_{2}, C_{4}) \oplus 0 \oplus 0)S^{-1}=k_{c_{1}, c_{2}, c_{4}, }(T_{1}, T_{2}, T_{4})T^{2}_{1}T^{2}_{2}(I_{n}-T^{2}_{3}), \end{eqnarray} $

(2.16)

$\begin{eqnarray} \label{k2} S(0 \oplus 0 \oplus 0 \oplus k_{c_{1}, c_{3}, c_{4}}(D_{1}, D_{3}, D_{4}) \oplus 0)S^{-1}=k_{c_{1}, c_{3}, c_{4}}(T_{1}, T_{3}, T_{4})T^{2}_{1}(I_{n}-T^{2}_{2}), \end{eqnarray} $

(2.17)

$\begin{eqnarray} \label{k3}S(0 \oplus 0 \oplus 0 \oplus 0 \oplus k_{c_{2}, c_{3}, c_{4}}(E_{2}, E_{3}, E_{4}))S^{-1}=k_{c_{2}, c_{3}, c_{4}}(T_{2}, T_{3}, T_{4})(I_{n}-T^{2}_{1}).\end{eqnarray} $

(2.18)

Considering (2.14)-(2.18) finishes the proof.

Corollary 2.2  Let $T_{1}, T_{2}, T_{3}, T_{4}\in \mathcal{M}_{n}\setminus \{0\}$ be four mutually commuting tripotent matrices and $c_{1}, c_{2}, c_{3}, c_{4}\in\mathbb{C^{\ast}}$, if $T^{2}_{1}+T^{2}_{2}+T^{2}_{3}+T^{2}_{4}$ is nonsingular, then $c_{1}T_{1}+c_{2}T_{2}+c_{3}T_{3}+c_{4}T_{4}$ is nonsingular and $(c_{1}T_{1}+c_{2}T_{2}+c_{3}T_{3}+c_{4}T_{4})^{-1}$ is given by (2.5).

Theorem 2.2  Let $T_{1}, T_{2}, T_{3}, T_{4}\in \mathcal{M}_{n}\setminus \{0\}$ be four mutually commuting tripotent matrices, then $T_{1}+T_{2}+T_{3}+T_{4}$ is nonsingular if and if only

$ I_{n}+T_{1}T_{2}+T_{1}T_{3}+T_{1}T_{4}+T_{2}T_{3}+T_{2}T_{4}+T_{3}T_{4}+T_{1}T_{2}T_{3}\\ +T_{1}T_{2}T_{4}+T_{1}T_{3}T_{4}+T_{2}T_{3}T_{4}+ T_{1}T_{2}T_{3}T_{4} $

and $T^{2}_{1}+T^{2}_{2}+T^{2}_{3}+T^{2}_{4}$ are nonsingular.

Proof  Since $T_{1}, T_{2}, T_{3}$ and $T_{4}$ are tripotent and mutually commutating, they are simultaneously diagonalizable (see, e.g. [11], p.52). Hence there is a single similarity matrix $ S \in \mathcal{M}_{n} $ such that $T_{1}=S {\rm diag}(\lambda_{1}, \cdots, \lambda_{n})S^{-1}, $ $T_{2}=S {\rm diag}(\mu_{1}, \cdots, \mu_{n})S^{-1}, $ $T_{3}=S {\rm diag}(\gamma_{1}, \cdots, \gamma_{n})S^{-1}$, $T_{4}=S {\rm diag}(\tau_{1}, \cdots, \tau_{n})S^{-1}$ being $\{\lambda_{i}\}_{i=1}^{n}$, $\{\mu_{i}\}_{i=1}^{n} $, $ \{\gamma_{i}\}_{i=1}^{n}$ and $\{\tau_{i}\}_{i=1}^{n}$ are the ordered sets of eigenvalues of $T_{1}, T_{2}, T_{3}$ and $T_{4}$ with proper multiplicities, respectively. On the other hand,

$ \begin{eqnarray}&& \label{Ti} T_{1}+T_{2}+T_{3}+T_{4}=S {\rm diag}(\lambda_{1}+\mu_{1}+\gamma_{1}+\tau_{1}, \cdots, \lambda_{n}+ \mu_{n}+\gamma_{n}+\tau_{n})S^{-1}\end{eqnarray} $

(2.19)

$\begin{eqnarray} &&I_{n}+T_{1}T_{2}+T_{1}T_{3}+T_{1}T_{4}+T_{2}T_{3}+T_{2}T_{4}+T_{3}T_{4}+T_{1}T_{2}T_{3}\nonumber\\ &&+T_{1}T_{2}T_{4}+T_{1}T_{3}T_{4}+T_{2}T_{3}T_{4}+ T_{1}T_{2}T_{3}T_{4}\nonumber\\ && \label{IT}= S {\rm diag}(g(\lambda_{1}+\mu_{1}+\gamma_{1}+\tau_{1}), \cdots, g(\lambda_{n}+ \mu_{n}+\gamma_{n}+\tau_{n}))S^{-1} \end{eqnarray} $

(2.20)

and

$ \begin{equation}\label{T2} T^{2}_{1}+T^{2}_{2}+T^{2}_{3}+T^{2}_{4}=S {\rm diag}(\lambda_{1}^{2}+\mu_{1}^{2}+\gamma_{1}^{2}+\tau_{1}^{2}, \cdots, \lambda_{n}^{2}+ \mu_{n}^{2}+\gamma_{n}^{2}+\tau_{n}^{2})S^{-1}, \end{equation} $

(2.21)

where $g: \mathbb{C}^{4}\rightarrow \mathbb{C}$ is given by

$ g(x, y, z, w) = 1 + xy + xz + xw + yz+yw+zw\\ +xyz+xyw+xzw+yzw+xyzw. $

Assume that $T_{1}+T_{2}+T_{3}+T_{4}$ is nonsingular. From (2.19) we get $\lambda_{i}+\mu_{i}+\gamma_{i}+\tau_{i}\neq 0$ for any $i=1, 2, \cdots, n $ and hence $(\lambda_{i}, \mu_{i}, \gamma_{i}, \tau_{i})\in \Phi^{4}\setminus \Omega$ for all $i=1, 2, \cdots, n $, where

$ \begin{eqnarray*}\Phi&=& \{-1, 0, 1\}, \\ \Omega&=& \{(0, 0, 1, -1), (0, 0, -1, 1), (0, 1, 0, -1), (0, -1, 0, 1), \\ && (0, 1, -1, 0), (0, -1, 1, 0), \\ && (1, 0, 0, -1), (-1, 0, 0, 1), (1, 0, -1, 0), (-1, 0, 1, 0), (1, -1, 0, 0), \\ && (-1, 1, 0, 0), (1, 1, -1, -1), \\ && (1, -1, 1, -1), (1, -1, -1, 1), (-1, 1, 1, -1), (-1, 1, -1, 1), \\ && (-1, -1, 1, 1), (0, 0, 0, 0)\}.\end{eqnarray*} $

Therefore, it is obtained that $g(\lambda_{i}+ \mu_{i}+\gamma_{i}+\tau_{i})\neq 0 $ and $\lambda_{i}^{2}+ \mu_{i}^{2}+\gamma_{i}^{2}+\tau_{i}^{2}\neq 0 $ for all $i=1, 2, \cdots, n $. In view of (2.20) and (2.21) it is seen that

$ \begin{eqnarray*}&& I_{n}+T_{1}T_{2}+T_{1}T_{3}+T_{1}T_{4}+T_{2}T_{3}+T_{2}T_{4}+T_{3}T_{4}\\ &&+T_{1}T_{2}T_{3}+T_{1}T_{2}T_{4}+T_{1}T_{3}T_{4}+T_{2}T_{3}T_{4}+ T_{1}T_{2}T_{3}T_{4}\end{eqnarray*} $

and $T^{2}_{1}+T^{2}_{2}+T^{2}_{3}+T^{2}_{4}$ are nonsingular.

Now, assume that

$ \begin{eqnarray*}&& I_{n}+T_{1}T_{2}+T_{1}T_{3}+T_{1}T_{4}+T_{2}T_{3}+T_{2}T_{4}+T_{3}T_{4}\\ &&+T_{1}T_{2}T_{3}+T_{1}T_{2}T_{4}+T_{1}T_{3}T_{4}+T_{2}T_{3}T_{4}+ T_{1}T_{2}T_{3}T_{4}\end{eqnarray*} $

and $T^{2}_{1}+T^{2}_{2}+T^{2}_{3}+T^{2}_{4}$ are nonsingular. From the nonsingularity of the first matrix, we get

$ 1+\lambda_{i}\mu_{i}+\lambda_{i}\gamma_{i}+\lambda_{i}\tau_{i}+\mu_{i}\gamma_{i}+\mu_{i}\tau_{i}+\gamma_{i}\tau_{i}+ \lambda_{i}\mu_{i}\gamma_{i}+\lambda_{i}\mu_{i}\tau_{i}\\ +\mu_{i}\gamma_{i}\tau_{i}+\lambda_{i}\mu_{i}\gamma_{i}\tau_{i}\neq 0 $

for all $i=1, 2, \cdots, n $. If $T_{1}+T_{2}+T_{3}+T_{4}$ were singular, then there would exist some $j\in \{1, 2, \cdots, n\}$ such that $\lambda_{i}+ \mu_{i}+\gamma_{i}+\tau_{i}= 0$. So, the unique solution satisfying simultaneously these two equations would be $(\lambda_{i}, \mu_{i}, \gamma_{i}, \tau_{i})= (0, 0, 0, 0)$. Hence

$ \lambda_{i}^{2}+ \mu_{i}^{2}+\gamma_{i}^{2}+\tau_{i}^{2}= 0, $

which would contradict to the assumption of the nonsingularity of

$ T^{2}_{1}+T^{2}_{2}+T^{2}_{3}+T^{2}_{4}. $

So the proof is completed.