# The Nullity, Rank, and Invertibility of Linear Combinations of k-Potent Matrices

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^{2}

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## Abstract

**:**

## 1. Introduction

## 2. The Nullity and Rank of a Linear Combination of Two k-Potent Matrices

**Theorem**

**1.**

**Proof.**

**Corollary**

**1.**

**Proof.**

**Corollary**

**2.**

**Proof.**

**Remark**

**1.**

## 3. Invertibility of A Linear Combination of Two k-Potent Matrices

**Theorem**

**2.**

- (i)
- If $AB={A}^{s},$$s\in \{0,1,3,\dots ,k-2\},$ and ${c}_{1}^{k-1}+{(-1)}^{k}{c}_{2}^{k-1}\ne 0,$ then ${c}_{1}A+{c}_{2}B$ is nonsingular if and only if ${A}^{k-1}+B({I}_{n}-{A}^{k-1})$ is nonsingular. Furthermore,$$\begin{array}{ccc}& & {({c}_{1}A+{c}_{2}B)}^{-1}={A}_{1}-{B}^{k-2}{({I}_{n}-{A}^{k-1})}^{2}B{A}^{k-1}{A}_{1}\hfill \\ & & \phantom{{({c}_{1}A+{c}_{2}B)}^{-1}=}+\frac{1}{{c}_{2}}{B}^{k-2}({I}_{n}-{A}^{k-1}),\hfill \end{array}$$$${A}_{1}=\frac{1}{{c}_{1}^{k-1}+{(-1)}^{k}{c}_{2}^{k-1}}\sum _{i=0}^{k-2}{(-1)}^{i}{c}_{1}^{k-2-i}{c}_{2}^{i}{A}^{k-2-i+(s-1)i}.$$
- (ii)
- If $AB={A}^{2}$ and ${c}_{1}+{c}_{2}\ne 0,$ then ${c}_{1}A+{c}_{2}B$ is nonsingular if and only if ${A}^{k-1}+B({I}_{n}-{A}^{k-1})$ is nonsingular. Furthermore,$$\begin{array}{ccc}& & {({c}_{1}A+{c}_{2}B)}^{-1}=\frac{1}{{c}_{1}+{c}_{2}}{A}^{k-2}-\frac{1}{{c}_{1}+{c}_{2}}{B}^{k-2}{({I}_{n}-{A}^{k-1})}^{2}B{A}^{k-1}{A}^{k-2}\hfill \\ & & \phantom{{({c}_{1}A+{c}_{2}B)}^{-1}=}+\frac{1}{{c}_{2}}{B}^{k-2}({I}_{n}-{A}^{k-1}).\hfill \end{array}$$

**Proof.**

- (i)
- By $AB={A}^{s}$, $s\in \{0,1,3,\dots ,k-2\}$, it follows that B has the form:$$B=U\left[\begin{array}{cc}{K}^{s-1}& 0\\ E& G\end{array}\right]{U}^{-1},$$$${B}^{m}=U\left[\begin{array}{cc}{K}^{m(s-1)}& 0\\ {E}_{m}& {G}^{m}\end{array}\right]{U}^{-1},\phantom{\rule{5.0pt}{0ex}}\forall m\in \mathbb{N}.$$Since ${B}^{k}=B$, we get that ${G}^{k}=G$. Thus, $G\in {\mathbb{C}}^{(n-r)\times (n-r)}$ is the k-potent matrix. Now, ${c}_{1}A+{c}_{2}B$ has the form:$${c}_{1}A+{c}_{2}B=U\left[\begin{array}{cc}{c}_{1}K+{c}_{2}{K}^{s-1}& 0\\ {c}_{2}E& {c}_{2}G\end{array}\right]{U}^{-1}.$$Using $\left(3\right)$, we have:$$\begin{array}{ccc}& & ({c}_{1}K+{c}_{2}{K}^{s-1})\sum _{i=0}^{k-2}{(-1)}^{i}{c}_{1}^{k-2-i}{c}_{2}^{i}{K}^{k-2-i}{\left({K}^{s-1}\right)}^{i}\hfill \\ & & ={c}_{1}^{k-1}{K}^{k-1}+{(-1)}^{k}{c}_{2}^{k-1}{K}^{(s-1)(k-1)}\hfill \\ & & =({c}_{1}^{k-1}+{(-1)}^{k}{c}_{2}^{k-1}){I}_{r}\hfill \end{array}$$$${({c}_{1}K+{c}_{2}{K}^{s-1})}^{-1}=\frac{1}{{c}_{1}^{k-1}+{(-1)}^{k}{c}_{2}^{k-1}}\sum _{i=0}^{k-2}{(-1)}^{i}{c}_{1}^{k-2-i}{c}_{2}^{i}{K}^{k-2-i+(s-1)i}.$$Therefore, the linear combination ${c}_{1}K+{c}_{2}{K}^{s}$ is invertible for all constants ${c}_{1},{c}_{2}\in \mathbb{C}\backslash \left\{0\right\}$ such that ${c}_{1}^{k-1}+{(-1)}^{k}{c}_{2}^{k-1}\ne 0.$ Now, we conclude that ${c}_{1}A+{c}_{2}B$ is nonsingular if and only if G is nonsingular. Since:$${A}^{k-1}=U\left[\begin{array}{cc}{I}_{r}& 0\\ 0& 0\end{array}\right]{U}^{-1},$$$$B({I}_{n}-{A}^{k-1})=U\left[\begin{array}{cc}0& 0\\ 0& G\end{array}\right]{U}^{-1}$$$${A}^{k-1}+B({I}_{n}-{A}^{k-1})=U\left[\begin{array}{ll}{I}_{r}& 0\\ 0& 0\end{array}\right]{U}^{-1},$$$${({c}_{1}A+{c}_{2}B)}^{-1}=U\left[\begin{array}{cc}{({c}_{1}K+{c}_{2}{K}^{s-1})}^{-1}& 0\\ -{G}^{-1}E{({c}_{1}K+{c}_{2}{K}^{s-1})}^{-1}& {c}_{2}^{-1}{G}^{-1}\end{array}\right]{U}^{-1},$$$${({c}_{1}A+{c}_{2}B)}^{-1}=U\left[\begin{array}{cc}{({c}_{1}K+{c}_{2}{K}^{s-1})}^{-1}& 0\\ -{G}^{k-2}E{({c}_{1}K+{c}_{2}{K}^{s-1})}^{-1}& {c}_{2}^{-1}{G}^{k-2}\end{array}\right]{U}^{-1},$$$${B}^{k-2}({I}_{n}-{A}^{k-1})=U\left[\begin{array}{cc}0& 0\\ 0& {G}^{k-2}\end{array}\right]{U}^{-1}$$$$({I}_{n}-{A}^{k-1})B{A}^{k-1}=U\left[\begin{array}{cc}0& 0\\ E& 0\end{array}\right]{U}^{-1}.$$Thus, the inverse ${({c}_{1}A+{c}_{2}B)}^{-1}$ has the form $\left(4\right)$.
- (ii)
- From $AB={A}^{2},$ we can conclude that B has the form:$$B=U\left[\begin{array}{cc}K& 0\\ E& G\end{array}\right]{U}^{-1},$$$${B}^{m}=U\left[\begin{array}{cc}{K}^{m}& 0\\ {E}_{m}& {G}^{m}\end{array}\right]{U}^{-1},\phantom{\rule{5.0pt}{0ex}}\forall m\in \mathbb{N}.$$Since ${B}^{k}=B,$ we get that ${G}^{k}=G.$ Hence, $G\in {\mathbb{C}}^{(n-r)\times (n-r)}$ is the k-potent matrix. Therefore,$${c}_{1}A+{c}_{2}B=U\left[\begin{array}{cc}({c}_{1}+{c}_{2})K& 0\\ {c}_{2}E& {c}_{2}G\end{array}\right]{U}^{-1}.$$Since $({c}_{1}+{c}_{2})K$ is nonsingular for all ${c}_{1},{c}_{2}\in \mathbb{C}\backslash \left\{0\right\}$ such that ${c}_{1}+{c}_{2}\ne 0,$ then ${c}_{1}A+{c}_{2}B$ is nonsingular if and only if G is nonsingular.Analogously, as in Part $\left(i\right),$ we conclude that ${A}^{k-1}+B({I}_{n}-{A}^{k-1})$ has the form $\left(7\right)$ and that ${A}^{k-1}+B({I}_{n}-{A}^{k-1})$ is nonsingular if and only if G is nonsingular. Thus, the necessary and sufficient condition for the invertibility of ${c}_{1}A+{c}_{2}B$ is the invertibility of ${A}^{k-1}+B({I}_{n}-{A}^{k-1}).$Furthermore,$${({c}_{1}A+{c}_{2}B)}^{-1}=U\left[\begin{array}{cc}{({c}_{1}+{c}_{2})}^{-1}{K}^{-1}& 0\\ -{G}^{-1}E{({c}_{1}+{c}_{2})}^{-1}{K}^{-1}& {c}_{2}^{-1}{G}^{-1}\end{array}\right]{U}^{-1},$$

**Theorem**

**3.**

- (i)
- If $BA={A}^{s},$$s\in \{0,1,3,\dots ,k-2\},$ and ${c}_{1}^{k-1}+{(-1)}^{k}{c}_{2}^{k-1}\ne 0,$ then ${c}_{1}A+{c}_{2}B$ is nonsingular if and only if ${A}^{k-1}+({I}_{n}-{A}^{k-1})B$ is nonsingular. Furthermore,$$\begin{array}{c}\hfill {({c}_{1}A+{c}_{2}B)}^{-1}={A}_{1}-{A}_{1}{A}^{k-1}B{({I}_{n}-{A}^{k-1})}^{2}{B}^{k-2}+\frac{1}{{c}_{2}}({I}_{n}-{A}^{k-1}){B}^{k-2},\end{array}$$where:$${A}_{1}=\frac{1}{{c}_{1}^{k-1}+{(-1)}^{k}{c}_{2}^{k-1}}\sum _{i=0}^{k-2}{(-1)}^{i}{c}_{1}^{k-2-i}{c}_{2}^{i}{A}^{k-2-i+(s-1)i}.$$
- (ii)
- If $BA={A}^{2}$ and ${c}_{1}+{c}_{2}\ne 0,$ then ${c}_{1}A+{c}_{2}B$ is nonsingular if and only if ${A}^{k-1}+({I}_{n}-{A}^{k-1})B$ is nonsingular. Furthermore,$$\begin{array}{ccc}& & {({c}_{1}A+{c}_{2}B)}^{-1}=\frac{1}{{c}_{1}+{c}_{2}}{A}^{k-2}-\frac{1}{{c}_{1}+{c}_{2}}{A}^{k-2}{A}^{k-1}B{({I}_{n}-{A}^{k-1})}^{2}{B}^{k-2}\hfill \\ & & \phantom{{({c}_{1}A+{c}_{2}B)}^{-1}=}+\frac{1}{{c}_{2}}({I}_{n}-{A}^{k-1}){B}^{k-2}.\hfill \end{array}$$

**Proof.**

**Lemma**

**1.**

**Proof.**

**Theorem**

**4.**

**Proof.**

**Theorem**

**5.**

**Proof.**

**Corollary**

**3.**

**Remark**

**2.**

## 4. Invertibility of a Linear Combination of Three k-Potent Matrices

**Theorem**

**6.**

- (i)
- If $AB={A}^{s},$$s\in \{0,1,3,\dots ,k-2\},$ and ${c}_{1}^{k-1}+{(-1)}^{k}{c}_{2}^{k-1}\ne 0,$ then ${c}_{1}A+{c}_{2}B+{c}_{3}C$ is nonsingular if and only if ${A}^{k-1}+(B+C)({I}_{n}-{A}^{k-1})$ is nonsingular. Furthermore,$$\begin{array}{ccc}& & {({c}_{1}A+{c}_{2}B)}^{-1}={A}_{1}-\frac{1}{{c}_{2}}{\left[({c}_{2}B+{c}_{3}C)({I}_{n}-{A}^{k-1})\right]}^{\u266f}({I}_{n}-{A}^{k-1})B{A}^{k-1}{A}_{1}\hfill \\ & & \phantom{{({c}_{1}A+{c}_{2}B)}^{-1}=}+{\left[({c}_{2}B+{c}_{3}C)({I}_{n}-{A}^{k-1})\right]}^{\u266f},\hfill \end{array}$$$${A}_{1}=\frac{1}{{c}_{1}^{k-1}+{(-1)}^{k}{c}_{2}^{k-1}}\sum _{i=0}^{k-2}{(-1)}^{i}{c}_{1}^{k-2-i}{c}_{2}^{i}{A}^{k-2-i+(s-1)i}.$$
- (ii)
- If $AB={A}^{2},$ then ${c}_{1}A+{c}_{2}B+{c}_{3}C$ is nonsingular if and only if ${A}^{k-1}+(B+C)({I}_{n}-{A}^{k-1})$ is nonsingular. Furthermore,$$\begin{array}{ccc}& & {({c}_{1}A+{c}_{2}B)}^{-1}=\frac{1}{{c}_{1}+{c}_{2}}{A}^{k-2}\hfill \\ & & -\frac{1}{{c}^{2}({c}_{1}+{c}_{2})}{\left[({c}_{2}B+{c}_{3}C)({I}_{n}-{A}^{k-1})\right]}^{\u266f}({I}_{n}-{A}^{k-1})B{A}^{k-1}{A}^{k-2}\hfill \\ & & +{\left[({c}_{2}B+{c}_{3}C)({I}_{n}-{A}^{k-1})\right]}^{\u266f}.\hfill \end{array}$$

**Proof.**

- (i)
- Since $AB={A}^{s},$$s\in \{0,1,3,\dots ,k-2\},$ we can write B as in $\left(6\right).$ We note that ${c}_{1}A+{c}_{2}B+{c}_{3}C$ can be represented as:$${c}_{1}A+{c}_{2}B+{c}_{3}C=U\left[\begin{array}{cc}{c}_{1}K+{c}_{2}{K}^{s-1}& 0\\ {c}_{2}E& {c}_{2}G+{c}_{3}T\end{array}\right]{U}^{-1},$$Therefore, the linear combination ${c}_{1}A+{c}_{2}B+{c}_{3}C$ is nonsingular if and only if ${c}_{2}G+{c}_{3}T$ is nonsingular. Since $G,T\in {\mathbb{C}}^{(n-r)\times (n-r)}$ are commuting k-potent matrices, we deduce that ${c}_{2}G+{c}_{3}T$ is nonsingular if and only if $G+T$ is nonsingular for all constants ${c}_{2},{c}_{3}\in \mathbb{C}\backslash \left\{0\right\}$ such that ${c}_{2}^{k-1}+{(-1)}^{k}{c}_{3}^{k-1}\ne 0$ by Corollary 3. Furthermore,$${A}^{k-1}+(B+C)({I}_{n}-{A}^{k-1})=U\left[\begin{array}{cc}{I}_{r}& 0\\ 0& G+T\end{array}\right]{U}^{-1},$$By a direct computation, we get:$${({c}_{1}A+{c}_{2}B+{c}_{3}C)}^{-1}=U\left[\begin{array}{cc}{({c}_{1}K+{c}_{2}{K}^{s-1})}^{-1}& 0\\ -{c}_{2}{({c}_{2}G+{c}_{3}T)}^{-1}E{({c}_{1}K+{c}_{2}{K}^{s-1})}^{-1}& {({c}_{2}G+{c}_{3}T)}^{-1}\end{array}\right]{U}^{-1},$$$$({c}_{2}B+{c}_{3}C)({I}_{n}-{A}^{k-1})=U\left[\begin{array}{cc}0& 0\\ 0& {c}_{2}G+{c}_{3}T\end{array}\right]{U}^{-1}$$$$({I}_{n}-{A}^{k-1})B{A}^{k-1}=U\left[\begin{array}{cc}0& 0\\ E& 0\end{array}\right]{U}^{-1}.$$It is noteworthy that the group inverse of $({c}_{2}B+{c}_{3}C)({I}_{n}-{A}^{k-1})$ is:$${\left[({c}_{2}B+{c}_{3}C)({I}_{n}-{A}^{k-1})\right]}^{\u266f}=U\left[\begin{array}{cc}0& 0\\ 0& {({c}_{2}G+{c}_{3}T)}^{-1}\end{array}\right]{U}^{-1}.$$Hence, the formula $\left(20\right)$ holds.
- (ii)
- The proof is similar to the one in Item $\left(i\right).$ We use the form $\left(8\right)$ for $B\in {\mathbb{C}}^{n\times n}.$ Now,$${c}_{1}A+{c}_{2}B+{c}_{3}C=U\left[\begin{array}{cc}({c}_{1}+{c}_{2})K& 0\\ {c}_{2}E& {c}_{2}G+{c}_{3}T\end{array}\right]{U}^{-1},$$$${({c}_{1}A+{c}_{2}B+{c}_{3}C)}^{-1}=U\left[\begin{array}{cc}{({c}_{1}+{c}_{2})}^{-1}{K}^{-1}& 0\\ -{c}_{2}{({c}_{1}+{c}_{2})}^{-1}{({c}_{2}G+{c}_{3}T)}^{-1}E{K}^{-1}& {({c}_{2}G+{c}_{3}T)}^{-1}\end{array}\right]{U}^{-1},$$

**Theorem**

**7.**

**Proof.**

**Theorem**

**8.**

**Proof.**

**Remark**

**3.**

## 5. Conclusions

## Author Contributions

## Funding

## Acknowledgments

## Conflicts of Interest

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**MDPI and ACS Style**

Tošić, M.; Ljajko, E.; Kontrec, N.; Stojanović, V.
The Nullity, Rank, and Invertibility of Linear Combinations of *k*-Potent Matrices. *Mathematics* **2020**, *8*, 2147.
https://doi.org/10.3390/math8122147

**AMA Style**

Tošić M, Ljajko E, Kontrec N, Stojanović V.
The Nullity, Rank, and Invertibility of Linear Combinations of *k*-Potent Matrices. *Mathematics*. 2020; 8(12):2147.
https://doi.org/10.3390/math8122147

**Chicago/Turabian Style**

Tošić, Marina, Eugen Ljajko, Nataša Kontrec, and Vladica Stojanović.
2020. "The Nullity, Rank, and Invertibility of Linear Combinations of *k*-Potent Matrices" *Mathematics* 8, no. 12: 2147.
https://doi.org/10.3390/math8122147