Least R R R For Which ( A + B ) R (A+B)^r ( A + B ) R Is A Null Matrix If A M = 0 A^m=0 A M = 0 And B N = 0 B^n=0 B N = 0 And A B = B A AB=BA A B = B A

Least $r$ for which $(A+B)^r$ is a null matrix if $A^m=0$ and $B^n=0$ and $AB=BA$

In the realm of matrix algebra, the concept of nilpotent matrices plays a crucial role in understanding various properties and behaviors of matrices. A nilpotent matrix is a square matrix $N$ such that $N^k = 0$ for some positive integer $k$. In this article, we will explore the problem of finding the least positive integer $r$ for which $(A+B)^r$ is a null matrix, given that $A^m=0$ and $B^n=0$ and $AB=BA$.

Let $A$ and $B$ be two square matrices of the same order such that $AB = BA$, $A^m = 0$ and $B^n = 0$ for some positive integers $m$ and $n$ with $\gcd(m, n) = 1$. What is the least positive integer $r$ for which $(A+B)^r$ is a null matrix?

Before diving into the problem, let's briefly discuss the properties of nilpotent matrices. A nilpotent matrix $N$ satisfies the condition $N^k = 0$ for some positive integer $k$. This means that when we raise the matrix $N$ to the power of $k$, the resulting matrix is a zero matrix. The smallest positive integer $k$ for which $N^k = 0$ is called the index of nilpotency of $N$.

Properties of Nilpotent Matrices

Nilpotent matrices have several interesting properties that will be useful in solving the problem. Some of these properties include:

• Nilpotent matrices are singular: A nilpotent matrix $N$ is singular, meaning that its determinant is zero.
• Nilpotent matrices have a zero eigenvalue: The eigenvalues of a nilpotent matrix $N$ are all zero.
• Nilpotent matrices commute with each other: If $N_1$ and $N_2$ are nilpotent matrices, then $N_1N_2 = N_2N_1$.

To find the least positive integer $r$ for which $(A+B)^r$ is a null matrix, we need to consider the properties of nilpotent matrices and the given conditions $A^m=0$ and $B^n=0$ and $AB=BA$.

Case 1: $m=n=1$

If $m=n=1$, then $A$ and $B$ are both nilpotent matrices with index of nilpotency 1. In this case, we can show that $(A+B)^r$ is a null matrix for any positive integer $r$.

Case 2: $m=1$ and $n>1$

If $m=1$ and $n>1$, then $A$ is a nilpotent matrix with index of nilpotency 1, and $B$ is a nilpotent matrix with index of nilpotency $n$. In this case, we can show that $(A+B)^r$ is a null matrix for $r \geq n$.

Case 3: $m>1$ and $n=1$

If $m>1$ and $n=1$, then $A$ is a nilpotent matrix with index of nilpotency $m$, and $B$ is a nilpotent matrix with index of nilpotency 1. In this case, we can show that $(A+B)^r$ is a null matrix for $r \geq m$.

Case 4: $m>1$ and $n>1$

If $m>1$ and $n>1$, then $A$ is a nilpotent matrix with index of nilpotency $m$, and $B$ is a nilpotent matrix with index of nilpotency $n$. In this case, we can show that $(A+B)^r$ is a null matrix for $r \geq \text{lcm}(m,n)$.

In conclusion, the least positive integer $r$ for which $(A+B)^r$ is a null matrix is given by:

• If $m=n=1$, then $r$ can be any positive integer.
• If $m=1$ and $n>1$, then $r \geq n$.
• If $m>1$ and $n=1$, then $r \geq m$.
• If $m>1$ and $n>1$, then $r \geq \text{lcm}(m,n)$.

The final answer is $\boxed{\text{lcm}(m,n)}$.
Q&A: Least $r$ for which $(A+B)^r$ is a null matrix if $A^m=0$ and $B^n=0$ and $AB=BA$

Q: What is the problem asking for?

A: The problem is asking for the least positive integer $r$ for which $(A+B)^r$ is a null matrix, given that $A^m=0$ and $B^n=0$ and $AB=BA$.

Q: What are the conditions given in the problem?

A: The conditions given in the problem are:

• $A$ and $B$ are two square matrices of the same order.
• $AB = BA$, meaning that $A$ and $B$ commute.
• $A^m = 0$ and $B^n = 0$ for some positive integers $m$ and $n$ with $\gcd(m, n) = 1$.

Q: What is the significance of $\gcd(m, n) = 1$?

A: The significance of $\gcd(m, n) = 1$ is that $m$ and $n$ are coprime, meaning that they have no common factors other than 1. This is important because it allows us to use the properties of coprime numbers to solve the problem.

Q: What is the relationship between $A$ and $B$?

A: The relationship between $A$ and $B$ is that they commute, meaning that $AB = BA$. This is an important property that we will use to solve the problem.

Q: How do we find the least positive integer $r$?

A: To find the least positive integer $r$, we need to consider the properties of nilpotent matrices and the given conditions $A^m=0$ and $B^n=0$ and $AB=BA$. We will use the properties of coprime numbers and the relationship between $A$ and $B$ to find the least positive integer $r$.

Q: What are the different cases that we need to consider?

A: There are four different cases that we need to consider:

• Case 1: $m=n=1$
• Case 2: $m=1$ and $n>1$
• Case 3: $m>1$ and $n=1$
• Case 4: $m>1$ and $n>1$

Q: How do we solve each case?

A: To solve each case, we need to use the properties of nilpotent matrices and the given conditions $A^m=0$ and $B^n=0$ and $AB=BA$. We will use the properties of coprime numbers and the relationship between $A$ and $B$ to find the least positive integer $r$ for each case.

Q: What is the final answer?

A: The final answer is $\boxed{\text{lcm}(m,n)}$.

Q: What is the significance of the final answer?

A: The significance of the final answer is that it gives us the least positive integer $r$ for which $(A+B)^r$ is a null matrix, given the conditions $A^m=0$ and $^n=0$ and $AB=BA$. This is an important result that can be used in various applications of matrix algebra.