Finding A 5x5 Matrix M Where M² = 0: A Detailed Guide

Aug 9, 2025 by ADMIN 54 views

Generalizing the Concept of A² = I₂ to Find a 5x5 Matrix M Where M² = 0

Hey everyone! Let's dive into a fascinating problem in linear algebra that involves generalizing the idea of a matrix squared equaling the identity matrix. We'll explore how to find a 5x5 matrix, M, such that M² = 0, and also deal with some block matrix concepts. This is a common topic in linear algebra, and hopefully, by breaking it down, we can make it super clear. So, grab your thinking caps, and let's get started!

Understanding the Problem

Before we jump into the solution, let’s make sure we fully grasp what the problem is asking. We're essentially extending the concept from 2x2 matrices to a 5x5 matrix. The initial idea revolves around a 2x2 matrix A such that A² equals the 2x2 identity matrix (I₂). Now, we want to find a 5x5 matrix M, where when you multiply M by itself (M²), you get the zero matrix. This is interesting because it tells us M is a nilpotent matrix.

Additionally, there's a twist: we need to consider a non-zero 2x3 matrix C. This C matrix will play a role in constructing our final M matrix. Block matrices are going to be our friends here, allowing us to build larger matrices from smaller ones.

Breaking Down the Key Concepts

Identity Matrix (I₂): A square matrix with ones on the main diagonal and zeros everywhere else. For a 2x2 matrix, it looks like this:
```
| 1  0 |
| 0  1 |
```
Zero Matrix: A matrix where all elements are zero.
Nilpotent Matrix: A matrix M is nilpotent if there exists a positive integer k such that M^k = 0. In our case, we want M² = 0, so M is nilpotent of order 2.
Block Matrix: A matrix that is interpreted as having been broken down into sections called blocks or submatrices.

Constructing the 5x5 Matrix M

The main challenge here is figuring out how to build a 5x5 matrix M that, when squared, results in the zero matrix. Let's use a block matrix approach. This involves dividing our 5x5 matrix into smaller blocks that we can work with more easily. We'll use the given 2x3 matrix C as a building block.

The Block Matrix Idea

We can construct M as follows:

M = | A  B |
    | C  D |

Where A, B, C, and D are submatrices. Since we're aiming for a 5x5 matrix, let's define the dimensions:

A: 3x2 matrix
B: 3x3 matrix
C: 2x2 matrix
D: 2x3 matrix

However, there's a slight issue with these dimensions. If A is 3x2 and C is 2x2, we can't directly use the given 2x3 matrix C in our construction. We need to tweak our block structure to accommodate the given C. Let's rethink our block arrangement.

A better approach might be:

M = | A  B |
    | C  D |

Where:

A: 3x3 matrix
B: 3x2 matrix
C: 2x3 matrix (the one given)
D: 2x2 matrix

Now, our dimensions add up correctly for a 5x5 matrix.

The M² = 0 Condition

For M² to be the zero matrix, we need to ensure that:

M² = M * M = | A  B | * | A  B | = | A² + BC    AB + BD |
             | C  D |   | C  D |   | CA + DC    CB + D² |

So, we need the following conditions to be met:

A² + BC = 0 (3x3 zero matrix)
AB + BD = 0 (3x2 zero matrix)
CA + DC = 0 (2x3 zero matrix)
CB + D² = 0 (2x2 zero matrix)

This looks complex, but we can simplify it by making strategic choices for our submatrices.

A Clever Choice: A = 0

To simplify things, let's make A the 3x3 zero matrix. This immediately eliminates A² and CA from our equations, leaving us with:

BC = 0
BD = 0
DC = 0
CB + D² = 0

This is more manageable! Now, we need to find B and D that satisfy these conditions, given our non-zero 2x3 matrix C.

Building B and D

Since C is a non-zero 2x3 matrix, let's keep it general for now:

C = | c11  c12  c13 |
    | c21  c22  c23 |

We need to find matrices B (3x2) and D (2x2) such that the conditions above hold.

Condition 3: DC = 0

Let's focus on DC = 0 first. If we choose D to be the 2x2 zero matrix, this condition is immediately satisfied. So, let's set D = 0.

Now our conditions simplify further:

BC = 0
BD = 0 (automatically satisfied since D = 0)
DC = 0 (already satisfied)
CB + D² = 0 => CB = 0 (since D = 0)

So, we're left with needing to find a 3x2 matrix B such that BC = 0 and CB = 0.

Finding B

Let's express B as a general 3x2 matrix:

B = | b11  b12 |
    | b21  b22 |
    | b31  b32 |

Now, we need to solve the systems BC = 0 and CB = 0. This involves matrix multiplication and setting each resulting element to zero.

BC = 0 gives us a set of equations:

| b11  b12 | * | c11  c12  c13 | = | 0  0  0 |
| b21  b22 |   | c21  c22  c23 |   | 0  0  0 |
| b31  b32 |                   | 0  0  0 |

CB = 0 gives us another set of equations:

| c11  c12  c13 | * | b11  b12 | = | 0  0 |
| c21  c22  c23 |   | b21  b22 |   | 0  0 |
                   | b31  b32 |

These equations can be a bit tedious to solve directly, but we can make a clever observation.

A Simple Solution for B

Since we need both BC = 0 and CB = 0, one way to ensure this is to choose B such that its columns are in the null space of the rows of C and vice versa. A simple way to achieve this is to make B a zero matrix as well. So, let's try B = 0.

The Final Matrix M

Now we have:

A = 0 (3x3 zero matrix)
B = 0 (3x2 zero matrix)
C = Non-zero 2x3 matrix (given)
D = 0 (2x2 zero matrix)

Our 5x5 matrix M becomes:

M = | 0  0 |
    | C  0 |

Where:

0 represents a zero matrix of the appropriate size.

Let's check if M² = 0:

M² = | 0  0 | * | 0  0 | = | 0*0 + 0*C   0*0 + 0*0 | = | 0  0 |
     | C  0 |   | C  0 |   | C*0 + 0*C   C*0 + 0*0 |   | 0  0 |

So, M² = 0, as required! We've successfully constructed a 5x5 matrix M that satisfies the condition.

Example

Let's consider a specific example to solidify our understanding. Suppose our non-zero 2x3 matrix C is:

C = | 1  2  3 |
    | 4  5  6 |

Then, our 5x5 matrix M would be:

M = | 0  0  0  0  0 |
    | 0  0  0  0  0 |
    | 0  0  0  0  0 |
    | 1  2  3  0  0 |
    | 4  5  6  0  0 |

It's easy to verify that M² = 0 for this example.

Conclusion

Guys, we did it! We've successfully generalized the concept of A² = I₂ to find a 5x5 matrix M such that M² = 0. We used the powerful technique of block matrices to break down the problem into manageable parts. By strategically choosing our submatrices, particularly setting A and D to zero matrices, we simplified the conditions for M² = 0 and were able to construct a suitable matrix B. Remember, the key is often in simplifying the problem by making smart choices and leveraging the properties of matrices. I hope this breakdown was helpful, and remember to keep exploring the fascinating world of linear algebra!