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Prove that a \(2 \times 2\) matrix \(A=\left[\begin{array}{ll}a_{11} & a_{12} \\\ a_{21} & a_{22}\end{array}\right]\) is reducible if and only if \(a_{12}=0\) or \(a_{21}=0\)

Short Answer

Expert verified
A 2x2 matrix \(A\) is reducible if and only if \(a_{12}=0\) or \(a_{21}=0\).

Step by step solution

01

Understanding Reducibility

A square matrix is said to be reducible if there exists a permutation matrix (that basically 'shuffles' rows and columns) such that the permutation of the matrix contains a zero block in the top-right or bottom-left corner. For a 2x2 matrix, this essentially means that one of the off-diagonal elements must be zero for it to be reducible.
02

Matrix Structure Analysis

The given matrix is: \[ A = \begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} \end{bmatrix} \]. Following the reducibility criteria, we are looking to find conditions where either \(a_{12} = 0\) or \(a_{21} = 0\).
03

Case 1: Zeroing Top-Right Element

When \(a_{12} = 0\), the matrix becomes \[ A = \begin{bmatrix} a_{11} & 0 \ a_{21} & a_{22} \end{bmatrix} \]. This can be seen as having a zero in the right block after permuting rows (essentially doing nothing here since zero is already top-right), meeting the reducibility criterion.
04

Case 2: Zeroing Bottom-Left Element

When \(a_{21} = 0\), the matrix becomes: \[ A = \begin{bmatrix} a_{11} & a_{12} \ 0 & a_{22} \end{bmatrix} \]. This shows a bottom-left zero block, which also meets the reducibility condition as it again matches the reduction through potential permutations.
05

Combined Conclusion

Thus, by analyzing these conditions, a \(2 \times 2\) matrix is reducible if and only if one of its off-diagonal elements, \(a_{12}\) or \(a_{21}\), is zero.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

2x2 matrix
A \(2 \times 2\) matrix is a simple type of matrix with two rows and two columns. This small matrix size is often the first example students encounter when learning about matrices. The general form of a \(2 \times 2\) matrix is \(\begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} \end{bmatrix}\), where each \(a_{ij}\) represents an element in the matrix. Here, the indices \(i\) and \(j\) denote the row and column positions of each element.

These matrices are fundamental in linear algebra, used to describe simple transformations, solve systems of linear equations, and serve as building blocks for larger matrices.
  • They are compact and easy to manipulate, making them ideal for illustrating key concepts like reducibility.
  • Operations such as addition, multiplication, and finding determinants are straightforward with \(2 \times 2\) matrices.
Understanding \(2 \times 2\) matrices is critical as they illustrate many foundational principles of matrix algebra, such as matrix transposition and inverses.
permutation matrix
A permutation matrix is a special type of matrix that arises from performing a permutation—basically a reordering—of rows or columns of another matrix. These matrices are square, meaning they have the same number of rows and columns, and consist solely of ones and zeros with exactly one "1" in each row and each column.

Pondering on their work:
  • Permutations are essential because they can rearrange the elements of a matrix without altering the fundamental properties of matrix algebra.
  • In the context of matrix reducibility, permutation matrices help in finding a form of the original matrix that contains a zero block, thereby proving reducibility.
Permutation matrices are essentially identity matrices that have had their rows or columns swapped around. They are crucial for computational purposes and make understanding and implementing matrix operations much more versatile.
off-diagonal elements
Off-diagonal elements in a matrix are those that are not located on its main diagonal. For a \(2 \times 2\) matrix \(\begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} \end{bmatrix}\), the elements \(a_{12}\) and \(a_{21}\) are considered off-diagonal.

These elements are critical to understanding matrix properties such as reducibility because:
  • In a diagonally dominant matrix, off-diagonal elements have lesser influence, but they are pivotal in other contexts.
  • For reducibility in a \(2 \times 2\) matrix, having one of these off-diagonal elements equal to zero allows the application of permutation matrices to transform the matrix.
Detecting zeros in these positions is a key diagnostic in proving whether a matrix is reducible, highlighting the importance of off-diagonal elements in matrix theory.
zero block
A zero block in a matrix is essentially a section where all the elements are zero. For a \(2 \times 2\) matrix, a zero block could appear as either the top-right or the bottom-left section.

Zero blocks are significant because:
  • They indicate that a matrix might be reducible, as having a zero block after permutation signifies that it's possible to partition the matrix into simpler forms.
  • In the context of a \(2 \times 2\) matrix, showing one off-diagonal element as zero creates a zero block, revealing the matrix's reducibility condition.
Understanding zero blocks is crucial in linear algebra because they help simplify complex problems by breaking them into smaller, more tractable parts. They also illustrate how certain properties of matrices can be detected visually, aiding in problem-solving and comprehension.

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Most popular questions from this chapter

The power method does not converge to the dominant eigenvalue and eigenvector. Verify this, using the given initial vector \(\mathbf{x}_{0}\). Compute the exact eigenvalues and eigenvectors and explain what is happening. $$A=\left[\begin{array}{rr} 2 & 1 \\ -2 & 5 \end{array}\right], \mathbf{x}_{0}=\left[\begin{array}{l} 1 \\ 1 \end{array}\right]$$

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It can be shown that a nonnegative \(n \times n\) matrix is irreducible if and only if \((I+A)^{n-1}>0 .\) Use this criterion to determine whether the matrix \(A\) is irreducible. If \(A\) is reducible, find a permutation of its rows and columns that puts \(A\) into the block form \\[ \left[\begin{array}{ll} B & C \\ O & D \end{array}\right] \\] $$A=\left[\begin{array}{lllll} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 0 & 0 \end{array}\right]$$

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