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Chapter 6: Problem 10

Prove that a \(2 \times 2\) matrix \(A\) is reducible if and only if \(a_{12} a_{21}=0\)

Short Answer

Expert verified
A \(2 \times 2\) matrix \(A\) is reducible if and only if \(a_{12} a_{21} = 0\). To prove this, we first showed that if \(A\) is reducible, then \(a_{12} a_{21} = 0\). We demonstrated this by finding an elementary matrix \(E\) whose product with \(A\) resulted in a matrix with a zero in the first row or first column. Next, we proved that if \(a_{12} a_{21} = 0\), then \(A\) is reducible, by showing two separate cases where either \(a_{12} = 0\) or \(a_{21} = 0\) or both. In both cases, we found an elementary matrix \(E\) such that the product \(EA\) had a zero in the first row or first column, concluding that \(A\) is reducible.

Step by step solution

01

Define reducible matrix and the given condition

A \(2 \times 2\) matrix \(A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}\) is reducible if there exists an elementary matrix \(E\) such that the product \(EA\) has a zero in either the first row or the first column. The given condition states that the product of the two off-diagonal elements of \(A\) is zero, that is \(a_{12} a_{21} = 0\).
02

Prove that if \(A\) is reducible, then \(a_{12} a_{21} = 0\)

Given a reducible matrix \(A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}\), we need to find an elementary matrix \(E\) such that \(EA\) has a zero either in the first row or the first column. Let \(E = \begin{pmatrix} 1 & -\frac{a_{12}}{a_{11}} \\ 0 & 1 \end{pmatrix}\) then, \(EA = \begin{pmatrix} 1 & -\frac{a_{12}}{a_{11}} \\ 0 & 1 \end{pmatrix} \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} = \begin{pmatrix} a_{11} - \frac{a_{12}a_{21}}{a_{11}} & 0 \\ a_{21} & a_{22} \end{pmatrix}\) Since \(A\) is reducible, it must have a zero in the first position of either the first row or first column. Here, the first position of the first row is \(\left(a_{11} - \frac{a_{12}a_{21}}{a_{11}}\right)\). For this expression to be zero, \(a_{12} a_{21} = 0\).
03

Prove that if \(a_{12} a_{21} = 0\), then \(A\) is reducible

Given a \(2 \times 2\) matrix \(A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}\) with the condition \(a_{12} a_{21} = 0\). We know that either \(a_{12} = 0\) or \(a_{21} = 0\) or both. Case 1: If \(a_{12} = 0\), then the matrix \(E = \begin{pmatrix} 1 & 0 \\ -\frac{a_{21}}{a_{11}} & 1 \end{pmatrix}\) is an elementary matrix such that \(EA = \begin{pmatrix} a_{11} & 0 \\ 0 & a_{22} - \frac{a_{21}^2}{a_{11}} \end{pmatrix}\). Matrix \(EA\) has a zero in the first column. Case 2: If \(a_{21} = 0\), then the matrix \(E = \begin{pmatrix} 1 & -\frac{a_{12}}{a_{11}} \\ 0 & 1 \end{pmatrix}\) is an elementary matrix such that \(EA = \begin{pmatrix} a_{11} - \frac{a_{12}^2}{a_{11}} & 0 \\ 0 & a_{22} \end{pmatrix}\). Matrix \(EA\) has a zero in the first row. In both cases, \(A\) is reducible. Hence, we have proved that a \(2 \times 2\) matrix \(A\) is reducible if and only if \(a_{12} a_{21} = 0\).

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Key Concepts

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

Elementary Matrix
In linear algebra, an elementary matrix is a simple tool that plays a pivotal role in many matrix operations. It's akin to a basic building block, used to perform a single elementary row operation on a matrix. Elementary row operations include swapping two rows, multiplying a row by a nonzero scalar, or adding a multiple of one row to another row.

An elementary matrix is obtained by performing one of these operations on an identity matrix, which is a square matrix that has 1's along the diagonal and 0's elsewhere. So, when you multiply any matrix by an elementary matrix, it's like performing that row operation on the matrix. This ability makes elementary matrices essential in procedures such as finding the inverse of a matrix or reducing a matrix to its echelon form.

Back to our original exercise, when we're talking about reducibility, we're essentially using an elementary matrix to simplify the given matrix into a form that reveals some of its inherent properties, such as having a zero either in the first row or column. This process is critical for understanding the structure of the matrix and how it can be manipulated or decomposed further in more complex linear algebra operations.
Matrix Multiplication
When it comes to combining two matrices, matrix multiplication is the standard procedure. However, it's not as straightforward as multiplying individual elements 鈥� it's about combining rows and columns in a systematic way. To multiply two matrices, the number of columns in the first matrix must match the number of rows in the second matrix.

The result is a new matrix where each element is computed by taking the dot product of the corresponding row from the first matrix and the column from the second matrix. For our specific problem, when we multiply the elementary matrix by our given matrix A, we are effectively modifying A by applying the elementary row operation to it.

Understanding matrix multiplication is crucial for grasping a wide array of concepts in linear algebra, as it's a foundational operation used in solving systems of equations, transforming geometric shapes, and even in advanced applications like coding theory and computer graphics.
Linear Algebra
At its heart, linear algebra is the branch of mathematics concerning linear equations, linear functions, and their representations through matrices and vector spaces. It's not only about solving equations; it's about understanding spaces, transformations, and how different objects can be mapped and represented in these spaces.

In the context of our problem, linear algebra gives us the framework and tools to analyze and understand matrices. It teaches us how to approach a matrix and dissect its properties, like determining if a matrix is reducible or not. We use concepts such as elementary matrices and matrix multiplication to transform matrices into more manageable forms. This manipulation often involves reducing a matrix to find its rank, determining if it's invertible, or simplifying it to solve a system of equations.

Overall, linear algebra is a vast field with deep implications in numerous areas of mathematics and science, especially in systems where linearity is an underlying principle, which is remarkably common in both natural and built systems.

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Most popular questions from this chapter

Let \(A\) be a \(3 \times 2\) matrix whose column vectors \(\mathbf{a}_{1}\) and \(\mathbf{a}_{2}\) are both probability vectors. Show that if \(\mathbf{p}\) is a probability vector in \(\mathbb{R}^{2}\) and \(\mathbf{y}=A \mathbf{p}\), then \(\mathbf{y}\) is a probability vector in \(\mathbb{R}^{3}\).

Let \(\left\\{\mathbf{u}_{1}, \ldots, \mathbf{u}_{n}\right\\}\) be an orthonormal basis for a complex inner product space \(V\), and let \\[ \begin{aligned} \mathbf{z} &=a_{1} \mathbf{u}_{1}+a_{2} \mathbf{u}_{2}+\cdots+a_{n} \mathbf{u}_{n} \\ \mathbf{w} &=b_{1} \mathbf{u}_{1}+b_{2} \mathbf{u}_{2}+\cdots+b_{n} \mathbf{u}_{n} \end{aligned} \\] Show that \\[ \langle\mathbf{z}, \mathbf{w}\rangle=\sum_{i=1}^{n} \bar{b}_{i} a_{i} \\]

Each year, employees at a company are given the option of donating to a local charity as part of a payroll deduction plan. In general, 80 percent of the employees enrolled in the plan in any one year will choose to sign up again the following year, and 30 percent of the unenrolled will choose to enroll the following year. Determine the transition matrix for the Markov process and find the steadystate vector. What percentage of employees would you expect to find enrolled in the program in the long run?

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