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Chapter 6: Problem 8

An \(n \times n\) matrix \(A\) is said to be idempotent if \(A^{2}=A .\) Show that if \(\lambda\) is an eigenvalue of an idempotent matrix, then \(\lambda\) must be either 0 or 1

Short Answer

Expert verified
To show that if \(\lambda\) is an eigenvalue of an idempotent matrix, it must be either 0 or 1, we start with the eigenvector equation \(Av = \lambda v\), where A is an idempotent matrix (\(A^2 = A\)). After a series of substitutions and simplifications, we arrive at \((\lambda^2 - \lambda)v = 0\). Since v is a nonzero eigenvector, the factor \((\lambda^2-\lambda)\) must be zero, so \(\lambda\) must be either 0 or 1.

Step by step solution

01

Write down the given information and definitions

We are given that \(A^2 = A\) for an idempotent matrix A and want to show that any eigenvalue \(\lambda\) of A must be either 0 or 1. Let's assume \(Av = \lambda v\) where v is a nonzero eigenvector corresponding to the eigenvalue \(\lambda\).
02

Multiply both sides of the eigenvector equation by A

To make use of the idempotent property, we multiply both sides of the equation \(Av = \lambda v\) by A: \(A(Av) = A(\lambda v)\)
03

Apply associativity of matrix multiplication

Using the associativity property of matrix multiplication, we can rewrite the equation from step 2 as follows: \((A^2)v = \lambda(Av)\)
04

Substitute the idempotent property and the definition of eigenvalue

Since \(A^2 = A\) for an idempotent matrix, we can substitute this back into the equation from step 3: \(Av = \lambda(Av)\) And since \(Av = \lambda v\), we can also substitute this into the equation, resulting in: \(\lambda v = \lambda(\lambda v)\)
05

Factor out eigenvalue 位 from the right side of the equation

Next, we factor out \(\lambda\) from the right side of the equation: \(\lambda v = \lambda^2 v\)
06

Subtract 位v from both sides of the equation

Subtract \(\lambda v\) from both sides to get the difference of the two sides: \((\lambda^2 - \lambda)v = 0\)
07

Conclude that 位 must be 0 or 1

Since v is a nonzero eigenvector, and as the product is zero, we conclude that the factor \((\lambda^2-\lambda)\) must be zero. Therefore, \(\lambda\) must be either 0 or 1.

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Key Concepts

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

Eigenvector
An eigenvector of a matrix is a non-zero vector that changes only by a scalar factor when that matrix is multiplied by it. This scalar is known as the eigenvalue. The relationship can be represented by the equation \( Av = \(lambda v \), where \( A \) is a matrix, \( v \) is an eigenvector, and \( lambda \) is the corresponding eigenvalue.

Understanding eigenvectors and eigenvalues is crucial because they reveal important properties about the matrix such as its invertibility, determine the axes of a transformation, and are foundational in fields like Quantum Mechanics and Principal Component Analysis (PCA) in statistics.

Practical Implication of Eigenvectors

Consider if we have a matrix representing a transformation in a physical space, eigenvectors can represent the directions in which this transformation acts uniformly, either stretching or compressing by a certain factor (the eigenvalue), without altering the direction.
Matrix Multiplication
Matrix multiplication is a way to combine two matrices together to produce a new matrix. Unlike scalar multiplication, the process is more complex and involves taking the rows of the first matrix and the columns of the second matrix to produce the entries of the resulting matrix. The rule to remember is that the number of columns in the first matrix must match the number of rows in the second matrix for multiplication to be possible.

In computation, for matrix multiplication with matrices \( A \) and \( B \), the entry in the \( i \)-th row and \( j \)-th column of the product matrix \( AB \) is the sum of the products of the corresponding elements of the \( i \)-th row of \( A \) and the \( j \)-th column of \( B \).

Importance of Matrix Multiplication

Matrix multiplication is not just an abstract mathematical operation but is widely used in computer graphics, data science, economics, and other fields to model and compute complex relationships.
Associativity in Linear Algebra
The principle of associativity in matrix multiplication allows us to group matrices in different orders without affecting the result. This means that for any three matrices \( A \), \( B \), and \( C \), if the product is defined, then \( (AB)C = A(BC) \). This associative property is essential in simplifying the computation of matrix products and is used extensively in proving various algebraic properties.

Associativity is particularly important in computational contexts where the order of multiplication can significantly impact the efficiency and even the feasibility of calculations. Choosing the right order can reduce the computations required.

Application of Associativity

Suppose you have a chain of matrix transformations applied to a geometric object. Using associativity, these transformations can be precomputed into a single matrix, thus optimizing the number of operations necessary for rendering in computer graphics or animation.

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Most popular questions from this chapter

Find the output vector \(\mathbf{x}\) in the open version of the Leontief input- output model if \(A=\left(\begin{array}{ccc}0.2 & 0.4 & 0.4 \\ 0.4 & 0.2 & 0.2 \\\ 0.0 & 0.2 & 0.2\end{array}\right)\) and \(\quad \mathbf{d}=\left(\begin{array}{r}16,000 \\ 8,000 \\\ 24,000\end{array}\right)\)

The transition matrix in Example 5 has the property that both its rows and its columns add up to 1 In general, a matrix \(A\) is said to be doubly stochastic if both \(A\) and \(A^{T}\) are stochastic. Let \(A\) be an \(n \times n\) doubly stochastic matrix whose eigenvalues satisfy \\[ \lambda_{1}=1 \quad \text { and } \quad\left|\lambda_{j}\right|<1 \text { for } j=2,3, \ldots, n \\] Show that if \(\mathbf{e}\) is the vector in \(\mathbb{R}^{n}\) whose entries are all equal to \(1,\) then the Markov chain will converge to the steady-state vector \(\mathbf{x}=\frac{1}{n} \mathbf{e}\) for any starting vector \(\mathbf{x}_{0} .\) Thus, for a doubly stochastic transition matrix, the steady-state vector will assign equal probabilities to all possible outcomes.

Let \(A\) be a \(n \times n\) matrix with real entries and let \(\lambda_{1}=a+b i\) (where \(a\) and \(b\) are real and \(b \neq 0\) ) be an eigenvalue of \(A .\) Let \(\mathbf{z}_{1}=\mathbf{x}+i \mathbf{y}\) (where \(\mathbf{x}\) and \(\mathbf{y}\) both have real entries) be an eigenvector belonging to \(\lambda_{1}\) and let \(\mathbf{z}_{2}=\mathbf{x}-i \mathbf{y}\) (a) Explain why \(\mathbf{z}_{1}\) and \(\mathbf{z}_{2}\) must be linearly independent. (b) Show that \(\mathbf{y} \neq \mathbf{0}\) and that \(\mathbf{x}\) and \(\mathbf{y}\) are linearly independent.

The city of Mawtookit maintains a constant population of 300,000 people from year to year. A political science study estimated that there were 150,000 Independents, 90,000 Democrats, and 60,000 Republicans in the town. It was also estimated that each year 20 percent of the Independents become Democrats and 10 percent become \(\operatorname{Re}\) publicans. Similarly, 20 percent of the Democrats become Independents and 10 percent become Republicans, while 10 percent of the Republicans defect to the Democrats and 10 percent become Independents each year. Let \\[ \mathbf{x}=\left(\begin{array}{r} 150,000 \\ 90,000 \\ 60,000 \end{array}\right) \\] and let \(\mathbf{x}^{(1)}\) be a vector representing the number of people in each group after one year. (a) Find a matrix \(A\) such that \(A \mathbf{x}=\mathbf{x}^{(1)}\) (b) Show that \(\lambda_{1}=1.0, \lambda_{2}=0.5,\) and \(\lambda_{3}=0.7\) are the eigenvalues of \(A,\) and factor \(A\) into a product \(X D X^{-1},\) where \(D\) is diagonal. (c) Which group will dominate in the long run? Justify your answer by computing \(\lim A^{n} \mathbf{x}\)

Show that if a matrix \(U\) is both unitary and Hermitian then any eigenvalue of \(U\) must equal either 1 or -1
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