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Chapter 4: Problem 48

Let \(A\) be a nilpotent matrix (that is, \(A^{m}=O\) for some \(m > 1\) ). Prove that if \(A\) is diagonalizable, then \(A\) must be the zero matrix.

Short Answer

Expert verified
If a nilpotent matrix is diagonalizable, it must be the zero matrix.

Step by step solution

01

Understand Nilpotent Matrix

A matrix is nilpotent if there exists some positive integer \(m\) such that \(A^m = O\), where \(O\) is the zero matrix.
02

Define Diagonalizable Matrix

A matrix is diagonalizable if there exists an invertible matrix \(P\) and a diagonal matrix \(D\) such that \(A = PDP^{-1}\).
03

Eigenvalues of Nilpotent Matrix

Since \(A\) is nilpotent, its only eigenvalue is 0. This is because the minimal polynomial of \(A\), which divides \(x^m\), must be of the form \(x^k\), forcing all eigenvalues to be zero.
04

Implication of Diagonalizability and Nilpotency

For a diagonalizable matrix, its diagonal elements (eigenvalues) on \(D\) are its eigenvalues. Since all eigenvalues of a nilpotent matrix are 0, \(D\) becomes the zero matrix.
05

Conclude A is the Zero Matrix

Given \(A = PDP^{-1}\) and \(D = O\), it follows that \(A = P \times O \times P^{-1} = O\). Thus, if \(A\) is diagonalizable and nilpotent, \(A\) must be the zero matrix.

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Key Concepts

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

Diagonalizable Matrix
A diagonalizable matrix is a special type of matrix that can be expressed in a particular form known as diagonal form. This means it can be written as the product of three matrices: an invertible matrix \( P \), a diagonal matrix \( D \), and the inverse of \( P \). The relationship can be denoted as \( A = PDP^{-1} \).

This property is useful because diagonal matrices are straightforward to work with, allowing for simpler computations, especially when it comes to raising matrices to a power.
  • Diagonalization simplifies operations such as computing powers of a matrix.
  • Not all matrices are diagonalizable, but if they are, it means they behave more predictably under certain operations.
  • Being diagonalizable implies that a matrix has enough linearly independent eigenvectors to form the basis of its space.
Eigenvalues
Eigenvalues are fundamental to understanding the nature of a matrix. To put it simply, an eigenvalue is a scalar that indicates how a linear transformation scales a vector, known as an eigenvector.

For a matrix \( A \), if there exists a non-zero vector \( \mathbf{v} \) such that \( A\mathbf{v} = \lambda\mathbf{v} \), then \( \lambda \) is called an eigenvalue of \( A \).

Eigenvalues can tell a lot about the behavior of a matrix:
  • If all eigenvalues of a matrix are zero, as in the case of nilpotent matrices, it indicates that the matrix tends toward nothing under repeated multiplication.
  • Eigenvalues are visible directly on a matrix's diagonal when it is in diagonal form.
  • The sum and product of a matrix's eigenvalues are related to the trace and determinant respectively, giving clues about the matrix's properties.
Zero Matrix
The zero matrix is a matrix in which every element is zero. It is denoted by \( O \) and serves as the additive identity in matrix operations. This means that for any matrix \( A \) of the same size, \( A + O = A \).

The zero matrix is critical in understanding concepts like nilpotency:
  • When a nilpotent matrix \( A \) is raised to a power, at some point, the result will be the zero matrix (i.e., \( A^m = O \) for some \( m > 1 \)).
  • In the case where a nilpotent matrix is also diagonalizable, it can only be the zero matrix since its diagonal form has all eigenvalues as zeros, leading to the matrix \( D \) being a zero matrix.
  • The zero matrix's properties make it a crucial identity element that aids in understanding matrix operations and structures.
Invertible Matrix
An invertible matrix, also known as a non-singular or non-degenerate matrix, is a matrix that possesses an inverse. The inverse of a matrix \( A \) is denoted as \( A^{-1} \), and it has the property that \( AA^{-1} = A^{-1}A = I \), where \( I \) is the identity matrix.

The concept of invertibility is important in the study of matrices:
  • A matrix must be invertible to be diagonalized because we need the inverse of the matrix \( P \) in the diagonalization process.
  • Not every matrix is invertible. A matrix is invertible only if it has a full rank, meaning none of its rows or columns are linearly dependent.
  • The determinant of an invertible matrix is non-zero, offering a quick way to verify its invertibility.
Invertibility gives a matrix the ability to 'undo' the transformations it performs, making it crucial in systems of equations and transformations.

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Most popular questions from this chapter

Verify the Cayley-Hamilton Theorem for \(A=\) \(\left[\begin{array}{rrr}1 & 1 & 0 \\\ -1 & 0 & 1 \\ -2 & 1 & 0\end{array}\right]\) The Cayley-Hamilton Theorem can be used to calculate powers and inverses of matrices. For example, if \(A\) is a \(2 \times 2\) matrix with characteristic polynomial \(c_{A}(\lambda)=\lambda^{2}+a \lambda+b\) \(\operatorname{then} A^{2}+a A+b I=O, s o\) \\[ A^{2}=-a A-b I \\] and \\[ \begin{aligned} A^{3} &=A A^{2}=A(-a A-b I) \\ &=-a A^{2}-b A \\ &=-a(-a A-b I)-b A \\ &=\left(a^{2}-b\right) A+a b I \end{aligned} \\] It is easy to see that by continuing in this fashion we can express any positive power of \(A\) as a linear combination of I and \(A .\) From \(A^{2}+a A+b I=O,\) we also obtain \(A(A+a I)=-b I, s o\) \\[ A^{-1}=-\frac{1}{b} A-\frac{a}{b} I \\] provided \(b \neq 0\)

If \(A\) is an \(n \times n\) matrix, prove that \\[ \operatorname{det}(\operatorname{adj} A)=(\operatorname{det} A)^{n-1} \\]

A matrix \(A\) is given along with an iterate \(\mathbf{x}_{5},\) produced as in Example 4.30 (a) Use these data to approximate a dominant eigenvector whose first component is 1 and a corresponding dominant eigenvalue. (Use three-decimal-place accuracy.) (b) Compare your approximate eigenvalue in part (a) with the actual dominant eigenvalue. $$A=\left[\begin{array}{rr}7 & 4 \\ -3 & -1\end{array}\right], \mathbf{x}_{5}=\left[\begin{array}{r}7811 \\ -3904\end{array}\right]$$

Prove that if \(A\) is a diagonalizable matrix such that every eigenvalue of \(A\) is either 0 or 1 , then \(A\) is idempotent (that is, \(A^{2}=A\) ).

Use the power method to approximate the dominant eigenvalue and eigenvector of \(A .\) Use the given initial vector \(\mathbf{x}_{0},\) the specified number of iterations \(k,\) and three-decimal-place accuracy. $$A=\left[\begin{array}{ll}7 & 2 \\ 2 & 3\end{array}\right], \mathbf{x}_{0}=\left[\begin{array}{l}1 \\ 0\end{array}\right], k=6$$
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