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Chapter 7: Problem 40

Prove that if a symmetric matrix \(A\) has only one eigenvalue \(\lambda\) then \(A=\lambda I\).

Short Answer

Expert verified
From the premise that \( A \) only has one eigenvalue \( \lambda \), we infer that all vectors are eigenvectors satisfying \( Ax = \lambda x \), i.e., \( A \) performs scalar multiplication to all vectors. This implies \( A \) is equivalent to \( \lambda I \). Hence, the assertion \( A = \lambda I \) is proven true.

Step by step solution

01

Understand the properties of a symmetric matrix

A symmetric matrix is a matrix that equals its own transpose, meaning that \( A = A^T \). This property will be important in realizing that symmetric matrices are diagonalizable.
02

Understand the nature of eigenvalues

The eigenvalue of a matrix is a number which, when used with its corresponding eigenvector in a certain linear transformation, causes the direction of the eigenvector to remain the same, although the eigenvector may be scaled by the eigenvalue.
03

Understand the nature of the identity matrix

The identity matrix is a square matrix where the main diagonal elements are ones and all the off-diagonal elements are zeros. When any matrix is multiplied by an identity matrix of suitable order, the original matrix is obtained.
04

Investigation of the given composition

Next, it's important to consider the equation \( A = \lambda I \) that needs to be validated. Given that \( A \) only has one eigenvalue \( \lambda \), this implies that any eigenvector \( x \) of \( A \) satisfies \( Ax = \lambda x \). This indicates that every vector \( x \) is an eigenvector because every vector satisfies the equality on substitution into the equation \( Ax = \lambda x \). This means \( A \) essentially performs scalar multiplication to all vectors in the space. Therefore, \( A \) can be represented by the matrix \( \lambda I \) since \( I \) is the identity operator in matrix representation.
05

The Conclusion

The above argumentation shows that for a symmetric matrix \( A \) with only one eigenvalue \( \lambda\), then \( A \) should be a scalar multiple of the identity matrix \( I \). Hence, the assertion \( A = \lambda I \) is proved. The proof relies on the characteristics of symmetric matrices, eigenvectors, eigenvalues and the identity matrix.

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Key Concepts

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

Symmetric Matrix
A symmetric matrix is a fascinating mathematical concept characterized by its unique structure, where the matrix is equal to its transpose, i.e., \( A = A^T \). This means the elements are symmetrical about the main diagonal. For symmetric matrices, each element \( a_{ij} \) is equal to \( a_{ji} \). This property ensures that the matrix 'mirrors' itself, creating a balanced layout.

Symmetric matrices have several important features:
  • They are always square - this is because the transpose operation implies a reflection over the main diagonal.
  • All their eigenvalues are real numbers, a property derived from their symmetric nature.
  • They are diagonalizable, meaning they can be expressed as \( PDP^{-1} \), where \( P \) is an orthogonal matrix, and \( D \) is a diagonal matrix.
These characteristics make symmetric matrices particularly useful in various scientific fields, from physics to computer science, and they're often used to simplify computations, especially in the context of eigenvalues and eigenvectors.
Identity Matrix
The identity matrix, often denoted as \( I \), is a special type of square matrix. Its defining feature is that all the elements on the main diagonal are ones, while all other elements are zeros. For a \( n \times n \) identity matrix, the entries will look like this: \( a_{ii} = 1 \) (for diagonal entries) and \( a_{ij} = 0 \) for all \( i eq j \).

Identity matrices have several key properties:
  • When any matrix \( A \) is multiplied by an identity matrix of suitable size, \( I \), the original matrix \( A \) is obtained, i.e., \( AI = IA = A \). This is somewhat similar to multiplying a number by one in arithmetic.
  • The identity matrix serves as the neutral element in matrix multiplication.
  • In the context of linear transformations, an identity matrix represents a transformation that leaves vectors unchanged.
Understanding the identity matrix is fundamental for grasping more complex concepts like matrix diagonalization and eigenvalues, as it acts as a reference or baseline matrix in many computations.
Diagonalization
Diagonalization is a powerful method used in linear algebra to simplify complex matrix operations. It's the process of transforming a given matrix into a diagonal form. A diagonal matrix is one where all the off-diagonal elements are zero. The process of diagonalization hinges on expressing a matrix \( A \) as \( PDP^{-1} \), where \( D \) is a diagonal matrix and \( P \) is an orthogonal matrix whose columns are the eigenvectors of \( A \).

The benefits of diagonalization include:
  • Speeding up matrix computations by transforming them into simpler forms.
  • Facilitating the computation of powers of matrices and exponential of matrices.
  • Providing insight into the geometric and algebraic structure of transformations described by matrices.
For diagonalization to be possible, a matrix must have enough linearly independent eigenvectors. Symmetric matrices are always diagonalizable, which is one of their many useful properties, allowing simplification of complex mathematical problems.

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Most popular questions from this chapter

For an invertible matrix \(A\), prove that \(A\) and \(A^{-1}\) have the same eigenvectors. How are the eigenvalues of \(A\) related to the eigenvalues of \(A^{-1} ?\)

Prove that a triangular matrix is nonsingular if and only if its eigenvalues are real and nonzero. Getting Started: Because this is an "if and only if" statement, you must prove that the statement is true in both directions. Review Theorems 3.2 and 3.7. (i) To prove the statement in one direction, assume that the triangular matrix \(A\) is nonsingular. Use your knowledge of nonsingular and triangular matrices and determinants to conclude that the entries on the main diagonal of \(A\) are nonzero. (ii) Because \(A\) is triangular, you can use Theorem 7.3 and part (i) to conclude that the eigenvalues are real and nonzero. (iii) To prove the statement in the other direction, assume that the eigenvalues of the triangular matrix \(A\) are real and nonzero. Repeat parts (i) and (ii) in reverse order to prove that \(A\) is nonsingular.

Let \(A\) be an \(n \times n\) matrix such that the sum of the entries in each row is a fixed constant \(r .\) Prove that \(r\) is an eigenvalue of \(A\) Illustrate this result with a specific example.

Demonstrate the Cayley-Hamilton Theorem for the given matrix. The Cayley- Hamilton Theorem states that a matrix satisfies its characteristic equation. For example, the characteristic equation of \(A=\left[\begin{array}{rr}1 & -3 \\\ 2 & 5\end{array}\right]\) is \(\lambda^{2}-6 \lambda+11=0,\) and by the theorem you have \(A^{2}-6 A+11 I_{2}=O\). $$\left[\begin{array}{lll} 3 & 1 & 4 \\ 2 & 4 & 0 \\ 5 & 5 & 6 \end{array}\right]$$

Prove that nonzero nilpotent matrices are not diagonalizable. Getting Started: From Exercise 73 in Section \(7.1,\) you know that 0 is the only eigenvalue of the nilpotent matrix \(A\). Show that it is impossible for \(A\) to be diagonalizable. (i) Assume \(A\) is diagonalizable, so there exists an invertible matrix \(P\) such that \(P^{-1} A P=D,\) where \(D\) is the zero matrix. (ii) Find \(A\) in terms of \(P, P^{-1},\) and \(D\) (iii) Find a contradiction and conclude that nonzero nilpotent matrices are not diagonalizable.
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