Question

If all \(n\) eigenvalues of the operator \(A: R^{n} \rightarrow R^{n}\) are real and distinct, then the operator is diagonalizable.

Short answer

Question: Show that if all \(n\) eigenvalues of the operator \(A: R^{n} \rightarrow R^{n}\) are real and distinct, then the operator is diagonalizable. Answer: To show that the operator A is diagonalizable, we first show that if all eigenvalues are real and distinct, then for each eigenvalue, there exists a linearly independent eigenvector. Our proof consists of three steps: Step 1: We understand the concept of eigenvalues and eigenvectors, where an eigenvalue \(\lambda\) of the operator \(A\) is a scalar and its associated eigenvector \(x\) satisfies \(Ax = \lambda x\). Step 2: We prove that distinct eigenvalues imply linearly independent eigenvectors. We assume that there exists a non-trivial linear combination of eigenvectors equal to zero and derive a contradiction. Step 3: With linearly independent eigenvectors, we can construct an invertible matrix \(P\) and a diagonal matrix \(D\) such that \(A = PDP^{-1}\), showing that the operator \(A\) is diagonalizable.

Step-by-step solution

Understanding Eigenvalues and Eigenvectors

An eigenvalue \(\lambda\) of the operator \(A\) is a scalar such that there exists a non-zero vector \(x \in R^n\) satisfying \(Ax = \lambda x\). The vector \(x\) associated with this eigenvalue is called an eigenvector.

Distinct Eigenvalues Imply Linearly Independent Eigenvectors

Suppose \(\lambda_1, \lambda_2, ... , \lambda_n\) are \(n\) distinct eigenvalues of \(A\). For each eigenvalue \(\lambda_i\), let \(x_i\) be a non-zero eigenvector. We want to show that the set \(\{x_1, x_2, ..., x_n\}\) is linearly independent. Assume the contrary, i.e., assume there exists a non-trivial linear combination of eigenvectors such that \(\sum_{i=1}^n c_ix_i=0\), where not all \(c_i\) are zero. Without loss of generality, let \(c_{n}\neq 0\). Then \(x_n = -\frac{c_1}{c_n}x_1 - \cdots - \frac{c_{n-1}}{c_n}x_{n-1}\). Now, apply the operator \(A\) to both sides of the equation and use the fact that \(Ax_i = \lambda_i x_i\) for each \(i\). We get \(A x_n = -\frac{c_1}{c_n}A x_1 - \cdots - \frac{c_{n-1}}{c_n}A x_{n-1}\) \(\lambda_{n} x_n = -\frac{c_1}{c_n}\lambda_1 x_1 - \cdots - \frac{c_{n-1}}{c_n}\lambda_{n-1} x_{n-1}\) Now, multiply both sides of the equation we derived for \(x_n\) by \(\lambda_n\): \(\lambda_{n} x_n = -\frac{c_1}{c_n}\lambda_{n} x_1 - \cdots - \frac{c_{n-1}}{c_n}\lambda_{n} x_{n-1}\) Subtract the last equation from the previous one: \(0 = \left(\frac{c_1}{c_n}(\lambda_1 - \lambda_n)\right)x_1 + \cdots + \left(\frac{c_{n-1}}{c_n}(\lambda_{n-1} - \lambda_n)\right)x_{n-1}\) Since \(\lambda_1 - \lambda_n, \cdots, \lambda_{n-1} - \lambda_n\) are all non-zero (since their corresponding eigenvalues are distinct), this implies that \(x_1, x_2, ..., x_{n-1}\) are linearly dependent, which is a contradiction. Thus, \(\{x_1, x_2, ..., x_n\}\) must be linearly independent.

Concluding Diagonalizability

Since we have shown that there exists a linearly independent set of eigenvectors \(\{x_1, x_2, ..., x_n\}\), it is possible to construct an invertible matrix \(P\) whose columns are these eigenvectors, and a diagonal matrix \(D\) with the corresponding eigenvalues on the diagonal such that \(A = PDP^{-1}\). Therefore, the operator \(A\) is diagonalizable.

Key concepts

Eigenvalues

Eigenvalues are a fundamental concept in linear algebra that help us understand linear transformations. When we talk about eigenvalues of an operator (which can be represented as a matrix in a finite-dimensional space), we are looking for scalars \( \lambda \) such that there exists a non-zero vector \( x \), which when multiplied by the operator \( A \), equals \( \lambda \) times the vector \( x \). This relationship is expressed in the equation \( A\mathbf{x} = \lambda \mathbf{x} \). Here are some important points to remember about eigenvalues:

• Each matrix has a characteristic polynomial, and its roots are the eigenvalues of the matrix.
• Eigenvalues can be real or complex numbers, depending on the matrix.
• In the context of diagonalizability, having distinct eigenvalues is crucial, as it often implies that the matrix is diagonalizable.

Understanding eigenvalues helps us analyze the stability and behavior of dynamic systems, among various other applications.

Eigenvectors

While eigenvalues provide us with the scale of transformation, eigenvectors give us the direction. Given an eigenvalue \( \lambda \), an eigenvector \( \mathbf{x} \) satisfies \( A\mathbf{x} = \lambda \mathbf{x} \). This means that when the transformation represented by \( A \) is applied to \( \mathbf{x} \), the resulting vector remains parallel to \( \mathbf{x} \).

• Eigenvectors corresponding to different eigenvalues are linearly independent.
• The eigenvector associated with an eigenvalue defines the directions in which a particular transformation stretches or compresses space.

It is important to note that eigenvectors are not unique; if \( \mathbf{x} \) is an eigenvector, any scalar multiple of \( \mathbf{x} \) is also an eigenvector. Together, eigenvalues and eigenvectors reveal the geometric properties of linear transformations.

Diagonalizable Operators

A matrix or operator is considered diagonalizable if it can be expressed in the form \( A = PDP^{-1} \), where \( D \) is a diagonal matrix containing the eigenvalues of \( A \), and \( P \) is an invertible matrix whose columns are the linearly independent eigenvectors of \( A \). Diagonalizable operators have several advantages:

• They simplify computations, especially powers of matrices, since working with a diagonal matrix is much easier.
• They provide insight into the structure and behavior of the transformation.

For a matrix with distinct eigenvalues, it is typically diagonalizable, meaning that we can find a complete set of linearly independent eigenvectors. This is the basis for diagonalization and facilitates many practical applications in physics, engineering, and computer science.

Linear Independence

Linear independence is a key concept in linear algebra which ensures that vectors in a set are unique in terms of direction. Vectors \( \mathbf{v_1}, \mathbf{v_2}, ..., \mathbf{v_n} \) are said to be linearly independent if the only solution to the equation \( c_1\mathbf{v_1} + c_2\mathbf{v_2} + \cdots + c_n\mathbf{v_n} = \mathbf{0} \) is when all coefficients \( c_1, c_2, ..., c_n \) are zero.

• Linear independence is crucial for the dimensions of a vector space, defining its basis.
• If eigenvectors associated with eigenvalues of a matrix are linearly independent, it implies that the matrix is diagonalizable.
• Linearly independent vectors form a basis of the vector space they span, enabling representation of any vector in that space as a unique combination.

Understanding linear independence helps in solving system of equations, determining solutions, and greatly aids in the simplification of problems in linear algebra.