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Chapter 11: Problem 268

Prove that \(\mathrm{n} \times \mathrm{n}\) matrix \(\mathrm{A}\) is diagonalizable if and only if it has n linearly independent eigenvectors. In this case \(\mathrm{A}\) is similar to a matrix D whose diagonal elements are the eigenvalues of \(\mathrm{A}\).

Short Answer

Expert verified
An n x n matrix A is diagonalizable if and only if it has n linearly independent eigenvectors. If A is diagonalizable, it can be factored as A = PDP^(-1), where P is an invertible matrix with eigenvectors as its columns and D is a diagonal matrix with eigenvalues on its diagonal. Conversely, if A has n linearly independent eigenvectors, they form the columns of an invertible matrix P, and A = PDP^(-1) holds, making A diagonalizable and similar to the diagonal matrix D with its eigenvalues on the diagonal.

Step by step solution

01

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are an important consideration when working with matrices. If for a given matrix A and a non-zero vector x, we have Ax = 位x, where 位 is a scalar, then 位 is called the eigenvalue of A, and x is called the eigenvector corresponding to 位.
02

Diagonalizable Matrices

A matrix A is diagonalizable if it can be written as A = PDP^(-1), where P is an invertible matrix and D is a diagonal matrix with the eigenvalues of A on its diagonal. The columns of P are the eigenvectors corresponding to the eigenvalues on the diagonal of D.
03

Similar Matrices

Two matrices A and B are similar if there exists an invertible matrix P such that B = P^(-1)AP. Now let us proceed with the proof.
04

1. If A is diagonalizable, it has n linearly independent eigenvectors

Assume A is diagonalizable. Then by definition, A can be factored as A = PDP^(-1), where P is an invertible matrix and D is a diagonal matrix with eigenvalues on its diagonal. Since P is invertible, it has n linearly independent columns, which are eigenvectors of A. Hence, if A is diagonalizable, it must have n linearly independent eigenvectors.
05

2. If A has n linearly independent eigenvectors, it is diagonalizable and similar to matrix D

Assume A has n linearly independent eigenvectors, which we denote as v_1, v_2, ..., v_n. These eigenvectors form the columns of an invertible matrix P. For each eigenvector v_i, we have Av_i = 位_iv_i, where 位_i is the eigenvalue corresponding to v_i. Now, let's consider the matrix product AP. The i-th column of AP will be Av_i = 位_iv_i. Therefore, AP = [位_1v_1, 位_2v_2, ..., 位_nv_n]. On the other hand, let's form the diagonal matrix D with 位_i's on its diagonal elements, i.e., D = diag(位_1, 位_2,..., 位_n). When we multiply PD, the i-th column of PD is 位_iv_i, and the result is the same as before: PD = [位_1v_1, 位_2v_2, ..., 位_nv_n]. Since both AP and PD give the same result, we have AP = PD. Multiplying both sides by P^(-1) on the right, we get A = PDP^(-1). This shows that A is diagonalizable, and it is similar to the diagonal matrix D, with its eigenvalues on the diagonal. In conclusion, an n x n matrix A is diagonalizable if and only if it has n linearly independent eigenvectors, and in this case, A is similar to a diagonal matrix D with its eigenvalues on the diagonal.

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Key Concepts

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

Eigenvalues
The concept of eigenvalues is fundamental when dealing with matrices. Eigenvalues are special scalars associated with a matrix. When we multiply the matrix by a certain non-zero vector (called an eigenvector), and the result is simply the original vector scaled by a scalar, this scalar is the eigenvalue. Mathematically, if we have a matrix \textbf{A} and a vector \textbf{x}, the relationship can be expressed as \(\textbf{Ax} = \lambda\textbf{x}\), where \(\lambda\) is the eigenvalue. Understanding eigenvalues gives us insight into the properties of a matrix, including whether it is diagonalizable or not.

In the context of diagonalization, eigenvalues populate the diagonal of the resulting diagonal matrix. They play a crucial role as they directly impact the matrix's behavior, like its determinant and trace, which are the product and sum of the eigenvalues respectively. Let's not forget that finding eigenvalues is often the first step in determining the diagonalizability of a matrix, making them a cornerstone in linear algebra.
Eigenvectors
Eigenvectors are vectors that, when transformed by a matrix, result in a vector that is parallel to the original, scaled by the corresponding eigenvalue. The equation \(\textbf{Ax} = \lambda\textbf{x}\) showcases this phenomenon, with \(\textbf{x}\) being the eigenvector. For a matrix to be diagonalizable, it must have a full set of linearly independent eigenvectors. This means each eigenvector points in a unique direction and cannot be expressed as a combination of the others.

These independent eigenvectors can be used to form a matrix鈥攐ften denoted by \textbf{P}鈥攖hat plays a key role in transforming the original matrix into a diagonal matrix. The importance of identifying these eigenvectors cannot be overstated, as they are the building blocks for the matrix \textbf{P} that defines the space in which the original matrix acts more simply.
Similar Matrices
The term 'similar matrices' refers to a pair of matrices that represent the same linear transformation, but in different bases. Two matrices, \textbf{A} and \textbf{B}, are similar if you can find an invertible matrix \textbf{P} such that \(\textbf{B} = \textbf{P}^{-1}\textbf{AP}\). Why is this interesting? Because similar matrices have the same eigenvalues, determinant, rank, and other intrinsic properties. They are essentially different expressions of the same linear operation.

In our discussion on diagonalization, the similarity is crucial as it links the original matrix \textbf{A} to a simpler, diagonal form \textbf{D}. The transformation by \textbf{P} doesn鈥檛 change the essence of \textbf{A}, but provides a clearer view of its features through \textbf{D}. Thus, understanding the concept of similarity in matrices becomes vital for both theoretical and practical applications in linear algebra.
Invertible Matrix
An invertible matrix, also known as a non-singular or reversible matrix, has a unique matrix inverse. In the world of matrices, invertibility is akin to a function having an inverse in standard mathematics. If a matrix \textbf{A} is invertible, there exists another matrix \textbf{A}^{-1} such that \(\textbf{AA}^{-1} = \textbf{A}^{-1}\textbf{A} = \textbf{I}\), where \textbf{I} is the identity matrix. This property is critical when discussing diagonalization because the matrix \textbf{P}, which contains our eigenvectors, must be invertible for us to transition from \textbf{A} to its diagonal form.

The existence of an inverse implies that the matrix鈥檚 columns are linearly independent, hence the matrix \textbf{P} reflects a change of basis that is reversible. In other words, invertibility guarantees that there is a reliable path back to the original matrix from its diagonal form. For a matrix to be a candidate for diagonalization, checking for invertibility is therefore an essential step.

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Most popular questions from this chapter

Let \(\mathrm{T}\) be the linear operator on \(\mathrm{R}^{2}\) such that and only if Compute the matrix of \(\mathrm{T}\) with respect to the basis \(\\{(1,1),(1,-1)\\}\)

Verify the following rules by giving examples: (a) If \(\mathrm{A}\) is an \(\mathrm{n} \times \mathrm{n}\) diagonal matrix and \(\mathrm{B}\) is an \(\mathrm{n} \times \mathrm{n}\) matrix, each row of \(\mathrm{AB}\) is then just the product of the diagonal entry of \(A\) times the corresponding row of \(B\). (b) If \(\mathrm{B}\) is a diagonal matrix, each column of \(\mathrm{AB}\) is just the product of the corresponding column of \(\mathrm{A}\) with the corresponding diagonal entry of \(\mathrm{B}\).

Find an orthogonal matrix P that diagonalizes $$ \mathrm{A}=\begin{array}{rrr} 14 & 2 & 2 \mid \\ 2 & 4 & 2 \mid \\ \mid 2 & 2 & 4 \end{array} $$

Given: $$ \begin{array}{rl} \mathrm{A}=\mid 1 & 1 \mid \\ \mid 0 & 1 \mid \end{array} \text { and } \mathrm{P}=\mid \begin{array}{cc} 1 & 1 \mid \\ 1 & -1 \mid \end{array} $$ (a) Find \(\mathrm{P}^{-1}\). (b) Find \(\mathrm{P}^{-1 \mathrm{AP}}\). (c) Verify that, if \(\mathrm{B}\) is similar to \(\mathrm{A}\), then \(\mathrm{A}\) is similar to \(\mathrm{B}\). (d) Show that \(\mathrm{B}^{\mathrm{k}}=\mathrm{P}^{-1} \mathrm{~A}^{\mathrm{k}} \mathrm{P}\) if \(\mathrm{B}=\mathrm{P}^{-1} \mathrm{AP}\) where \(\mathrm{k}\) is any positive integer.

Show that the matrix \(\mathrm{A}\) is diagonalizable where $$ \mathrm{A}=\begin{array}{cll} 10 & 0 & 0 \\ 0 & 1 & 0 \\ \mid 1 & 0 & 1 \mid \end{array} $$
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