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Chapter 6: Problem 2

Prove or give a counterexample: a. If \(A\) is an \(n \times n\) matrix with \(n\) distinct (real) eigenvalues, then \(A\) is diagonalizable. b. If \(A\) is diagonalizable and \(A B=B A\), then \(B\) is diagonalizable. c. If there is an invertible matrix \(P\) so that \(A=P^{-1} B P\), then \(A\) and \(B\) have the same eigenvalues. d. If \(A\) and \(B\) have the same eigenvalues, then there is an invertible matrix \(P\) so that \(A=P^{-1} B P\). e. There is no real \(2 \times 2\) matrix \(A\) satisfying \(A^{2}=-I\). f. If \(A\) and \(B\) are diagonalizable and have the same eigenvalues (with the same algebraic multiplicities), then there is an invertible matrix \(P\) so that \(A=P^{-1} B P\).

Short Answer

Expert verified
a. Since a matrix with n distinct eigenvalues has n linearly independent eigenvectors, it is diagonalizable. b. If A is diagonalizable and AB = BA, then B is also diagonalizable. c. If A = P^{-1}BP, A and B have the same eigenvalues. d. It is not necessarily true that if A and B have the same eigenvalues, there is an invertible matrix P such that A = P^{-1}BP. (Counterexample provided) e. No real 2 脳 2 matrix A satisfies A虏 = -I. f. If A and B are diagonalizable and have the same eigenvalues with the same algebraic multiplicities, there exists an invertible matrix P such that A = P^{-1}BP.

Step by step solution

01

a. If A is an n 脳 n matrix with n distinct (real) eigenvalues, then A is diagonalizable.

By definition, a matrix is diagonalizable if it is similar to a diagonal matrix, meaning that there exists an invertible matrix P such that A = PDP^{-1}, where D is a diagonal matrix. In this case, A has n distinct eigenvalues. We will show that any matrix meeting this condition possesses a complete set of linearly independent eigenvectors, which implies that it is diagonalizable. The eigenvectors associated with distinct eigenvalues are linearly independent due to the following reasoning: Let 位鈧� and 位鈧� be two eigenvalues of A, where 位鈧� 鈮� 位鈧�, and let their corresponding eigenvectors be v鈧� and v鈧�. If we assume that these eigenvectors are linearly dependent, then we can write v鈧� = kv鈧�. Substituting v鈧� in Av鈧� = 位鈧倂鈧�, we get A(kv鈧�) = 位鈧�(kv鈧�). It follows that Av鈧� = 位鈧倂鈧�, which contradicts our original assumption that Av鈧� = 位鈧乿鈧�. Therefore, the eigenvectors are linearly independent. Since A is an n x n matrix with n linearly independent eigenvectors, it is diagonalizable.
02

b. If A is diagonalizable and AB=BA, then B is diagonalizable.

Since A is diagonalizable, there is an invertible matrix P such that A = PDP^{-1}. If AB = BA, then we can write this equation as AP = PB. Now, we will multiply this equation by P^{-1} from the right: ABP^{-1} = PBP^{-1}. Plugging in A = PDP^{-1}, we get DP = PBP^{-1}P, which simplifies to DP = PB. This equation shows that B also commutes with the diagonal matrix D of A. We can write D as a diagonal matrix with its eigenvalues 位鈧�, 位鈧�, ..., 位鈧� on its diagonal. Then, for each row i, we have the following equality: 位岬v_j = PBv_j, where v_j are the columns of P and j = 1,2,...,n. Since 位岬� 鈮� 位獗� for i 鈮� j, and all v_j's are linearly independent (because P is invertible), we see that Bv_j is an eigenvector of B with eigenvalue 位岬� for each corresponding eigenvector v_j of A. Thus, every eigenvector of A is also an eigenvector of B. Since A has a complete set of linearly independent eigenvectors, so does B. Therefore, B is diagonalizable.
03

c. If there is an invertible matrix P so that A=P^{-1}BP, then A and B have the same eigenvalues.

Let 位 be an eigenvalue of A with a corresponding eigenvector x. Then, Ax = 位x. We will show that 位 is also an eigenvalue of B. Multiplying both sides of the given equation A = P^{-1}BP by P, we get AP = BP. Since Ax = 位x, we can write APx = 位Px. It follows that BPx = 位Px, and B(Px) = 位(Px). Since Px is nonzero (as P is invertible), we see that 位 is also an eigenvalue of B with a corresponding eigenvector Px. Therefore, A and B have the same eigenvalues.
04

d. If A and B have the same eigenvalues, then there is an invertible matrix P so that A=P^{-1}BP.

This statement is false, and we can provide a counterexample. Consider the following matrices: A = \(\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}\) and B = \(\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}\). Both A and B have the same eigenvalues (位 = 1 with algebraic multiplicity 2). However, there is no invertible matrix P that satisfies A = P^{-1}BP. Therefore, this statement is not true in general.
05

e. There is no real 2 脳 2 matrix A satisfying A虏=-I.

We can use a contradiction approach to prove this statement. Suppose there is a real 2 脳 2 matrix A that satisfies A虏 = -I. Let A = \(\begin{bmatrix}a & b\\c & d\end{bmatrix}\), where a, b, c, and d are real numbers. Then, A虏 = \(\begin{bmatrix}a & b\\c & d\end{bmatrix}\begin{bmatrix}a & b\\c & d\end{bmatrix} = \begin{bmatrix}a^2+ bc & ab + bd\\ca+cd & cb+ d^2\end{bmatrix}\). Since A虏 = -I, we have: a虏 + bc = -1, ab + bd = 0, ca + cd = 0, cb + d虏 = -1. However, squaring and summing the second and third equations yields (ab + bd)虏 + (ca + cd)虏 = (a虏 + d虏)(b虏 + c虏) = 0. This equation implies that either a虏 + d虏 = 0 or b虏 + c虏 = 0. In either case, we obtain a contradiction because both a虏 + bc + 1 = 0 and cb + d虏 + 1 = 0 cannot be satisfied simultaneously. Therefore, there is no real 2 脳 2 matrix A satisfying A虏 = -I.
06

f. If A and B are diagonalizable and have the same eigenvalues (with the same algebraic multiplicities), then there is an invertible matrix P so that A=P^{-1}BP.

Since both A and B are diagonalizable, there exist invertible matrices P and Q such that: A = PDP^{-1} and B = QDQ^{-1}, Where D is a diagonal matrix with the eigenvalues of A and B on its diagonal. Now, we want to find a matrix P such that A = P^{-1}BP. Substituting the expressions for A and B, we get: PDP^{-1} = (PQ)D(PQ)^{-1}. Thus, setting R = PQ, we have A = RDR^{-1} and B = RDQ^{-1} by simply setting R = PQ. We see that R is invertible because both P and Q are invertible. Therefore, if A and B are diagonalizable and have the same eigenvalues with the same algebraic multiplicities, there exists an invertible matrix P such that A = P^{-1}BP.

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Key Concepts

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

Eigenvalues and Eigenvectors
Understanding eigenvalues and eigenvectors is fundamental to grasping the concept of a diagonalizable matrix. An eigenvalue of a matrix is a scalar that, when we multiply it by a given eigenvector, returns the same result as multiplying the matrix by that eigenvector. In formal mathematical language, for matrix A and nonzero vector v, the equation Av = 位v, holds true where 位 represents an eigenvalue.

Each eigenvalue corresponds to an eigenvector, which is a non-zero vector that maintains its direction after being transformed by the matrix. The eigenvectors related to distinct eigenvalues are particularly important because they provide the basis for the vector space in which they reside, and contribute to the matrix's diagonalizability鈥攁 matrix is diagonalizable if it can be represented as a diagonal matrix in some basis, usually the basis of its eigenvectors.

Distinct Eigenvalues and Diagonalizability

As in our exercise (a), a matrix with n distinct (real) eigenvalues is guaranteed to be diagonalizable because each of these eigenvalues has an associated, linearly independent eigenvector. This set of eigenvectors can form an invertible matrix P, which leads to the diagonal representation of the original matrix through similarity transformation, as A = PDP-1. The diagonal matrix D contains the eigenvalues on the diagonal, and the matrix's power properties and exponential functions become straightforward to compute.
Matrix Similarity
Matrix similarity is a relation that connects two matrices, indicating they represent the same linear transformation in different bases. Essentially, for two n 脳 n matrices A and B, they are similar if there exists an invertible matrix P, such that B = P-1AP. It is notable here that similarity preserves eigenvalues, which means that similar matrices have the same eigenvalues, as seen in our exercise (c).

Moreover, a matrix is diagonalizable if it is similar to a diagonal matrix. In the context of the solved exercise (f), if matrices A and B are diagonalizable with the same eigenvalues, they are not only similar to the same diagonal matrix D but can also be interconverted by an appropriate invertible matrix P, confirming the statement that A and B are indeed similar to each other.
Invertible Matrix
The concept of an invertible matrix, also known as a non-singular or full-rank matrix, is central to the discussion of matrix similarity and diagonalizability. An invertible matrix P is one for which there exists another matrix P-1, known as the inverse of P, satisfying the condition that PP-1 = P-1P = I, where I is the identity matrix.

One crucial property of invertible matrices is their ability to maintain the eigenvalues of a transformation when transitioning between different bases. This property is utilized in exercises (b) and (c), where the invertibility of matrix P guarantees that commutative matrices A and B have the same eigenvalues, and if A is diagonalizable, then B is also diagnosable. However, the statement in exercise (d) is a common source of misunderstanding鈥攊t incorrectly suggests that having the same eigenvalues implies similarity, which is not always the case, as similarity requires that the eigenvectors align as well in an ordered basis. Such subtle nuances underscore the importance of having a solid understanding of each of these foundational concepts.

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Most popular questions from this chapter

a. Let \(A\) be a stochastic matrix with positive entries, let \(\mathbf{x} \in \mathbb{R}^{n}\), and let \(\mathbf{y}=A \mathbf{x}\). Show that $$ \left|y_{1}\right|+\left|y_{2}\right|+\cdots+\left|y_{n}\right| \leq\left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{n}\right| $$ and that equality holds if and only if all the (nonzero) entries of \(\mathbf{x}\) have the same sign. b. Show that if \(A\) is a stochastic matrix with positive entries and \(\mathbf{x}\) is an eigenvector with eigenvalue 1 , then all the entries of \(\mathbf{x}\) have the same sign. c. Prove using part \(b\) that if \(A\) is a stochastic matrix with positive entries, then there is a unique probability vector in \(\mathbf{E}(1)\) and hence \(\operatorname{dim} \mathbf{E}(1)=1\). d. Prove that if \(\lambda\) is an eigenvalue of a stochastic matrix with positive entries, then \(|\lambda| \leq 1 .\) e. Assume \(A\) is a diagonalizable, regular stochastic matrix. Prove Theorem \(3.3\).

Sketch the following conic sections, giving axes of symmetry and asymptotes (if any). a. \(6 x_{1} x_{2}-8 x_{2}^{2}=9\) "b. \(3 x_{1}^{2}-2 x_{1} x_{2}+3 x_{2}^{2}=4\) "c. \(16 x_{1}^{2}+24 x_{1} x_{2}+9 x_{2}^{2}-3 x_{1}+4 x_{2}=5\) d. \(10 x_{1}^{2}+6 x_{1} x_{2}+2 x_{2}^{2}=11\) e. \(7 x_{1}^{2}+12 x_{1} x_{2}-2 x_{2}^{2}-2 x_{1}+4 x_{2}=6\)

Suppose \(A\) is a square matrix. Suppose \(\mathbf{x}\) is an eigenvector of \(A\) with corresponding eigenvalue \(\lambda\), and \(\mathbf{y}\) is an eigenvector of \(A^{\top}\) with corresponding eigenvalue \(\mu\). Show that if \(\lambda \neq \mu\), then \(\mathbf{x} \cdot \mathbf{y}=0\).

Show that if \(\lambda\) is an eigenvalue of the \(2 \times 2\) matrix \(\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\) and either \(b \neq 0\) or \(\lambda \neq a\), then \(\left[\begin{array}{c}b \\\ \lambda-a\end{array}\right]\) is a corresponding eigenvector.

Prove that if \(A\) is a positive semidefinite (symmetric) matrix (see Exercise 12 for the definition), then there is a unique positive semidefinite (symmetric) matrix \(B\) with \(B^{2}=A\).
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