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Let \(\lambda\) be an eigenvalue of \(A\) with corresponding eigenvector \(\mathbf{x}\). If \(B=P^{-1} A P\) is similar to \(A\), show that \(P^{-1} \mathbf{x}\) is an eigenvector of \(B\) corresponding to \(\lambda\)

Short Answer

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\( P^{-1} \mathbf{x} \) is an eigenvector of \( B \) with eigenvalue \( \lambda \).

Step by step solution

01

Understanding the Similarity

Matrix \( B \) is similar to \( A \), meaning \( B = P^{-1} A P \), where \( P \) is an invertible matrix. This similarity implies that \( A \) and \( B \) have the same eigenvalues. Therefore, if \( \lambda \) is an eigenvalue of \( A \), it is also an eigenvalue of \( B \).
02

Eigenvector Relationship for A

Given that \( \lambda \) is an eigenvalue of \( A \) with corresponding eigenvector \( \mathbf{x} \), we have the equation \( A \mathbf{x} = \lambda \mathbf{x} \).
03

Applying Similarity Transformation

By the properties of similar matrices, substituting \( A \) in terms of \( B \), we have \( A = P B P^{-1} \). Using this, and given \( A \mathbf{x} = \lambda \mathbf{x} \), we can express \( A \mathbf{x} = \lambda \mathbf{x} \) as \( P B P^{-1} \mathbf{x} = \lambda \mathbf{x} \).
04

Transforming the Eigenvector

Let \( \mathbf{y} = P^{-1} \mathbf{x} \). Then, the equation \( P B \mathbf{y} = \lambda P \mathbf{y} \) can be derived from substituting \( \mathbf{y} \) in Step 3, which simplifies to \( B \mathbf{y} = \lambda \mathbf{y} \) because \( P \) is invertible, canceling with its inverse.
05

Conclusion of Eigenvector for B

From the equation \( B \mathbf{y} = \lambda \mathbf{y} \), it follows that \( \mathbf{y} = P^{-1} \mathbf{x} \) is an eigenvector of \( B \) corresponding to eigenvalue \( \lambda \).

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Key Concepts

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

Matrix Similarity
Matrix similarity is an important concept in linear algebra that helps us understand how matrices relate to each other. Two matrices, say \( A \) and \( B \), are said to be similar if there exists an invertible matrix \( P \) such that \( B = P^{-1} A P \). Similar matrices share several key properties:
  • They have the same eigenvalues, although their eigenvectors may differ.
  • Since they represent the same linear transformation in different bases, we can think of similarity as a way of translating one matrix into another form that might be easier to understand or compute with.
  • Similarity preserves the determinant and trace of the matrices, meaning these values remain unchanged as well.
Understanding matrix similarity is crucial when solving problems involving eigenvalues and eigenvectors, as it allows us to find solutions and truths about a matrix by studying a related matrix.
Linear Transformation
A linear transformation is a fundamental concept in linear algebra, representing a function between two vector spaces that preserves the operations of vector addition and scalar multiplication. When considering a matrix \( A \), applying \( A \) to a vector \( \mathbf{x} \) in a vector space \( V \) results in another vector in \( V \):
  • This operation is often written as \( A \mathbf{x} \), representing how \( A \) transforms \( \mathbf{x} \).
  • The matrix \( A \) itself can be seen as providing the instructions for this transformation.
  • Linear transformations can be rotations, reflections, scaling, or any combination of these.
These transformations are crucial in various fields such as computer graphics and systems of differential equations. By understanding how matrices effectuate such transformations, one can manipulate and predict the behaviors of complex systems.
Invertible Matrix
An invertible matrix, also known as a non-singular or non-degenerate matrix, has a crucial property: it can be reversed. Specifically, a square matrix \( P \) is invertible if there exists another matrix \( P^{-1} \) such that:
  • \( P P^{-1} = P^{-1} P = I \), where \( I \) is the identity matrix.
  • The identity matrix \( I \) serves as the multiplicative identity in matrix algebra, having the property that multiplying any matrix by \( I \) leaves the original matrix unchanged.
  • An invertible matrix has a non-zero determinant, making its inverse well-defined.
In the context of eigenvalues and eigenvectors, invertible matrices are often used to change the basis in which a problem is considered, facilitating the computation or understanding of a problem. This is why they are integral to the concept of matrix similarity, providing the bridge through which one can connect two similar matrices.

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Most popular questions from this chapter

In each case find bases for the row and column spaces of \(A\) and determine the rank of \(A\). a. \(\left[\begin{array}{rrrr}2 & -4 & 6 & 8 \\ 2 & -1 & 3 & 2 \\ 4 & -5 & 9 & 10 \\ 0 & -1 & 1 & 2\end{array}\right]\) b. \(\left[\begin{array}{rrr}2 & -1 & 1 \\ -2 & 1 & 1 \\ 4 & -2 & 3 \\ -6 & 3 & 0\end{array}\right]\) c. \(\left[\begin{array}{rrrrr}1 & -1 & 5 & -2 & 2 \\ 2 & -2 & -2 & 5 & 1 \\\ 0 & 0 & -12 & 9 & -3 \\ -1 & 1 & 7 & -7 & 1\end{array}\right]\) d. \(\left[\begin{array}{rrrr}1 & 2 & -1 & 3 \\ -3 & -6 & 3 & -2\end{array}\right]\)

Use the Cauchy inequality to prove that: a. \(r_{1}+r_{2}+\cdots+r_{n} \leq n\left(r_{1}^{2}+r_{2}^{2}+\cdots+r_{n}^{2}\right)\) for all \(r_{i}\) in \(\mathbb{R}\) and all \(n \geq 1\) b. \(r_{1} r_{2}+r_{1} r_{3}+r_{2} r_{3} \leq r_{1}^{2}+r_{2}^{2}+r_{3}^{2}\) for all \(r_{1}, r_{2},\) and \(r_{3}\) in \(\mathbb{R}\). [Hint: See part (a).]

Let \(A\) denote an \(n \times n\) upper triangular matrix. a. If all the main diagonal entries of \(A\) are distinct, show that \(A\) is diagonalizable. b. If all the main diagonal entries of \(A\) are equal, show that \(A\) is diagonalizable only if it is already diagonal. c. Show that \(\left[\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 2\end{array}\right]\) is diagonalizable but that \(\left[\begin{array}{lll}1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2\end{array}\right]\) is not diagonalizable.

In each case, decide whether the matrix \(A\) is diagonalizable. If so, find \(P\) such that \(P^{-1} A P\) is diagonal. $$ \text { a. }\left[\begin{array}{lll} 1 & 0 & 0 \\ 1 & 2 & 1 \\ 0 & 0 & 1 \end{array}\right] \quad \text { b. }\left[\begin{array}{rrr} 3 & 0 & 6 \\ 0 & -3 & 0 \\ 5 & 0 & 2 \end{array}\right] $$ c. \(\left[\begin{array}{rrr}3 & 1 & 6 \\ 2 & 1 & 0 \\ -1 & 0 & -3\end{array}\right]\) d. \(\left[\begin{array}{lll}4 & 0 & 0 \\ 0 & 2 & 2 \\ 2 & 3 & 1\end{array}\right]\)

We often write vectors in \(\mathbb{R}^{n}\) as rows. If \(a_{1}, a_{2}, \ldots, a_{k}\) are nonzero scalars, show that \(\operatorname{span}\left\\{a_{1} \mathbf{x}_{1}, a_{2} \mathbf{x}_{2}, \ldots, a_{k} \mathbf{x}_{k}\right\\}=\) \(\operatorname{span}\left\\{\mathbf{x}_{1}, \mathbf{x}_{2}, \ldots, \mathbf{x}_{k}\right\\}\) for any vectors \(\mathbf{x}_{i}\) in \(\mathbb{R}^{n}\)

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