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If \(A\) is invertible, show that \(A B\) is similar to \(B A\) for all \(B\).

Short Answer

Expert verified
Matrices \(AB\) and \(BA\) are similar if \(A\) is invertible, using \(P = A\).

Step by step solution

01

Understand the Definition of Similar Matrices

Two matrices, say \(X\) and \(Y\), are similar if there exists an invertible matrix \(P\) such that \(X = PYP^{-1}\). Our goal is to express \(AB\) as similar to \(BA\) by finding such an invertible matrix \(P\).
02

Use the Property of Inverse Matrices

Since \(A\) is given as invertible, there exists \(A^{-1}\) such that \(A A^{-1} = I\), where \(I\) denotes the identity matrix. We will use \(A\) and \(A^{-1}\) to construct our similarity relation.
03

Constructing the Similarity Relation

For the similarity of \(AB\) to \(BA\), use the matrix \(P = A\). We need to show \(AB = P(BAP^{-1})\). Substituting \(P = A\) and \(P^{-1} = A^{-1}\), verify the expression by calculation: \[(AB = A(BA)A^{-1} = BAA A^{-1} = BAB)\].
04

Confirm the Similarity

It's proven that \(AB = A (BA) A^{-1}\), meaning \(AB\) and \(BA\) are similar matrices based on the definition of similarity. Since \(A\) is invertible, the similarity condition is satisfied with \(P = A\) and \(P^{-1} = A^{-1}\).

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Key Concepts

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

Invertible Matrices
When working with matrices, particularly in linear algebra, we often encounter the concept of invertible matrices. But what does it mean for a matrix to be invertible? Simply put, a matrix is considered invertible if there exists another matrix, known as the inverse, which when multiplied with the original matrix, results in the identity matrix. This is much like how a number has a reciprocal. For example, the reciprocal of 2 is \(\frac{1}{2}\), and when you multiply 2 by \(\frac{1}{2}\), you get 1.
In mathematical terms, a square matrix \(A\) is invertible if there exists a matrix \(A^{-1}\) such that \(A A^{-1} = A^{-1} A = I\), where \(I\) is the identity matrix. Identity matrices are special matrices with 1's on the diagonal and 0's elsewhere, acting as the "1" in matrix calculations.
Invertible matrices are also called non-singular or non-degenerate. Only square matrices (same number of rows and columns) can possibly be invertible. If a matrix is not invertible, we say it is singular, and it does not have an inverse.
Inverse Matrix
The inverse of a matrix is an essential concept in mathematics, especially in solving systems of linear equations and in various applications involving transformations. If you have a matrix \(A\), its inverse \(A^{-1}\) is the matrix you multiply it by to get the identity matrix. When a matrix is multiplied by its inverse, it essentially "undoes" the effect of the original matrix.
Finding the inverse involves a specific process which usually requires that the matrix is square and has a non-zero determinant. The determinant is a special number that can be calculated from a square matrix, and if it is zero, the matrix is not invertible.
For a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\), the inverse is calculated as:
  • First, find the determinant, \(ad - bc\).
  • If the determinant is non-zero, compute the inverse using the formula: \[A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]
In larger matrices, finding the inverse often involves more complex methods such as row operations or utilizing specific algorithms like the Gauss-Jordan method.
Matrix Multiplication
Matrix multiplication is a fundamental operation in linear algebra and is not performed in the same way as regular arithmetic multiplication. To multiply two matrices, you perform a specific operation that involves rows and columns. To multiply matrix \(A\) by \(B\), you must follow these rules:
  • Make sure the number of columns in the first matrix is equal to the number of rows in the second matrix.
  • The result will be a new matrix with dimensions determined by the rows of the first matrix and the columns of the second matrix.
  • Each element in the resulting matrix is obtained by taking the dot product of the rows of the first matrix with the columns of the second.
For example, if \(A\) is a 2x3 matrix and \(B\) is a 3x2 matrix, the resulting matrix \(C = AB\) will be 2x2. The \(i, j\)-th element of matrix \(C\) is calculated by summing the products of the corresponding elements from the \(i\)-th row of \(A\) and the \(j\)-th column of \(B\).
It's important to remember that matrix multiplication is not commutative, which means that \(AB\) does not necessarily equal \(BA\). However, properties of invertible matrices and matrix similarity can sometimes show special relationships between \(AB\) and \(BA\), as seen in this exercise.

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Most popular questions from this chapter

We write vectors \(\mathbb{R}^{n}\) as rows. Find a basis and calculate the dimension of the following subspaces of \(\mathbb{R}^{4}\). a. \(\text {span}\\{(1,-1,2,0),(2,3,0,3),(1,9,-6,6)\\}\) b. \(\text {span}\\{(2,1,0,-1),(-1,1,1,1),(2,7,4,1)\\}\) c. \(\text {span}\\{(-1,2,1,0),(2,0,3,-1),(4,4,11,-3), (3,-2,2,-1)\\}\) d. span \(\\{(-2,0,3,1),(1,2,-1,0),(-2,8,5,3), (-1,2,2,1)\\}\)

The yield \(y\) of wheat in bushels per acre appears to be a linear function of the number of days \(x_{1}\) of sunshine, the number of inches \(x_{2}\) of rain, and the number of pounds \(x_{3}\) of fertilizer applied per acre. Find the best fit to the data in the table by an equation of the form \(y=r_{0}+r_{1} x_{1}+r_{2} x_{2}+r_{3} x_{3} .\) [Hint: If a calculator for inverting \(A^{T} A\) is not available, the inverse is given in the answer.] $$ \begin{array}{|c|c|c|c|} \hline y & x_{1} & x_{2} & x_{3} \\ \hline 28 & 50 & 18 & 10 \\ 30 & 40 & 20 & 16 \\ 21 & 35 & 14 & 10 \\ 23 & 40 & 12 & 12 \\ 23 & 30 & 16 & 14 \\ \hline \end{array} $$

The following table gives IQ scores for 10 fathers and their eldest sons. Calculate the means, the variances, and the correlation coefficient \(r\). (The data scaling formula is useful.) $$ \begin{array}{|l|c|c|c|c|c|c|c|c|c|c|} \hline & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline \text { Father's IQ } & 140 & 131 & 120 & 115 & 110 & 106 & 100 & 95 & 91 & 86 \\ \text { Son's IQ } & 130 & 138 & 110 & 99 & 109 & 120 & 105 & 99 & 100 & 94 \\\ \hline \end{array} $$

a. Show that two diagonalizable matrices are similar if and only if they have the same eigenvalues with the same multiplicities. b. If \(A\) is diagonalizable, show that \(A \sim A^{T}\). c. Show that \(A \sim A^{T}\) if \(A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\)

We write vectors \(\mathbb{R}^{n}\) as rows. Suppose that \(\\{\mathbf{x}, \mathbf{y}, \mathbf{z}, \mathbf{w}\\}\) is a basis of \(\mathbb{R}^{4}\). Show that: a. \(\\{\mathbf{x}+a \mathbf{w}, \mathbf{y}, \mathbf{z}, \mathbf{w}\\}\) is also a basis of \(\mathbb{R}^{4}\) for any choice of the scalar \(a\). b. \(\\{\mathbf{x}+\mathbf{w}, \mathbf{y}+\mathbf{w}, \mathbf{z}+\mathbf{w}, \mathbf{w}\\}\) is also a basis of \(\mathbb{R}^{4}\). c. \(\\{\mathbf{x}, \mathbf{x}+\mathbf{y}, \mathbf{x}+\mathbf{y}+\mathbf{z}, \mathbf{x}+\mathbf{y}+\mathbf{z}+\mathbf{w}\\}\) is also a basis of \(\mathbb{R}^{4}\)

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