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Chapter 9: Problem 6

Show that a matrix \(A\) and its transpose \(A^{T}\) have the same characteristic polynomial.

Short Answer

Expert verified
To show that a matrix A and its transpose \(A^T\) have the same characteristic polynomial, we first define the characteristic polynomial of A as \(p_A(\lambda) = \det(A - \lambda I)\), and the characteristic polynomial of \(A^T\) as \(p_{A^T}(\lambda) = \det(A^T - \lambda I)\). Then, using the property that the transpose of a matrix does not change its determinant, we have \(\det((A - \lambda I)^T) = \det(A - \lambda I)\). Finally, we arrive at the conclusion that \(p_{A^T}(\lambda) = p_A(\lambda)\), showing that A and \(A^T\) have the same characteristic polynomial.

Step by step solution

01

Definition of matrix transpose

By definition, if A is an arbitrary m脳n matrix, then the transpose of A, denoted by \(A^T\), is formed by interchanging the rows and the columns of A. In other words, the row i in A becomes the column i in \(A^T\).
02

Definition of characteristic polynomial

For an n脳n square matrix A, the characteristic polynomial p(位) is defined as the polynomial in 位 obtained by evaluating the determinant of the matrix (A - 位I), where I is the identity matrix of the same order as A, and 位 is an eigenvalue. In mathematical terms, \(p(\lambda) = \det(A - \lambda I)\).
03

Finding the characteristic polynomial of A

To find the characteristic polynomial of A, we need to calculate the determinant of (A - 位I). Let's assume A is an n脳n matrix, with elements denoted by a(i,j). Then the matrix (A - 位I) will have elements equal to (a(i,j) - 位未(i,j)), where 未(i, j)=1 if i=j and 未(i, j)=0 if i鈮爅. Now, we can evaluate the determinant of (A - 位I): \[p_A(\lambda) = \det(A - \lambda I)\]
04

Finding the characteristic polynomial of \(A^T\)

Now, we need to find the characteristic polynomial of \(A^T\). First, we note that the transpose of (A - 位I) is equal to \((A^T - \lambda I)\): \[((A - \lambda I)^T = A^T - (\lambda I^T) = A^T - \lambda I)]\] Now we compute the determinant of \((A^T - 位I)\): \[p_{A^T}(\lambda) = \det(A^T - \lambda I)\]
05

Showing that the characteristic polynomials of A and \(A^T\) are the same

Recall that the transpose of a matrix does not change its determinant, meaning that: \[\det((A - \lambda I)^T) = \det(A - \lambda I)\] So, \[p_{A^T}(\lambda) = \det(A^T - \lambda I) = \det((A - \lambda I)^T) = \det(A - \lambda I) = p_A(\lambda)\] Therefore, we have shown that the matrix A and its transpose \(A^T\) have the same characteristic polynomial.

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Key Concepts

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

Matrix Transpose
A matrix transpose is an operation where we swap the rows and columns of a given matrix. Imagine you have a matrix \(A\). If it has rows labeled as 1, 2, 3, and columns labeled as A, B, C, then its transpose, denoted \(A^T\), will have rows A, B, C and columns 1, 2, 3. This operation is like flipping the matrix along its diagonal.
If matrix \(A\) is an \(m \times n\) matrix (meaning it has \(m\) rows and \(n\) columns), its transpose \(A^T\) becomes an \(n \times m\) matrix.
This is especially simple in square matrices which have the same number of rows and columns because their transpose remains the same size.
  • Example: If \(A = \begin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix}\), then \(A^T = \begin{pmatrix} 1 & 3 \ 2 & 4 \end{pmatrix}\).
  • This simple reordering provides significant implications in various applications such as solving linear equations, computer graphics, and more!
Determinant
The determinant is a special number that we can compute from a square matrix. It鈥檚 a bit like a fingerprint for matrices, as it holds critical information about them.
For a 2x2 matrix \(A = \begin{pmatrix} a & b \ c & d \end{pmatrix}\), the determinant is calculated as \(ad - bc\). For larger matrices, the calculation is more complex, involving a process called cofactor expansion.
The determinant has a few key properties worth noting:
  • If the determinant of a matrix is zero, the matrix is said to be singular, meaning it does not have an inverse.
  • For the characteristic polynomial, we evaluate the determinant of \(A - \lambda I\) to find the roots that are the eigenvalues of the matrix.
This concept is fundamental in linear algebra because it provides ways to solve systems of linear equations, among other things.
Eigenvalues
Eigenvalues are key scalar values associated with a square matrix. They reveal important characteristics about the matrix.
To find eigenvalues, you typically solve the equation \(\det(A - \lambda I) = 0\). Here, \(A\) is a square matrix, \(I\) is the identity matrix of the same dimension, and \(\lambda\) represents the eigenvalues.
What makes eigenvalues fascinating?
  • When a matrix operates on an eigenvector, the result is simply the eigenvector scaled by the eigenvalue.
  • Eigenvalues tell us if a matrix's system is stable, indicating whether solutions grow, decay, or oscillate over time.
Knowing the eigenvalues can inform you about the geometric properties and behavior of the matrix in various applications, from vibration analysis to facial recognition technology.
Square Matrix
A square matrix is one with the same number of rows and columns. These matrices hold special importance in linear algebra, as many important properties and operations apply uniquely to them.
For example, the identity matrix and invertible matrices are always square. Additionally, only square matrices have determinants and can be used to find eigenvalues and eigenvectors.
  • Example: A \(2 \times 2\) matrix is square, such as \(\begin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix}\).
  • Non-square matrices don鈥檛 have determinants or eigenvalues.
Understanding square matrices is foundational because they are the point of study for matrix algebra, crucial in systems of linear equations, and they have rich interactions with transformations.

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Most popular questions from this chapter

Show that a matrix \(A\) and its transpose \(A^{T}\) have the same minimal polynomial.

For each of the following matrices, find all eigenvalues and a maximum set \(S\) of linearly independent eigenvectors: (a) \(A=\left[\begin{array}{lll}1 & -3 & 3 \\ 3 & -5 & 3 \\ 6 & -6 & 4\end{array}\right]\) (b) \(\quad B=\left[\begin{array}{lll}3 & -1 & 1 \\ 7 & -5 & 1 \\ 6 & -6 & 2\end{array}\right]\) (c) \(C=\left[\begin{array}{rrr}1 & 2 & 2 \\ 1 & 2 & -1 \\ -1 & 1 & 4\end{array}\right]\) Which matrices can be diagonalized, and why?

Find the characteristic and minimal polynomials of each of the following matrices: (a) \(A=\left[\begin{array}{lllll}2 & 5 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 4 & 2 & 0 \\ 0 & 0 & 3 & 5 & 0 \\ 0 & 0 & 0 & 0 & 7\end{array}\right]\), (b) \(\quad B=\left[\begin{array}{rrrrr}4 & -1 & 0 & 0 & 0 \\ 1 & 2 & 0 & 0 & 0 \\ 0 & 0 & 3 & 1 & 0 \\ 0 & 0 & 0 & 3 & 1 \\ 0 & 0 & 0 & 0 & 3\end{array}\right]\) (c) \(C=\left[\begin{array}{lllll}3 & 2 & 0 & 0 & 0 \\ 1 & 4 & 0 & 0 & 0 \\ 0 & 0 & 3 & 1 & 0 \\ 0 & 0 & 1 & 3 & 0 \\ 0 & 0 & 0 & 0 & 4\end{array}\right]\)

Find the minimal polynomial \(m(t)\) of each of the following matrices: (a) \(A=\left[\begin{array}{ll}5 & 1 \\ 3 & 7\end{array}\right]\) (b) \(B=\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 3 \\ 0 & 0 & 3\end{array}\right]\) (c) \(\quad C=\left[\begin{array}{rr}4 & -1 \\ 1 & 2\end{array}\right]\)

Prove Theorem 9.10 : The geometric multiplicity of an eigenvalue \(\lambda\) of \(T\) does not exceed its algebraic multiplicity.
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