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Chapter 8: Problem 13

Show that \(A^{2}=P D^{2} P^{-1},\) where \(P\) is a matrix whose columns are the eigenvectors of \(A,\) and \(D\) is a diagonal matrix with the corresponding eigenvalues.\(A=\left[ \begin{array}{ll}{a} & {0} \\ {0} & {b}\end{array}\right] \quad\) with \(a \neq b\)

Short Answer

Expert verified
The equality \(A^2 = P D^2 P^{-1}\) is verified as both sides equal \(\begin{bmatrix} a^2 & 0 \\ 0 & b^2 \end{bmatrix}\).

Step by step solution

01

Understand the Problem Statement

We need to prove the equality \(A^2 = P D^2 P^{-1}\), given that \(A\) is a diagonal matrix with distinct elements \(a\) and \(b\), \(P\) is the matrix of eigenvectors of \(A\), and \(D\) is the diagonal matrix of eigenvalues.
02

Identify Eigenvalues and Eigenvectors of Matrix A

For the matrix \(A = \begin{bmatrix} a & 0 \ 0 & b \end{bmatrix}\), the eigenvalues are the diagonal elements, \(a\) and \(b\). The corresponding eigenvectors are \(\begin{bmatrix} 1 \ 0 \end{bmatrix}\) and \(\begin{bmatrix} 0 \ 1 \end{bmatrix}\). Hence, \(P = I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\) and \(P^{-1} = P\) because \(P\) is the identity matrix.
03

Construct the Diagonal Matrix D

The diagonal matrix \(D\), based on the eigenvalues, is \(D = \begin{bmatrix} a & 0 \ 0 & b \end{bmatrix}\).
04

Calculate D^2

Since \(D\) is diagonal, \(D^2 = D \times D\) which results in \(D^2 = \begin{bmatrix} a^2 & 0 \ 0 & b^2 \end{bmatrix}\).
05

Evaluate P D^2 P^{-1}

We need to compute \(P D^2 P^{-1}\). Since \(P\) and \(P^{-1}\) are identity, \(P D^2 P^{-1} = D^2 = \begin{bmatrix} a^2 & 0 \ 0 & b^2 \end{bmatrix}\).
06

Calculate A^2

Since \(A\) is already diagonal, \(A^2 = A \times A = \begin{bmatrix} a & 0 \ 0 & b \end{bmatrix} \times \begin{bmatrix} a & 0 \ 0 & b \end{bmatrix} = \begin{bmatrix} a^2 & 0 \ 0 & b^2 \end{bmatrix}\).
07

Verify the Equality

From Steps 5 and 6, we observe that \(A^2 = \begin{bmatrix} a^2 & 0 \ 0 & b^2 \end{bmatrix}\) and \(P D^2 P^{-1} = \begin{bmatrix} a^2 & 0 \ 0 & b^2 \end{bmatrix}\). Thus, \(A^2 = P D^2 P^{-1}\) is verified.

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Key Concepts

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

Eigenvectors
Eigenvectors are crucial components in the understanding of matrix diagonalization. When you have a square matrix, such as matrix \( A \) mentioned in the exercise, eigenvectors are the special vectors associated with this matrix. These vectors, when multiplied by the matrix, result in a scalar multiple of themselves.
In the given problem, the eigenvectors for matrix \( A \) are \( \begin{bmatrix} 1 \ 0 \end{bmatrix} \) and \( \begin{bmatrix} 0 \ 1 \end{bmatrix} \). These are column vectors that point along the coordinate axes.
The process of finding eigenvectors usually involves solving the characteristic equation, which is obtained by setting the determinant of \( (A - \lambda I) \) to zero, where \( \lambda \) is the eigenvalue and \( I \) is the identity matrix. However, in this simplified case, the matrix \( A \) is already in a form that clearly reveals the eigenvectors.
Diagonal Matrix
A diagonal matrix is a matrix in which all the elements outside the main diagonal are zero. This special property makes diagonal matrices particularly useful in computations, such as simplifying the raising of a matrix to a power.
In our exercise, both the matrix \( A \) and the matrix \( D \) are diagonal matrices. The matrix \( D \) is constructed from the eigenvalues of \( A \). For our matrix \( A = \begin{bmatrix} a & 0 \ 0 & b \end{bmatrix} \), the diagonal matrix \( D \) is \( \begin{bmatrix} a & 0 \ 0 & b \end{bmatrix} \).
Diagonal matrices have properties that simplify many mathematical operations. For example, the square of a diagonal matrix, \( D^2 \), can be found by simply squaring each of the diagonal elements. This results in \( \begin{bmatrix} a^2 & 0 \ 0 & b^2 \end{bmatrix} \).
Eigenvalues
Eigenvalues are scalar values that are associated with eigenvectors. For each eigenvector of a matrix, there is a corresponding eigenvalue. In the context of the given exercise, the eigenvalues of a matrix \( A = \begin{bmatrix} a & 0 \ 0 & b \end{bmatrix} \) are the numbers \( a \) and \( b \).
These values are found on the diagonal of the matrix. They indicate how much the eigenvector is stretched or compressed when the matrix is applied.
To derive these eigenvalues from a matrix, one would typically solve the polynomial equation derived from \( \text{det}(A-\lambda I) = 0 \). For our given diagonal matrix \( A \), this step is straightforward as the eigenvalues are the same as the elements \( a \) and \( b \) of the matrix.

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Most popular questions from this chapter

Suppose that all sides of a quadrilateral are equal in length and opposite sides are parallel. Use vector methods to show that the diagonals are perpendicular.

Suppose that \(A^{2}=0\) for some matrix \(A .\) Prove that the only possible eigenvalues of \(A\) are then \(0 .\)

$$\begin{array}{l}{\text { The magnitude of a velocity vector is called speed. Suppose }} \\ {\text { that a wind is blowing from the direction } \mathrm{N} 45^{\circ} \mathrm{W} \text { at a speed }} \\ {\text { of } 50 \mathrm{km} / \mathrm{h} \text { . (This means that the direction from which the }} \\\ {\text { wind blows is } 45^{\circ} \text { west of the northerly direction.) A pilot is }}\end{array}$$ $$ \begin{array}{l}{\text { steering a plane in the direction } \mathrm{N} 60^{\circ} \mathrm{E} \text { at an airspeed (speed }} \\ {\text { in still air) of } 250 \mathrm{km} / \mathrm{h} \text { . The true course, or track, of the }} \\ {\text { plane is the direction of the resultant of the velocity vectors }}\end{array} $$ $$\begin{array}{l}{\text { of the plane and the wind. The ground speed of the plane is }} \\ {\text { the magnitude of the resultant. Find the true course and the }} \\ {\text { ground speed of the plane. }}\end{array}$$

\(\begin{array}{l}{\text { Suppose that } \mathbf{x}=\mathbf{p} \text { and } \mathbf{x}=\mathbf{q} \text { are both solutions to the }} \\ {\text { inhomogeneous system of equations } A \mathbf{x}=\mathbf{b} . \text { Show that }} \\ {\mathbf{z}=\alpha \mathbf{p}+(1-\alpha) \mathbf{q} \text { is then also a solution, where } \alpha \text { is a }} \\ {\text { scalar. }}\end{array}\)

Show that the characteristic polynomials of \(A\) and \(A^{T}\) are the same.
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