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Chapter 8: Problem 1

\(1-8\) Show that \(A=P D P^{-1},\) where \(P\) is a matrix whose columns are the eigenvectors of \(A,\) and \(D\) is a diagonal matrix with the corresponding eigenvalues. \(A=\left[ \begin{array}{ll}{2} & {0} \\ {0} & {1}\end{array}\right]\)

Short Answer

Expert verified
Matrix \(A\) can be expressed as \(A = PDP^{-1}\) with the given \(P\) and \(D\).

Step by step solution

01

Find Eigenvalues of A

Calculate the eigenvalues of matrix \(A\) by solving the characteristic equation \(\det(A - \lambda I) = 0\). Here, \(A\) minus \(\lambda I\) looks like \(\begin{bmatrix} 2-\lambda & 0 \ 0 & 1-\lambda \end{bmatrix}\). The determinant is \((2-\lambda)(1-\lambda) = 0\), so the eigenvalues are \(\lambda_1 = 2\) and \(\lambda_2 = 1\).
02

Find Eigenvectors of A

For each eigenvalue, find the corresponding eigenvector. For \(\lambda_1 = 2\), solve \((A - 2I)\mathbf{v} = 0\). So, \(\begin{bmatrix} 0 & 0 \ 0 & -1 \end{bmatrix}\begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}\). The eigenvector for \(\lambda_1 = 2\) is \(\begin{bmatrix} 1 \ 0 \end{bmatrix}\). Similarly, for \(\lambda_2 = 1\), solve \((A - 1I)\mathbf{v} = 0\). \(\begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix}\begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}\). The eigenvector for \(\lambda_2 = 1\) is \(\begin{bmatrix} 0 \ 1 \end{bmatrix}\).
03

Construct P and D Matrices

The matrix \(P\) is formed by placing the eigenvectors as columns. So, \(P = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\). Matrix \(D\) is a diagonal matrix with eigenvalues on the diagonal: \(D = \begin{bmatrix} 2 & 0 \ 0 & 1 \end{bmatrix}\).
04

Verify A = PDP^{-1}

Calculate \(PDP^{-1}\) and check if it equals \(A\). First, find \(P^{-1}\) which is \(\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\), because \(P\) is the identity matrix. Thus, \(PDP^{-1} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\begin{bmatrix} 2 & 0 \ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 0 \ 0 & 1 \end{bmatrix}\), which is equal to \(A\).

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Key Concepts

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

Matrix Diagonalization
Matrix diagonalization is a process to simplify matrices, making calculations easier. In linear algebra, it involves expressing a matrix in terms of a diagonal matrix and two other matrices. A diagonal matrix is special because it has non-zero elements only on its main diagonal. This makes operations like raising the matrix to a power much simpler. To diagonalize a matrix, you need three matrices: original matrix \( A \), a matrix \( P \), composed of the eigenvectors, and a diagonal matrix \( D \), which holds eigenvalues on its diagonal.Here鈥檚 why matrix diagonalization is useful:
  • It simplifies many matrix computations.
  • Makes matrix functions easier to compute.
  • Helps in understanding the properties of the matrix structure.
The expression \( A = PDP^{-1} \) stands at the heart of matrix diagonalization. To achieve this, the matrix must have as many independent eigenvectors as the dimensions of the matrix. Not all matrices are diagonalizable, which means some matrices do not have enough independent eigenvectors.
Eigenvalues and Eigenvectors
Eigenvalues and eigenvectors are key concepts in linear algebra and are vital in matrix diagonalization. An eigenvector is a non-zero vector that only changes by a scalar factor when a linear transformation is applied. This scalar is called the eigenvalue. You find these by solving the characteristic equation: \( \det(A - \lambda I) = 0 \).Why are they significant?
  • They provide insight into the matrix鈥� properties, like stability and symmetry.
  • Help in solving systems of linear equations.
  • Important in many fields like physics, engineering, and computer science.
In practice, each eigenvalue corresponds to at least one eigenvector. In the exercise, the matrix \( A \) has eigenvalues \( \lambda_1 = 2 \) and \( \lambda_2 = 1 \). The corresponding eigenvectors were found by solving \( (A - \lambda I)\mathbf{v} = 0 \) for each \( \lambda \). These eigenvectors help form the matrix \( P \) used in diagonalization.
Invertible Matrices
An invertible matrix, also known as a non-singular or non-degenerate matrix, is a square matrix that has an inverse. A crucial requirement for matrix diagonalization using the equation \( A = PDP^{-1} \) is that the matrix \( P \) must be invertible. This ensures that you can reverse the transformation made by the matrix.Characteristics of invertible matrices include:
  • Having a non-zero determinant.
  • The existence of a matrix \( P^{-1} \) such that \( PP^{-1} = I \), where \( I \) is the identity matrix.
  • Ability to solve matrix equations uniquely.
In our exercise, the matrix \( P \), which is the identity matrix \( \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \), is perfectly invertible since its own inverse is itself. This property allows \( P \) to serve efficiently in the transformation \( A = PDP^{-1} \). Remember, if a matrix is non-invertible, or singular, this means its determinant is zero and diagonalization cannot proceed as outlined.

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Most popular questions from this chapter

\(20-22\) Write the system of equations and construct the cor- responding matrix model for each of the following exercises. DNA methylation DNA methylation is a process by which methyl chemical groups are attached to the DNA. Thus each gene in the genome can be methylated or unmethylated. Suppose that 80\(\%\) of methylated genes remain methylated and 50\(\%\) of unmethylated genes remain unmethylated each generation.

Find all solutions to the system of linear equations. \(\begin{aligned} 5 x_{1}+x_{2} &=0 \\ 25 x_{1}+5 x_{2} &=0 \end{aligned}\)

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

\(\begin{array}{c}{\text { Leslie matrices Consider the following model for the }} \\ {\text { population size } \mathbf{n}_{t} \text { of an age-structured population with two }} \\ {\text { age classes: }} \\\ {\mathbf{n}_{t+1}=\left[ \begin{array}{cc}{b} & {2} \\ {\frac{1}{2}} & {0}\end{array}\right] \mathbf{n}_{t}}\end{array}\) \(\begin{array}{c}{\text { An equilibrium is a value of the vector for which no change }} \\ {\text { occurs (that is, } \mathbf{n}_{t+1}=\mathbf{n}_{1} ) . \text { Denoting such values by } \hat{\mathbf{n}}, \text { they }} \\ {\text { must therefore satisfy the equation }} \\ {\hat{\mathbf{n}}=\left[ \begin{array}{cc}{b} & {2} \\ {\frac{1}{2}} & {0}\end{array}\right] \hat{\mathbf{n}}}\end{array}\) \(\begin{array}{l}{\text { (a) Suppose that } b \neq 0 . \text { Find all possible equilibrium }} \\ {\text { values. }} \\ {\text { (b) Suppose that } b=0 . \text { Find all possible equilibrium }} \\ {\text { values. }}\end{array}\)

Determine whether or not \(\mathbf{x}\) is an eigenvector of \(A .\) If it is, determine its associated eigenvalue. $$(\mathrm{a})A=\left[ \begin{array}{rr}{3} & {-1} \\ {2} & {0}\end{array}\right] \quad \mathbf{x}=\left[ \begin{array}{l}{1} \\\ {2}\end{array}\right]$$ $$(\mathrm{b})A=\left[ \begin{array}{rrr}{-3} & {-1} & {5} \\ {-2} & {1} & {2} \\ {-2} & {-1} & {4}\end{array}\right] \quad \mathbf{x}=\left[ \begin{array}{l}{2} \\ {1} \\ {1}\end{array}\right]$$ $$(\mathrm{c})A=\left[ \begin{array}{rr}{1} & {-1} \\ {-2} & {0}\end{array}\right] \quad \mathbf{x}=\left[ \begin{array}{l}{1} \\\ {0}\end{array}\right]$$ $$(\mathrm{d}) A=\left[ \begin{array}{rr}{1} & {2} \\ {-2} & {1}\end{array}\right] \quad \mathbf{x}=\left[ \begin{array}{l}{2} \\ {2 i}\end{array}\right]$$ $$(\mathrm{e}) A=\left[ \begin{array}{ll}{2+3 a} & {-2-2 a} \\ {3+3 a} & {-3-2 a}\end{array}\right] \quad \mathbf{x}=\left[ \begin{array}{l}{2} \\\ {3}\end{array}\right]$$ $$(\mathrm{f}) A=\left[ \begin{array}{rrr}{-9} & {4} & {6} \\ {-6} & {3} & {4} \\ {-9} & {4} & {6}\end{array}\right] \quad \mathbf{x}=\left[ \begin{array}{l}{2} \\ {0} \\ {3}\end{array}\right]$$
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