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Chapter 7: Problem 4

Determine whether the given matrix \(A\) is diagonalizable. Where possible, find a matrix \(S\) such that $$S^{-1} A S=\operatorname{diag}\left(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n}\right).$$ $$A=\left[\begin{array}{ll}1 & -8 \\\2 & -7\end{array}\right]$$

Short Answer

Expert verified
The given matrix \(A\) is diagonalizable with the matrix \(S = \begin{bmatrix} -8/3 & -8 \\ 1 & 1 \end{bmatrix}\) and its inverse \(S^{-1} = \begin{bmatrix} -1/8 & -3 \\ 1/8 & 1 \end{bmatrix}\). The diagonal matrix obtained from the diagonalization is \(\operatorname{diag}(4, 2)\).

Step by step solution

01

Find the eigenvalues of A

To find the eigenvalues of the matrix \(A\), let's solve the characteristic equation, \(\det(A-\lambda I) = 0\): \[ \det \left( \begin{bmatrix} 1-\lambda & -8 \\ 2 & -7 - \lambda \end{bmatrix} \right) = (1-\lambda)(-7-\lambda) - (-8)(2) = 0 \]
02

Solve the characteristic equation for eigenvalues

Let's solve the equation obtained in step 1: \[ \lambda^2 -6\lambda =(\lambda - 4)(\lambda -2)=0 \] This equation yields the eigenvalues \(\lambda_1 = 4\) and \(\lambda_2 = 2\).
03

Find the eigenvectors associated with each eigenvalue

Now, let's find the eigenvectors associated with each eigenvalue by solving the equation \((A - \lambda I)v = 0\) for each eigenvalue. For \(\lambda_1 = 4\): \[ \begin{bmatrix} 1-4 & -8 \\ 2 & -7-4 \end{bmatrix}v_1 = \begin{bmatrix} -3 & -8 \\ 2 & -11 \end{bmatrix}v_1 = 0 \] Row reducing to echelon form, we get: \[ \begin{bmatrix} 1 & 8/3 \\ 0 & 0 \end{bmatrix}v_1 = 0 \] Thus, \(v_1 = \begin{bmatrix} -8/3 \\ 1 \end{bmatrix}\) is the eigenvector associated with \(\lambda_1 = 4\). For \(\lambda_2 = 2\): \[ \begin{bmatrix} 1-2 & -8 \\ 2 & -7-2 \end{bmatrix}v_2 = \begin{bmatrix} -1 & -8 \\ 2 & -9 \end{bmatrix}v_2 = 0 \] Row reducing to echelon form, we get: \[ \begin{bmatrix} 1 & 8 \\ 0 & 1 \end{bmatrix}v_2 = 0 \] Thus, \(v_2 = \begin{bmatrix} -8 \\ 1 \end{bmatrix}\) is the eigenvector associated with \(\lambda_2 = 2\).
04

Check if the eigenvectors form a basis

Since we have two linearly independent eigenvectors and the dimension of the eigenspace is 2 (which is equal to the dimension of matrix A), we can form a basis using these eigenvectors. This implies that the matrix \(A\) is diagonalizable.
05

Form the matrix S and calculate its inverse

Now, let's form the matrix of eigenvectors \(S\): \[ S = \begin{bmatrix} -8/3 & -8 \\ 1 & 1 \end{bmatrix} \] The inverse of \(S\) can be found by using the formula \(S^{-1} = \frac{1}{\mathrm{det}(S)} \mathrm{adj}(S)\), where adj(S) is the adjoint of S. \[ S^{-1} = \frac{1}{(-(8/3)(1)-(1)(-8))} \begin{bmatrix} 1 & 8 \\ -1 & -8/3 \end{bmatrix} = \begin{bmatrix} -1/8 & -3 \\ 1/8 & 1 \end{bmatrix} \]
06

Diagonalize A using the matrix S

Finally, let's use the matrix \(S\) to diagonalize the matrix \(A\) using the formula \(S^{-1}AS = \operatorname{diag}(\lambda_1, \lambda_2)\): \[ \begin{bmatrix} -1/8 & -3 \\ 1/8 & 1 \end{bmatrix} \begin{bmatrix} 1 & -8 \\ 2 & -7 \end{bmatrix} \begin{bmatrix} -8/3 & -8 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 2 \end{bmatrix} \] Thus, the matrix \(A\) is diagonalizable with the matrix \(S\), and \(S^{-1}AS = \operatorname{diag}(4, 2)\).

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Key Concepts

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

Eigenvalues
Eigenvalues are fundamental to understanding how a matrix can be transformed into a simpler form. They are special numbers associated with a matrix that give insight into its behavior. Finding eigenvalues involves solving the characteristic equation, which is derived from the matrix given.

In short, when solving \( ext{det}(A - \lambda I) = 0\), where \(I\) is the identity matrix and \(\lambda\) is a scalar, the roots of this equation are the eigenvalues. For the matrix \(A\) from our exercise, we calculate its eigenvalues to determine whether it can be diagonalized. If a matrix has \(n\) distinct eigenvalues, it is diagonalizable.
  • They provide a scalar measure of transformation along a specific direction.
  • Play a critical role in simplifying complex matrix computations.
Eigenvectors
An eigenvector is a vector that remains in the same linear span even after a linear transformation. It is special because multiplying the matrix by the eigenvector results in a scaled version of the vector.

The equation \(A - \lambda I)v = 0\) is used to find eigenvectors, where \(v\) is the non-zero vector satisfying this equation. Each eigenvalue is associated with one or more eigenvectors. The eigenvectors corresponding to different eigenvalues are linearly independent.
  • Provide directions invariant under the linear transformation.
  • Useful in determining matrix scalability and orientation.
Matrix Algebra
Matrix Algebra encompasses operations with matrices essential for various computations in linear transformations. Understanding these operations is crucial for diagonalizing a matrix. Some of the key operations include finding the determinant and inverse of a matrix.

Using Matrix Algebra, you compute \(S^{-1}AS\) to determine if matrix \(A\) can be written as a diagonal matrix. Once determined, certain calculations become easier to perform.
  • Encompasses addition, multiplication, and inversion of matrices.
  • Essential for transforming matrices into diagonal form.
Linear Independence
Linear Independence is a concept that deals with whether a set of vectors is dependent or independent in a vector space. If no vector in the set is a linear combination of the others, they are linearly independent.

This concept is central when determining if a matrix can be diagonalized. For a matrix to be diagonalizable, its eigenvectors must form a linearly independent set. In the provided example, the matrix \(A\) has two linearly independent eigenvectors, thus ensuring diagonalizability.
  • A critical condition for a matrix to be diagonalizable.
  • Helps in forming a complete basis set for the vector space.

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Most popular questions from this chapter

If \(\mathbf{v}_{1}, \mathbf{v}_{2},\) and \(\mathbf{v}_{3}\) are linearly independent eigenvectors of \(A\) corresponding to the eigenvalue \(\lambda,\) and \(c_{1}, c_{2},\) and \(c_{3}\) are scalars (not all zero), show that \(c_{1} \mathbf{v}_{1}+c_{2} \mathbf{v}_{2}+c_{3} \mathbf{v}_{3}\) is also an eigenvector of \(A\) corresponding to the eigenvalue \(\lambda\).

Find the Jordan canonical form \(J\) for the matrix \(A_{1}\) and determine an invertible matrix \(S\) such that \(S^{-1} A S=J\). \(A=\left[\begin{array}{rr}4 & 4 \\ -4 & 12\end{array}\right]\)

Let \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\) be eigenvectors of \(A\) corresponding to the distinct eigenvalues \(\lambda_{1}\) and \(\lambda_{2},\) respectively. Prove that \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\) are linearly independent. [Hint: Model your proof on the general case considered in Theorem 7.2.5.]

Prove the following properties for similar matrices: (a) A matrix \(A\) is always similar to itself. (b) If \(A\) is similar to \(B,\) then \(B\) is similar to \(A\). (c) If \(A\) is similar to \(B\) and \(B\) is similar to \(C,\) then \(A\) is similar to \(C .\)

The characteristic polynomial \(p(\lambda)\) for a square matrix \(A\) is given. Write down a set \(S\) of matrices such that every square matrix with characteristic polynomial \(p(\lambda)\) is guaranteed to be similar to exactly one of the matrices in the set \(S\). \(p(\lambda)=(3-\lambda)^{2}(-2-\lambda)^{3} \lambda^{2}\).
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