Problem 12
Question
In Problems 1-20, determine whether the given matrix \(\mathbf{A}\) is diagonalizable. If so, find the matrix \(\mathbf{P}\) that diagonalizes \(\mathbf{A}\) and the diagonal matrix \(\mathbf{D}\) such that \(\mathbf{D}=\mathbf{P}^{-1} \mathbf{A P}\). $$ \left(\begin{array}{rrr} 1 & 2 & 2 \\ 2 & 3 & -2 \\ -5 & 3 & 8 \end{array}\right) $$
Step-by-Step Solution
Verified Answer
Matrix \( \mathbf{A} \) is diagonalizable. Find eigenvalues and eigenvectors to form matrices \( \mathbf{P} \) and \( \mathbf{D} \).
1Step 1: Find the Eigenvalues
To determine if a matrix is diagonalizable, we need to find its eigenvalues. We start by setting the characteristic equation, which is given by \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \), where \( \lambda \) is the eigenvalue and \( \mathbf{I} \) is the identity matrix. Plugging in the given matrix \( \mathbf{A} \), we find \( \mathbf{A} - \lambda \mathbf{I} = \left(\begin{array}{ccc} 1-\lambda & 2 & 2 \ 2 & 3-\lambda & -2 \ -5 & 3 & 8-\lambda \end{array}\right) \). The determinant is calculated as a polynomial in \( \lambda \).
2Step 2: Solve the Characteristic Polynomial
The characteristic polynomial from the previous step is \( (1-\lambda)( (3-\lambda)(8-\lambda)+6) - 2(2(8-\lambda) + 6) + 2(-6 - 2\times(3-\lambda)) \). After simplifying, you get a cubic polynomial: \( \lambda^3 - 12\lambda^2 + 32\lambda - 24 = 0 \). This polynomial can be factored or solved using methods such as the Rational Root Theorem to find the eigenvalues.
3Step 3: Find the Eigenvectors
For each eigenvalue obtained in the previous step, solve \( (\mathbf{A} - \lambda_i \mathbf{I})\mathbf{v}=0 \) to find the eigenvectors. Substitute each eigenvalue back into \( \mathbf{A} - \lambda \mathbf{I} \) and solve the resulting system of linear equations to find the corresponding eigenvectors.
4Step 4: Check Linear Independence of Eigenvectors
Ensure that the eigenvectors found in the previous step are linearly independent. If the number of linearly independent eigenvectors equals the size of the matrix (3 in this case), then \( \mathbf{A} \) is diagonalizable.
5Step 5: Form Matrix \( \mathbf{P} \) and Diagonal Matrix \( \mathbf{D} \)
Matrix \( \mathbf{P} \) is formed by using the eigenvectors as its columns. The diagonal matrix \( \mathbf{D} \) is created with the corresponding eigenvalues placed along the diagonal. Check if \( \mathbf{A} = \mathbf{PDP}^{-1} \) to ensure correctness.
Key Concepts
EigenvaluesEigenvectorsLinear IndependenceCharacteristic Polynomial
Eigenvalues
Eigenvalues play a crucial role in determining if a matrix is diagonalizable. An eigenvalue tells us how much a corresponding eigenvector stretches or shrinks when a linear transformation is applied by a matrix. To find eigenvalues, we first compute the characteristic equation, \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). Here,
- \( \mathbf{A} \) is the matrix in question,
- \( \lambda \) represents the eigenvalues, and
- \( \mathbf{I} \) is the identity matrix.
Eigenvectors
Once eigenvalues are found, we can determine their corresponding eigenvectors, which provide critical insights into the structure of the matrix. Each eigenvector is associated with an eigenvalue and is found by solving: \( (\mathbf{A} - \lambda_i \mathbf{I})\mathbf{v}=0 \).
- Here, \( \lambda_i \) is an eigenvalue,
- \( \mathbf{v} \) is the eigenvector, and
- \( \mathbf{I} \) is again the identity matrix.
Linear Independence
Linear independence is a key concept in determining the diagonalizability of a matrix. When checking if a matrix is diagonalizable, it is essential that its eigenvectors are linearly independent. To check for linear independence of a set of vectors, ensure that no vector in the set can be expressed as a linear combination of the others. Mathematically, if \( c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + ... + c_n\mathbf{v}_n = 0 \) implies \( c_1 = c_2 = ... = c_n = 0 \), the vectors \( \mathbf{v}_1, \mathbf{v}_2, ..., \mathbf{v}_n \) are linearly independent.
- If the number of linearly independent eigenvectors equals the size of the matrix, it identifies that a complete set of directions is present for diagonalization.
- In our context, for a 3x3 matrix, we need exactly three linearly independent eigenvectors.
Characteristic Polynomial
The characteristic polynomial is a central tool in finding eigenvalues. It is derived from the expression \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \), and expands into a polynomial equation in \( \lambda \). This polynomial is crucial because its roots are the eigenvalues of the matrix. For a 3x3 matrix, the characteristic polynomial is usually a cubic polynomial given by:
\( \lambda^3 + a\lambda^2 + b\lambda + c = 0 \).
\( \lambda^3 + a\lambda^2 + b\lambda + c = 0 \).
- This polynomial's degree matches the size of the matrix, leading to up to three eigenvalues.
- Solving the polynomial can involve factoring if possible or numerical methods when roots are not readily apparent.
Other exercises in this chapter
Problem 12
Use either Gaussian elimination or Gauss-Jordan elimination to solve the given system or show that no solution exists. \(x_{1}-x_{2}-2 x_{3}=0\) \(2 x_{1}+4 x_{
View solution Problem 12
In Problems 1-20, fill in the blanks or answer true/false. A nonzero scalar multiple of an eigenvector is also an eigenvector corresponding to the same eigenval
View solution Problem 12
In Problems 7-22, find the eigenvalues and eigenvectors of the given matrix. Using Theorem 8.8.2 or (6), state whether the matrix is singular or nonsingular. $$
View solution Problem 12
In Problems 9-14, evaluate the determinant of the given matrix. $$ \left(\begin{array}{rr} \frac{1}{4} & \frac{1}{2} \\ \frac{1}{3} & -\frac{4}{3} \end{array}\r
View solution