MA 265 Final Exam
- Thursday, 8/2
- 3:30pm-5:30pm
- FRNY G140
- 20 multiple choice problems
- Covers all lessons
- no special emphasis on material since exam 2
7.1 Diagonalization of Symmetric Matrices (continued)
If \( A \) is symmetric (\( A^T = A \)) then \( A = PDP^{-1} = PDP^T \) where \( P \) is an orthogonal matrix whose columns are orthonormal eigenvectors of \( A \) and \( D \) is a diagonal matrix with the eigenvalues on the main diagonal.
Example
\[ A = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix} \quad \lambda = 1, 4 \]
Note: \( \lambda = 1 \) is a repeated eigenvalue (\( 1, 1, 4 \)).
For \( \lambda = 1 \):
Solving \( (A - \lambda I)\vec{v} = \vec{0} \):
\[ \begin{bmatrix} 1 & -1 & 1 & 0 \\ -1 & 1 & -1 & 0 \\ 1 & -1 & 1 & 0 \end{bmatrix} \sim \begin{bmatrix} 1 & -1 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \]
\( x_2 \) is free, \( x_3 \) is free. From the first row: \( x_1 = x_2 - x_3 \).
\[ \vec{x} = x_2 \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} \quad \vec{v}_1 = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, \vec{v}_2 = \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} \]
Eigenvalue Calculation and Normalization
For the eigenvalue λ = 4, we consider the following matrix reduction:
\[ \begin{bmatrix} -2 & -1 & 1 & 0 \\ -1 & -2 & -1 & 0 \\ 1 & -1 & -2 & 0 \end{bmatrix} \sim \dots \implies \vec{v}_3 = \begin{bmatrix} -1 \\ -1 \\ 1 \end{bmatrix} \]
Normalize the vectors:
\[ \vec{u}_1 = \frac{\vec{v}_1}{\|\vec{v}_1\|} = \begin{bmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{bmatrix} \]
\[ \vec{u}_2 = \frac{\vec{v}_2}{\|\vec{v}_2\|} = \begin{bmatrix} -1/\sqrt{2} \\ 0 \\ 1/\sqrt{2} \end{bmatrix} \]
\[ \vec{u}_3 = \frac{\vec{v}_3}{\|\vec{v}_3\|} = \begin{bmatrix} 1/\sqrt{3} \\ -1/\sqrt{3} \\ 1/\sqrt{3} \end{bmatrix} \]
Orthogonalization
Make \( \vec{u}_2 \) orthogonal to the rest using the Gram-Schmidt process:
\[ \vec{w}_2 = \vec{u}_2 - \langle \vec{u}_2, \vec{u}_1 \rangle \vec{u}_1 - \langle \vec{u}_2, \vec{u}_3 \rangle \vec{u}_3 \]
\[ = \begin{bmatrix} -1/\sqrt{2} \\ 0 \\ 1/\sqrt{2} \end{bmatrix} - \frac{1}{2} \begin{bmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{bmatrix} = \begin{bmatrix} -3/2\sqrt{2} \\ -1/2\sqrt{2} \\ 1/\sqrt{2} \end{bmatrix} \]
Normalization and Matrix Construction
Normalize the resulting vector:
\[ \vec{w}_2 = \dots = \begin{bmatrix} 1/\sqrt{6} \\ 2/\sqrt{6} \\ 1/\sqrt{6} \end{bmatrix} \]
Constructing matrices \( P \) and \( D \):
\[ P = \begin{bmatrix} 1/\sqrt{2} & 1/\sqrt{3} & 1/\sqrt{6} \\ 1/\sqrt{2} & -1/\sqrt{3} & 2/\sqrt{6} \\ 0 & 1/\sqrt{3} & 1/\sqrt{6} \end{bmatrix} \quad D = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]
Some Properties of Symmetric Matrices
If \( A \) is symmetric, then \( (A\vec{x}) \cdot \vec{y} = \vec{x} \cdot (A\vec{y}) \).
Proof
If \( A \) is symmetric, then \( A = A^T \).
\[ (A\vec{x}) \cdot \vec{y} = (A\vec{x})^T \vec{y} = \vec{x}^T A^T \vec{y} \]
\[ = \vec{x}^T (A\vec{y}) \]
\[ = \vec{x} \cdot (A\vec{y}) \]
Properties of Symmetric and Orthogonally Diagonalizable Matrices
If \( A \) is symmetric, \( A^2 \) is symmetric.
Why?
\( A = A^T \)
\[ (A^2)^T = (AA)^T = A^T A^T = (A^T)^2 = A^2 \]
What about \( A^3 \)?
\[ (A^3)^T = (AA^2)^T = (A^2)^T A^T = A^2 A = A^3 \]
Inverses of Orthogonally Diagonalizable Matrices
If \( A \) is orthogonally diagonalizable and invertible, then \( A^{-1} \) is also orthogonally diagonalizable.
\[ A = PDP^{-1} = PDP^T \]
\[ A^{-1} = (PDP^{-1})^{-1} = (P^{-1})^{-1} D^{-1} P^{-1} \]\[ = P D^{-1} P^{-1} \]
Note: \( D^{-1} \) is also diagonal
e.g. \( D = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix} \)
\[ \begin{bmatrix} 1 & 0 & 0 & | & 1 & 0 & 0 \\ 0 & 2 & 0 & | & 0 & 1 & 0 \\ 0 & 0 & 3 & | & 0 & 0 & 1 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 0 & | & 1 & 0 & 0 \\ 0 & 1 & 0 & | & 0 & 1/2 & 0 \\ 0 & 0 & 1 & | & 0 & 0 & 1/3 \end{bmatrix} \]
Where \( D^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1/2 & 0 \\ 0 & 0 & 1/3 \end{bmatrix} \)
Spectral Decomposition
The set of eigenvalues of \( A \) is called the spectrum of \( A \).
If \( A \) is orthogonally diagonalizable, then:
\[ A = PDP^T \]\[ = \begin{bmatrix} \vec{u}_1 & \vec{u}_2 & \dots & \vec{u}_n \end{bmatrix} \begin{bmatrix} \lambda_1 & & 0 \\ & \ddots & \\ 0 & & \lambda_n \end{bmatrix} \begin{bmatrix} \vec{u}_1^T \\ \vdots \\ \vec{u}_n^T \end{bmatrix} \]\[ = \begin{bmatrix} \lambda_1 \vec{u}_1 & \lambda_2 \vec{u}_2 & \dots & \lambda_n \vec{u}_n \end{bmatrix} \begin{bmatrix} \vec{u}_1^T \\ \vdots \\ \vec{u}_n^T \end{bmatrix} \]\[ = \lambda_1 \vec{u}_1 \vec{u}_1^T + \lambda_2 \vec{u}_2 \vec{u}_2^T + \dots + \lambda_n \vec{u}_n \vec{u}_n^T \]
If \( A \) is \( n \times n \), then \( \vec{u}_i \) is \( n \times 1 \) and \( \vec{u}_i \vec{u}_i^T \) is \( n \times n \).
\[ A\vec{x} = \lambda_1 \vec{u}_1 \vec{u}_1^T \vec{x} + \lambda_2 \vec{u}_2 \vec{u}_2^T \vec{x} + \dots + \lambda_n \vec{u}_n \vec{u}_n^T \vec{x} \]
Note: The term \( \vec{u}_1 \vec{u}_1^T \vec{x} \) represents the projection of \( \vec{x} \) onto the subspace spanned by \( \vec{u}_1 \).
Example
\[ A = \begin{bmatrix} 7 & 4 & -4 \\ 4 & 5 & 0 \\ -4 & 0 & 9 \end{bmatrix} \]
\( \lambda_1 = 13, \quad \vec{u}_1 = \begin{bmatrix} -2/3 \\ -1/3 \\ 2/3 \end{bmatrix} \)
\( \lambda_2 = 7, \quad \vec{u}_2 = \begin{bmatrix} 1/3 \\ 2/3 \\ 2/3 \end{bmatrix} \)
\( \lambda_3 = 1, \quad \vec{u}_3 = \begin{bmatrix} -2/3 \\ 2/3 \\ -1/3 \end{bmatrix} \)
\[ P = \begin{bmatrix} \vec{u}_1 & \vec{u}_2 & \vec{u}_3 \end{bmatrix} \]
\[ D = \begin{bmatrix} 13 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]
\[ A = PDP^T = \lambda_1 \vec{u}_1 \vec{u}_1^T + \lambda_2 \vec{u}_2 \vec{u}_2^T + \lambda_3 \vec{u}_3 \vec{u}_3^T \]
\[ = 13 \begin{bmatrix} 4/9 & 2/9 & -4/9 \\ 2/9 & 1/9 & -2/9 \\ -4/9 & -2/9 & 4/9 \end{bmatrix} + 7 \begin{bmatrix} 1/9 & 2/9 & 2/9 \\ 2/9 & 4/9 & 4/9 \\ 2/9 & 4/9 & 4/9 \end{bmatrix} + 1 \begin{bmatrix} 4/9 & -4/9 & 2/9 \\ -4/9 & 4/9 & -2/9 \\ 2/9 & -2/9 & 1/9 \end{bmatrix} \]