Exampile: Let A be a matrix $$ \left(\begin{array}{ll}2 & 4 \ 4 & 3\end{array} ight) $$. Find an orthogonal basis of $$ R^{2} $$ O. consisting of eigenvector of $$ A $$

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Exampile: Let A be a matrix $$ \left(\begin{array}{ll}2 & 4 \ 4 & 3\end{array}ight) $$. Find an orthogonal basis of $$ R^{2} $$ O. consisting of eigenvector of $$ A $$

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To find an **orthogonal basis** of $$\mathbb{R}^{2}$$ consisting of eigenvectors of the matrix
$$
A = \begin{pmatrix} 2 & 4 \ 4 & 3 \end{pmatrix},
$$
we will follow these steps:

1. **Verify Symmetry of $$A$$:**

The matrix $$A$$ is symmetric since $$A = A^{T}$$. Symmetric matrices have real eigenvalues and their eigenvectors corresponding to distinct eigenvalues are orthogonal. This property ensures that we can find an orthogonal basis consisting of eigenvectors of $$A$$.

2. **Find the Eigenvalues of $$A$$:**

To find the eigenvalues $$\lambda$$ of $$A$$, solve the characteristic equation:
   $$
   \det(A - \lambda I) = 0.
   $$
   
   Compute the determinant:
   $$
   \det\begin{pmatrix} 2 - \lambda & 4 \ 4 & 3 - \lambda \end{pmatrix} = (2 - \lambda)(3 - \lambda) - 16.
   $$
   
   Expanding this:
   $$
   (2 - \lambda)(3 - \lambda) - 16 = (6 - 5\lambda + \lambda^{2}) - 16 = \lambda^{2} - 5\lambda - 10 = 0.
   $$
   
   Solving the quadratic equation:
   $$
   \lambda = \frac{5 \pm \sqrt{25 + 40}}{2} = \frac{5 \pm \sqrt{65}}{2}.
   $$
   
   Thus, the eigenvalues are:
   $$
   \lambda_{1} = \frac{5 + \sqrt{65}}{2}, \quad \lambda_{2} = \frac{5 - \sqrt{65}}{2}.
   $$

3. **Find the Eigenvectors Corresponding to Each Eigenvalue:**

For each eigenvalue $$\lambda$$, solve $$(A - \lambda I)\mathbf{v} = \mathbf{0}$$.

- **For $$\lambda_{1} = \frac{5 + \sqrt{65}}{2}$$:**
   
     The system becomes:
     $$
     \begin{cases}
     \left(2 - \lambda_{1}\right)x + 4y = 0, \
     4x + \left(3 - \lambda_{1}\right)y = 0.
     \end{cases}
     $$
     
     Substituting $$\lambda_{1}$$:
     $$
     \begin{cases}
     \frac{-1 - \sqrt{65}}{2}x + 4y = 0, \
     4x + \frac{1 - \sqrt{65}}{2}y = 0.
     \end{cases}
     $$
     
     From the first equation:
     $$
     x = \frac{8}{1 + \sqrt{65}} y.
     $$
     
     To simplify, multiply numerator and denominator by the conjugate $$1 - \sqrt{65}$$:
     $$
     x = \frac{8(1 - \sqrt{65})}{(1 + \sqrt{65})(1 - \sqrt{65})} y = \frac{8(1 - \sqrt{65})}{1 - 65} y = \frac{8(1 - \sqrt{65})}{-64} y = -\frac{1 - \sqrt{65}}{8} y.
     $$
     
     Let $$y = 8$$, then:
     $$
     x = -(1 - \sqrt{65}).
     $$
     
     Thus, an eigenvector corresponding to $$\lambda_{1}$$ is:
     $$
     \mathbf{v}_{1} = \begin{pmatrix} - (1 - \sqrt{65}) \ 8 \end{pmatrix}.
     $$
     
     To make calculations simpler, we can scale the eigenvector by a constant. However, since normalization isn't required for orthogonality, this form suffices.

- **For $$\lambda_{2} = \frac{5 - \sqrt{65}}{2}$$:**
   
     Similarly, the system becomes:
     $$
     \begin{cases}
     \frac{-1 + \sqrt{65}}{2}x + 4y = 0, \
     4x + \frac{1 + \sqrt{65}}{2}y = 0.
     \end{cases}
     $$
     
     From the first equation:
     $$
     x = \frac{8}{1 - \sqrt{65}} y.
     $$
     
     Multiply numerator and denominator by the conjugate $$1 + \sqrt{65}$$:
     $$
     x = \frac{8(1 + \sqrt{65})}{(1 - \sqrt{65})(1 + \sqrt{65})} y = \frac{8(1 + \sqrt{65})}{1 - 65} y = \frac{8(1 + \sqrt{65})}{-64} y = -\frac{1 + \sqrt{65}}{8} y.
     $$
     
     Let $$y = 8$$, then:
     $$
     x = -(1 + \sqrt{65}).
     $$
     
     Thus, an eigenvector corresponding to $$\lambda_{2}$$ is:
     $$
     \mathbf{v}_{2} = \begin{pmatrix} - (1 + \sqrt{65}) \ 8 \end{pmatrix}.
     $$

4. **Form the Orthogonal Basis:**

The eigenvectors $$\mathbf{v}_{1}$$ and $$\mathbf{v}_{2}$$ corresponding to distinct eigenvalues of a symmetric matrix are orthogonal. Therefore, they form an orthogonal basis for $$\mathbb{R}^{2}$$.

Thus, an **orthogonal basis** consisting of eigenvectors of $$A$$ is:
   $$
   \left\{\, \mathbf{v}_{1} = \begin{pmatrix} - (1 - \sqrt{65}) \ 8 \end{pmatrix}, \quad \mathbf{v}_{2} = \begin{pmatrix} - (1 + \sqrt{65}) \ 8 \end{pmatrix} \,\right\}.
   $$

If desired, these vectors can be **normalized** to obtain an orthonormal basis by dividing each vector by its magnitude. However, normalization is not required unless specifically requested.
An orthogonal basis of $$\mathbb{R}^{2}$$ consisting of eigenvectors of matrix $$A$$ is: $$ \left\{ \begin{pmatrix} - (1 - \sqrt{65}) \ 8 \end{pmatrix}, \begin{pmatrix} - (1 + \sqrt{65}) \ 8 \end{pmatrix} \right\} $$

Exampile: Let A be a matrix . Find an orthogonal basis of
O. consisting of eigenvector of

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