A= [1,0,1;01,1]. Find the eigenvalues and eigenvectors of the matrix G = A superscript T A, and orthonomalise them. Call them v subscript 1,v subscript 2, v subscript 3 according to the order of h subscript 1>or = h subscript 2>or= h subscript 3 of eigenvalues and G

To solve the problem, let's break it down step by step.

### 1. Matrix \( A \):
Given:
\[
A = \begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 1
\end{bmatrix}
\]

### 2. Calculate \( G = A^T A \):
First, we need to compute the transpose of \( A \) (\( A^T \)), and then multiply \( A^T \) with \( A \) to get the matrix \( G \).

\[
A^T = \begin{bmatrix}
1 & 0 \\
0 & 1 \\
1 & 1
\end{bmatrix}
\]

Now, multiply \( A^T \) with \( A \):
\[
G = A^T A = \begin{bmatrix}
1 & 0 \\
0 & 1 \\
1 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 1
\end{bmatrix}
= \begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 1 \\
1 & 1 & 2
\end{bmatrix}
\]

So, \( G \) is:
\[
G = \begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 1 \\
1 & 1 & 2
\end{bmatrix}
\]

### 3. Find the Eigenvalues and Eigenvectors of \( G \):
To find the eigenvalues, we solve the characteristic equation \( \text{det}(G - \lambda I) = 0 \), where \( \lambda \) represents the eigenvalues and \( I \) is the identity matrix.

The characteristic equation is:
\[
\text{det}\left(\begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 1 \\
1 & 1 & 2
\end{bmatrix} - \lambda \begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}\right) = 0
\]

This simplifies to:
\[
\text{det}\left(\begin{bmatrix}
1-\lambda & 0 & 1 \\
0 & 1-\lambda & 1 \\
1 & 1 & 2-\lambda
\end{bmatrix}\right) = 0
\]

Now we calculate the determinant of this matrix to find the characteristic polynomial.

### 4. Solve the Characteristic Polynomial:
Expanding the determinant, we have:
\[
\text{det}((1-\lambda)(1-\lambda)(2-\lambda) - 1(1) + 0) = 0
\]
The characteristic polynomial is:
\[
\lambda^3 - 4\lambda^2 + 5\lambda - 2 = 0
\]

Solving for \( \lambda \) (eigenvalues), we'll get three eigenvalues.

### 5. Find Eigenvectors:
For each eigenvalue \( \lambda_i \), solve the equation \( (G - \lambda_i I)v_i = 0 \) to find the corresponding eigenvector \( v_i \).

### 6. Orthonormalize the Eigenvectors:
Use the Gram-Schmidt process to orthonormalize the eigenvectors \( v_1, v_2, v_3 \).

### 7. Arrange Eigenvectors:
Finally, arrange the eigenvectors \( v_1, v_2, v_3 \) in the order corresponding to the eigenvalues \( h_1 \geq h_2 \geq h_3 \).

I will now calculate the eigenvalues and corresponding eigenvectors explicitly.The eigenvalues and the corresponding orthonormal eigenvectors of the matrix \( G = A^T A \) are as follows:

### Eigenvalues:
\[
\lambda_1 = 3.0, \quad \lambda_2 = 1.0, \quad \lambda_3 \approx 0
\]

### Orthonormal Eigenvectors:
\[
v_1 = \begin{bmatrix} -0.4082 \\ -0.4082 \\ -0.8165 \end{bmatrix}, \quad
v_2 = \begin{bmatrix} -0.7071 \\ 0.7071 \\ 0 \end{bmatrix}, \quad
v_3 = \begin{bmatrix} -0.5774 \\ -0.5774 \\ 0.5774 \end{bmatrix}
\]

These eigenvectors \( v_1, v_2, v_3 \) correspond to the eigenvalues \( \lambda_1, \lambda_2, \lambda_3 \) in descending order.

Would you like more details or have any questions?

Here are 5 related questions to expand your understanding:

1. How do we derive the characteristic polynomial for finding eigenvalues?
2. Why are eigenvectors orthonormalized, and what is the significance?
3. How does the Gram-Schmidt process work for orthonormalization?
4. What is the geometric interpretation of eigenvectors and eigenvalues in linear algebra?
5. How does the eigenvalue decomposition relate to diagonalizing a matrix?

**Tip:** Eigenvalues close to zero often indicate directions of minimal variance or influence in the matrix's transformation.

Learn how to find the eigenvalues and eigenvectors of the matrix G = A^T A, using matrix transposition, characteristic polynomial, and the Gram-Schmidt process for orthonormalization. Detailed step-by-step solution provided.

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