Questions: Find the eigenvalues and eigenvectors of the matrix A = [1 0; 1 2].

Solution Steps

Okay, I understand. I will solve the eigenvalue problem for the given matrix \( A = \begin{bmatrix} 1 & 0 \\ 1 & 2 \end{bmatrix} \).

First, I need to find the eigenvalues by solving the characteristic equation, which is given by \( \det(A - \lambda I) = 0 \), where \( \lambda \) represents the eigenvalues and \( I \) is the identity matrix.

So, \( A - \lambda I = \begin{bmatrix} 1-\lambda & 0 \\ 1 & 2-\lambda \end{bmatrix} \).

The determinant of this matrix is \( (1-\lambda)(2-\lambda) - (0)(1) = (1-\lambda)(2-\lambda) \).

Setting the determinant equal to zero, we get \( (1-\lambda)(2-\lambda) = 0 \).

This gives us the eigenvalues \( \lambda_1 = 1 \) and \( \lambda_2 = 2 \).

Next, I need to find the eigenvectors corresponding to each eigenvalue.

For \( \lambda_1 = 1 \), we solve the equation \( (A - \lambda_1 I)v_1 = 0 \), where \( v_1 \) is the eigenvector.

\( \begin{bmatrix} 1-1 & 0 \\ 1 & 2-1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \)

\( \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \)

This gives us the equation \( x + y = 0 \), so \( y = -x \). Thus, the eigenvector \( v_1 \) can be written as \( \begin{bmatrix} x \\ -x \end{bmatrix} = x \begin{bmatrix} 1 \\ -1 \end{bmatrix} \). We can choose \( x = 1 \), so \( v_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix} \).

For \( \lambda_2 = 2 \), we solve the equation \( (A - \lambda_2 I)v_2 = 0 \), where \( v_2 \) is the eigenvector.

\( \begin{bmatrix} 1-2 & 0 \\ 1 & 2-2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \)

\( \begin{bmatrix} -1 & 0 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \)

This gives us the equation \( -x = 0 \) and \( x = 0 \). So \( x = 0 \), and \( y \) can be any value. Thus, the eigenvector \( v_2 \) can be written as \( \begin{bmatrix} 0 \\ y \end{bmatrix} = y \begin{bmatrix} 0 \\ 1 \end{bmatrix} \). We can choose \( y = 1 \), so \( v_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \).

Final Answer: The eigenvalues are \( \lambda_1 = 1 \) and \( \lambda_2 = 2 \). The corresponding eigenvectors are \( v_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix} \) and \( v_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \).

To find the eigenvalues of matrix \( A = \begin{bmatrix} 1 & 0 \\ 1 & 2 \end{bmatrix} \), we solve the characteristic equation:
\[ \det(A - \lambda I) = 0 \]
where \( I \) is the identity matrix and \( \lambda \) is the eigenvalue.

Subtract \( \lambda \) from the diagonal elements and compute the determinant:
\[ A - \lambda I = \begin{bmatrix} 1 - \lambda & 0 \\ 1 & 2 - \lambda \end{bmatrix} \]
\[ \det(A - \lambda I) = (1 - \lambda)(2 - \lambda) - (0 \cdot 1) = (1 - \lambda)(2 - \lambda) \]

Set the determinant equal to zero and solve for \( \lambda \):
\[ (1 - \lambda)(2 - \lambda) = 0 \]
The solutions are:
\[ \lambda_1 = 1, \quad \lambda_2 = 2 \]

Substitute \( \lambda_1 = 1 \) into \( A - \lambda I \) and solve \( (A - \lambda I)\mathbf{v} = 0 \):
\[ \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \]
This gives the equation \( v_1 + v_2 = 0 \). A corresponding eigenvector is:
\[ \mathbf{v}_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix} \]

Substitute \( \lambda_2 = 2 \) into \( A - \lambda I \):
\[ \begin{bmatrix} -1 & 0 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \]
This gives \( -v_1 = 0 \) and \( v_1 = 0 \). A corresponding eigenvector is:
\[ \mathbf{v}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \]

(Note: Only the first three steps are provided as per the guidelines.)

Final Answer