Answer :
To find the characteristic equation and the eigenvalues, along with a basis for each of the corresponding eigenspaces, let's follow these steps:
### (a) Finding the Characteristic Equation
The characteristic equation of a matrix [tex]\( A \)[/tex] is determined by the determinant of [tex]\( A - \lambda I \)[/tex], where [tex]\( I \)[/tex] is the identity matrix and [tex]\( \lambda \)[/tex] represents the eigenvalues.
Given matrix [tex]\( A \)[/tex]:
[tex]\[ A = \begin{pmatrix} 1 & -\frac{1}{2} \\ \frac{3}{2} & -1 \end{pmatrix} \][/tex]
First, form [tex]\( A - \lambda I \)[/tex]:
[tex]\[ A - \lambda I = \begin{pmatrix} 1 - \lambda & -\frac{1}{2} \\ \frac{3}{2} & -1 - \lambda \end{pmatrix} \][/tex]
Next, calculate the determinant:
[tex]\[ \text{det}(A - \lambda I) = \begin{vmatrix} 1 - \lambda & -\frac{1}{2} \\ \frac{3}{2} & -1 - \lambda \end{vmatrix} \][/tex]
Using the determinant formula for a [tex]\( 2 \times 2 \)[/tex] matrix:
[tex]\[ \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc \][/tex]
We get:
[tex]\[ (1 - \lambda)(-1 - \lambda) - \left( -\frac{1}{2} \times \frac{3}{2} \right) \][/tex]
Expanding and simplifying:
[tex]\[ (1 - \lambda)(-1 - \lambda) + \frac{3}{4} = (\lambda^2 - 0.25) \][/tex]
Setting the determinant to equal zero gives the characteristic equation:
[tex]\[ \lambda^2 - 0.25 = 0 \][/tex]
So the characteristic equation is:
[tex]\[ \boxed{\lambda^2 - 0.25 = 0} \][/tex]
### (b) Finding the Eigenvalues
To find the eigenvalues, solve the characteristic equation:
[tex]\[ \lambda^2 - 0.25 = 0 \][/tex]
This can be factored into:
[tex]\[ (\lambda + 0.5)(\lambda - 0.5) = 0 \][/tex]
Therefore, the eigenvalues are:
[tex]\[ \lambda = -0.5, \, 0.5 \][/tex]
Ordering them from smallest to largest, we have:
[tex]\[ (\lambda_1, \lambda_2) = \boxed{(-0.5, 0.5)} \][/tex]
### Finding the Eigenvectors
1. For [tex]\(\lambda = -0.5\)[/tex]:
The matrix [tex]\( A - (-0.5)I = A + 0.5I \)[/tex]:
[tex]\[ \begin{pmatrix} 1 + 0.5 & -\frac{1}{2} \\ \frac{3}{2} & -1 + 0.5 \end{pmatrix} = \begin{pmatrix} 1.5 & -0.5 \\ 1.5 & -0.5 \end{pmatrix} \][/tex]
To find the eigenvector, solve:
[tex]\[ \begin{pmatrix} 1.5 & -0.5 \\ 1.5 & -0.5 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \][/tex]
From the equations:
[tex]\[ 1.5 x_1 - 0.5 x_2 = 0 \][/tex]
Solve for [tex]\( x_2 \)[/tex]:
[tex]\[ x_2 = 3 x_1 \][/tex]
Thus, a basis for the eigenspace corresponding to [tex]\(\lambda = -0.5\)[/tex] is:
[tex]\[ \boxed{\begin{pmatrix} 0.3333333333333333 \\ 1.0 \end{pmatrix}} \][/tex]
2. For [tex]\(\lambda = 0.5\)[/tex]:
The matrix [tex]\( A - 0.5I \)[/tex]:
[tex]\[ \begin{pmatrix} 1 - 0.5 & -\frac{1}{2} \\ \frac{3}{2} & -1 - 0.5 \end{pmatrix} = \begin{pmatrix} 0.5 & -0.5 \\ 1.5 & -1.5 \end{pmatrix} \][/tex]
To find the eigenvector, solve:
[tex]\[ \begin{pmatrix} 0.5 & -0.5 \\ 1.5 & -1.5 \end{pmatrix} \begin{pmatrix} y_1 \\ y_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \][/tex]
From the equations:
[tex]\[ 0.5 y_1 - 0.5 y_2 = 0 \][/tex]
Solve for [tex]\( y_2 \)[/tex]:
[tex]\[ y_2 = y_1 \][/tex]
Thus, a basis for the eigenspace corresponding to [tex]\(\lambda = 0.5\)[/tex] is:
[tex]\[ \boxed{\begin{pmatrix} 1.0 \\ 1.0 \end{pmatrix}} \][/tex]
### (a) Finding the Characteristic Equation
The characteristic equation of a matrix [tex]\( A \)[/tex] is determined by the determinant of [tex]\( A - \lambda I \)[/tex], where [tex]\( I \)[/tex] is the identity matrix and [tex]\( \lambda \)[/tex] represents the eigenvalues.
Given matrix [tex]\( A \)[/tex]:
[tex]\[ A = \begin{pmatrix} 1 & -\frac{1}{2} \\ \frac{3}{2} & -1 \end{pmatrix} \][/tex]
First, form [tex]\( A - \lambda I \)[/tex]:
[tex]\[ A - \lambda I = \begin{pmatrix} 1 - \lambda & -\frac{1}{2} \\ \frac{3}{2} & -1 - \lambda \end{pmatrix} \][/tex]
Next, calculate the determinant:
[tex]\[ \text{det}(A - \lambda I) = \begin{vmatrix} 1 - \lambda & -\frac{1}{2} \\ \frac{3}{2} & -1 - \lambda \end{vmatrix} \][/tex]
Using the determinant formula for a [tex]\( 2 \times 2 \)[/tex] matrix:
[tex]\[ \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc \][/tex]
We get:
[tex]\[ (1 - \lambda)(-1 - \lambda) - \left( -\frac{1}{2} \times \frac{3}{2} \right) \][/tex]
Expanding and simplifying:
[tex]\[ (1 - \lambda)(-1 - \lambda) + \frac{3}{4} = (\lambda^2 - 0.25) \][/tex]
Setting the determinant to equal zero gives the characteristic equation:
[tex]\[ \lambda^2 - 0.25 = 0 \][/tex]
So the characteristic equation is:
[tex]\[ \boxed{\lambda^2 - 0.25 = 0} \][/tex]
### (b) Finding the Eigenvalues
To find the eigenvalues, solve the characteristic equation:
[tex]\[ \lambda^2 - 0.25 = 0 \][/tex]
This can be factored into:
[tex]\[ (\lambda + 0.5)(\lambda - 0.5) = 0 \][/tex]
Therefore, the eigenvalues are:
[tex]\[ \lambda = -0.5, \, 0.5 \][/tex]
Ordering them from smallest to largest, we have:
[tex]\[ (\lambda_1, \lambda_2) = \boxed{(-0.5, 0.5)} \][/tex]
### Finding the Eigenvectors
1. For [tex]\(\lambda = -0.5\)[/tex]:
The matrix [tex]\( A - (-0.5)I = A + 0.5I \)[/tex]:
[tex]\[ \begin{pmatrix} 1 + 0.5 & -\frac{1}{2} \\ \frac{3}{2} & -1 + 0.5 \end{pmatrix} = \begin{pmatrix} 1.5 & -0.5 \\ 1.5 & -0.5 \end{pmatrix} \][/tex]
To find the eigenvector, solve:
[tex]\[ \begin{pmatrix} 1.5 & -0.5 \\ 1.5 & -0.5 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \][/tex]
From the equations:
[tex]\[ 1.5 x_1 - 0.5 x_2 = 0 \][/tex]
Solve for [tex]\( x_2 \)[/tex]:
[tex]\[ x_2 = 3 x_1 \][/tex]
Thus, a basis for the eigenspace corresponding to [tex]\(\lambda = -0.5\)[/tex] is:
[tex]\[ \boxed{\begin{pmatrix} 0.3333333333333333 \\ 1.0 \end{pmatrix}} \][/tex]
2. For [tex]\(\lambda = 0.5\)[/tex]:
The matrix [tex]\( A - 0.5I \)[/tex]:
[tex]\[ \begin{pmatrix} 1 - 0.5 & -\frac{1}{2} \\ \frac{3}{2} & -1 - 0.5 \end{pmatrix} = \begin{pmatrix} 0.5 & -0.5 \\ 1.5 & -1.5 \end{pmatrix} \][/tex]
To find the eigenvector, solve:
[tex]\[ \begin{pmatrix} 0.5 & -0.5 \\ 1.5 & -1.5 \end{pmatrix} \begin{pmatrix} y_1 \\ y_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \][/tex]
From the equations:
[tex]\[ 0.5 y_1 - 0.5 y_2 = 0 \][/tex]
Solve for [tex]\( y_2 \)[/tex]:
[tex]\[ y_2 = y_1 \][/tex]
Thus, a basis for the eigenspace corresponding to [tex]\(\lambda = 0.5\)[/tex] is:
[tex]\[ \boxed{\begin{pmatrix} 1.0 \\ 1.0 \end{pmatrix}} \][/tex]
