Math Snap

STEP 1

1. We are given a $2 \times 2$ matrix $A$.
2. Eigenvalues $\lambda$ are found by solving the characteristic equation $\det(A - \lambda I) = 0$.
3. Eigenvectors are found by solving $(A - \lambda I)\mathbf{v} = \mathbf{0}$ for each eigenvalue $\lambda$.

STEP 2

1. Set up the characteristic equation.
2. Solve the characteristic equation for eigenvalues.
3. Find eigenvectors for each eigenvalue.

STEP 3

Set up the characteristic equation for matrix $A$:
The characteristic equation is given by:
$$\det(A - \lambda I) = 0$$ Where $I$ is the identity matrix of the same size as $A$. For a $2 \times 2$ matrix, $I$ is:
$$I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$ Thus, $A - \lambda I$ is:
$$A - \lambda I = \begin{bmatrix} 3 & 5 \\ -1 & -3 \end{bmatrix} - \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix} = \begin{bmatrix} 3 - \lambda & 5 \\ -1 & -3 - \lambda \end{bmatrix}$$

STEP 4

Solve the characteristic equation $\det(A - \lambda I) = 0$:
Calculate the determinant:
$$\det(A - \lambda I) = (3 - \lambda)(-3 - \lambda) - (5)(-1)$$ Simplify the expression:
$$= (3 - \lambda)(-3 - \lambda) + 5$$ $$= (9 + 3\lambda - 3\lambda + \lambda^2) + 5$$ $$= \lambda^2 - 9 + 5$$ $$= \lambda^2 - 4$$ Set the determinant to zero:
$$\lambda^2 - 4 = 0$$ Solve for $\lambda$:
$$(\lambda - 2)(\lambda + 2) = 0$$ Thus, the eigenvalues are:
$$\lambda_1 = 2, \quad \lambda_2 = -2$$

Find eigenvectors for each eigenvalue:
For $\lambda_1 = 2$:
Solve $(A - 2I)\mathbf{v} = \mathbf{0}$:
$$A - 2I = \begin{bmatrix} 1 & 5 \\ -1 & -5 \end{bmatrix}$$ Set up the system of equations:
$$1v_1 + 5v_2 = 0$$ $$-v_1 - 5v_2 = 0$$ Both equations are equivalent, so we can solve:
$$v_1 = -5v_2$$ Let $v_2 = t$, then $v_1 = -5t$.
Thus, an eigenvector is:
$$\mathbf{v_1} = \begin{bmatrix} -5 \\ 1 \end{bmatrix}$$ For $\lambda_2 = -2$:
Solve $(A + 2I)\mathbf{v} = \mathbf{0}$:
$$A + 2I = \begin{bmatrix} 5 & 5 \\ -1 & -1 \end{bmatrix}$$ Set up the system of equations:
$$5v_1 + 5v_2 = 0$$ $$-v_1 - v_2 = 0$$ Both equations are equivalent, so we can solve:
$$v_1 = -v_2$$ Let $v_2 = s$, then $v_1 = -s$.
Thus, an eigenvector is:
$$\mathbf{v_2} = \begin{bmatrix} -1 \\ 1 \end{bmatrix}$$ The eigenvalues are $\lambda_1 = 2$ and $\lambda_2 = -2$. The corresponding eigenvectors are $\mathbf{v_1} = \begin{bmatrix} -5 \\ 1 \end{bmatrix}$ and $\mathbf{v_2} = \begin{bmatrix} -1 \\ 1 \end{bmatrix}$.