Matrix Eigenvalues and Eigenvectors Explained

To find the eigenvalues $(\lambda)$ and eigenvectors $(X)$ of a given matrix $A$, we perform calculations based on the definition $AX = \lambda X$. This involves solving the characteristic equation and then solving a system of linear equations.

Eigenvalues Calculation for Matrix A

The eigenvalues $(\lambda)$ are found by solving the characteristic equation, which is derived from $\det(A - \lambda I) = 0$. For the given matrix $A = \begin{bmatrix} -4.0 & 4.0 \\ -1.6 & 1.2 \end{bmatrix}$, where $I$ is the identity matrix, we first set up $(A - \lambda I)$: $ A - \lambda I = \begin{bmatrix} -4.0 - \lambda & 4.0 \\ -1.6 & 1.2 - \lambda \end{bmatrix} $

The determinant of this matrix is:

$ \det(A - \lambda I) = (-4.0 - \lambda)(1.2 - \lambda) - (4.0)(-1.6) $

Expanding this expression gives:

$ = (-4.8 + 4.0\lambda - 1.2\lambda + \lambda^2) + 6.4 $

$ = \lambda^2 + 2.8\lambda + 1.6 $

Setting the determinant to zero yields the characteristic equation: $\lambda^2 + 2.8\lambda + 1.6 = 0$. We solve this quadratic equation using the quadratic formula $\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$ \lambda = \frac{-2.8 \pm \sqrt{(2.8)^2 - 4(1)(1.6)}}{2(1)} $

$ \lambda = \frac{-2.8 \pm \sqrt{7.84 - 6.4}}{2} $

$ \lambda = \frac{-2.8 \pm \sqrt{1.44}}{2} $

$ \lambda = \frac{-2.8 \pm 1.2}{2} $

This results in two eigenvalues:

• $ \lambda_1 = \frac{-2.8 + 1.2}{2} = \frac{-1.6}{2} = -0.8 $
• $ \lambda_2 = \frac{-2.8 - 1.2}{2} = \frac{-4.0}{2} = -2.0 $

The eigenvalues are $-0.8$ and $-2.0$. This information helps identify the correct option among the choices provided.

Eigenvectors Finding for Matrix A

Now, we find the eigenvectors $(X)$ corresponding to each eigenvalue by solving $(A - \lambda I)X = 0$. Let $X = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$.

Eigenvector for $\lambda = -2.0$

Substitute $\lambda = -2.0$ into $(A - \lambda I)X = 0$. The matrix $(A - \lambda I)$ becomes:

$ A - (-2.0)I = \begin{bmatrix} -2.0 & 4.0 \\ -1.6 & 3.2 \end{bmatrix} $

The system of equations to solve is:

$ \begin{bmatrix} -2.0 & 4.0 \\ -1.6 & 3.2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $

From the first row: $ -2.0x_1 + 4.0x_2 = 0 $, which simplifies to $ x_2 = 0.5x_1 $. If we choose $x_1 = 2$, then $x_2 = 1$. An eigenvector is therefore proportional to $ \begin{bmatrix} 2 \\ 1 \end{bmatrix} $.

Eigenvector for $\lambda = -0.8$

Substitute $\lambda = -0.8$ into $(A - \lambda I)X = 0$. The matrix $(A - \lambda I)$ becomes:

$ A - (-0.8)I = \begin{bmatrix} -3.2 & 4.0 \\ -1.6 & 2.0 \end{bmatrix} $

The system of equations to solve is:

$ \begin{bmatrix} -3.2 & 4.0 \\ -1.6 & 2.0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $

From the first row: $ -3.2x_1 + 4.0x_2 = 0 $, which simplifies to $ x_2 = 0.8x_1 $. If we choose $x_1 = 1$, then $x_2 = 0.8$. An eigenvector is therefore proportional to $ \begin{bmatrix} 1 \\ 0.8 \end{bmatrix} $.

Eigenvalues and Eigenvectors Match

The calculated eigenvalues are $\lambda = -2.0$ and $\lambda = -0.8$. The corresponding eigenvectors are $X_1 = \begin{bmatrix} 2 \\ 1 \end{bmatrix}$ (for $\lambda = -2.0$) and $X_2 = \begin{bmatrix} 1 \\ 0.8 \end{bmatrix}$ (for $\lambda = -0.8$). This set of eigenvalues and eigenvectors matches option 4.