Find the eigenualue
and eigen uectors of
.

Ask by Christensen Boone. in Nepal

Jan 11,2025

Question

Find the eigenualue and eigen uectors of $$ A=\left[\begin{array}{cc}1 & -2 \ 1 & 3\end{array}ight] $$.

UpStudy AI · Accepted Answer

To find the **eigenvalues** and **eigenvectors** of the matrix

$$
A = \begin{bmatrix} 1 & -2 \ 1 & 3 \end{bmatrix},
$$

we follow these steps:

### 1. Finding the Eigenvalues

Eigenvalues $$ \lambda $$ satisfy the characteristic equation:

$$
\det(A - \lambda I) = 0,
$$

where $$ I $$ is the identity matrix.

Let's compute $$ A - \lambda I $$:

$$
A - \lambda I = \begin{bmatrix} 1 - \lambda & -2 \ 1 & 3 - \lambda \end{bmatrix}.
$$

The determinant is:

$$
\det(A - \lambda I) = (1 - \lambda)(3 - \lambda) - (-2)(1) = (1 - \lambda)(3 - \lambda) + 2.
$$

Expanding this:

$$
\det(A - \lambda I) = (1 \cdot 3) - 1 \cdot \lambda - 3 \cdot \lambda + \lambda^2 + 2 = \lambda^2 - 4\lambda + 5.
$$

Setting the determinant to zero gives the characteristic equation:

$$
\lambda^2 - 4\lambda + 5 = 0.
$$

Solving this quadratic equation:

$$
\lambda = \frac{4 \pm \sqrt{16 - 20}}{2} = \frac{4 \pm \sqrt{-4}}{2} = 2 \pm i.
$$

**Eigenvalues:**
$$
\lambda_1 = 2 + i, \quad \lambda_2 = 2 - i.
$$

### 2. Finding the Eigenvectors

For each eigenvalue, we find a corresponding eigenvector by solving:

$$
(A - \lambda I)\mathbf{v} = \mathbf{0}.
$$

#### a. Eigenvector for $$ \lambda_1 = 2 + i $$

Substitute $$ \lambda_1 $$ into $$ A - \lambda I $$:

$$
A - \lambda_1 I = \begin{bmatrix} 1 - (2 + i) & -2 \ 1 & 3 - (2 + i) \end{bmatrix} = \begin{bmatrix} -1 - i & -2 \ 1 & 1 - i \end{bmatrix}.
$$

This gives the system of equations:

$$
\begin{cases}
(-1 - i)v_1 - 2v_2 = 0, \
v_1 + (1 - i)v_2 = 0.
\end{cases}
$$

From the first equation:

$$
(-1 - i)v_1 = 2v_2 \quad \Rightarrow \quad v_1 = \frac{-2}{1 + i}v_2.
$$

Simplifying the fraction:

$$
\frac{-2}{1 + i} = \frac{-2(1 - i)}{(1 + i)(1 - i)} = \frac{-2 + 2i}{2} = -1 + i.
$$

Thus:

$$
v_1 = (-1 + i)v_2.
$$

Choosing $$ v_2 = 1 $$ for simplicity, an eigenvector is:

$$
\mathbf{v}_1 = \begin{bmatrix} -1 + i \ 1 \end{bmatrix}.
$$

#### b. Eigenvector for $$ \lambda_2 = 2 - i $$

Similarly, for $$ \lambda_2 = 2 - i $$, the eigenvector is the complex conjugate of $$ \mathbf{v}_1 $$:

$$
\mathbf{v}_2 = \begin{bmatrix} -1 - i \ 1 \end{bmatrix}.
$$

### 3. Summary

**Eigenvalues and Corresponding Eigenvectors:**

- **Eigenvalue:** $$ \lambda_1 = 2 + i $$
  
  **Eigenvector:**
  $$
  \mathbf{v}_1 = \begin{bmatrix} -1 + i \ 1 \end{bmatrix}.
  $$

- **Eigenvalue:** $$ \lambda_2 = 2 - i $$
  
  **Eigenvector:**
  $$
  \mathbf{v}_2 = \begin{bmatrix} -1 - i \ 1 \end{bmatrix}.
  $$

These eigenvectors are defined up to a non-zero scalar multiple. Any scalar multiple of $$ \mathbf{v}_1 $$ or $$ \mathbf{v}_2 $$ is also a valid eigenvector corresponding to $$ \lambda_1 $$ and $$ \lambda_2 $$, respectively.
The matrix $$ A = \begin{bmatrix} 1 & -2 \ 1 & 3 \end{bmatrix} $$ has eigenvalues $$ \lambda_1 = 2 + i $$ and $$ \lambda_2 = 2 - i $$. The corresponding eigenvectors are: - For $$ \lambda_1 = 2 + i $$: $$ \mathbf{v}_1 = \begin{bmatrix} -1 + i \ 1 \end{bmatrix} $$ - For $$ \lambda_2 = 2 - i $$: $$ \mathbf{v}_2 = \begin{bmatrix} -1 - i \ 1 \end{bmatrix} $$

Find the eigenualue
and eigen uectors of
.

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