1. Find eigenvalues and eigenvectors of the following matrices: $$ A=\left(\begin{array}{rr}-1 & 0 \ 1 & -1\end{array}\right), \quad E=\left(\begin{array}{lll}2 & 0 & 1 \ 0 & 3 & 4 \ 0 & 0 & 1\end{array}\right) $$

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Let's find the eigenvalues and eigenvectors for the given matrices $$ A $$ and $$ E $$.

## 1. Matrix $$ A $$

$$ 
A = \begin{pmatrix}
-1 & 0 \
1 & -1
\end{pmatrix} 
$$

### **a. Finding Eigenvalues**

To find the eigenvalues $$ \lambda $$, solve the characteristic equation:

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

Compute $$ A - \lambda I $$:

$$
A - \lambda I = \begin{pmatrix}
-1 - \lambda & 0 \
1 & -1 - \lambda
\end{pmatrix}
$$

Calculate the determinant:

$$
\det(A - \lambda I) = (-1 - \lambda)^2 - (0 \cdot 1) = (-1 - \lambda)^2 = 0
$$

Setting the determinant to zero:

$$
(-1 - \lambda)^2 = 0 \implies \lambda = -1 \quad \text{(with multiplicity 2)}
$$

### **b. Finding Eigenvectors**

For $$ \lambda = -1 $$, solve $$ (A - \lambda I)\mathbf{x} = 0 $$:

$$
(A + I)\mathbf{x} = \begin{pmatrix}
0 & 0 \
1 & 0
\end{pmatrix} \begin{pmatrix}
x_1 \
x_2
\end{pmatrix} = \begin{pmatrix}
0 \
x_1
\end{pmatrix} = \begin{pmatrix}
0 \
0
\end{pmatrix}
$$

This yields the system:

$$
\begin{cases}
0 \cdot x_1 + 0 \cdot x_2 = 0 \
1 \cdot x_1 + 0 \cdot x_2 = 0
\end{cases}
$$

From the second equation:

$$
x_1 = 0
$$

The first equation doesn't provide additional information. Therefore, $$ x_2 $$ is free.

**Eigenvectors corresponding to $$ \lambda = -1 $$:**

$$
\mathbf{x} = \begin{pmatrix}
0 \
t
\end{pmatrix}, \quad t \in \mathbb{R}
$$

---

## 2. Matrix $$ E $$

$$
E = \begin{pmatrix}
2 & 0 & 1 \
0 & 3 & 4 \
0 & 0 & 1
\end{pmatrix}
$$

### **a. Finding Eigenvalues**

Since $$ E $$ is an upper triangular matrix, its eigenvalues are the entries on its diagonal:

$$
\lambda_1 = 2, \quad \lambda_2 = 3, \quad \lambda_3 = 1
$$

### **b. Finding Eigenvectors**

#### **i. Eigenvalue $$ \lambda = 2 $$**

Solve $$ (E - 2I)\mathbf{x} = 0 $$:

$$
E - 2I = \begin{pmatrix}
0 & 0 & 1 \
0 & 1 & 4 \
0 & 0 & -1
\end{pmatrix}
$$

Set up the system:

$$
\begin{cases}
0 \cdot x_1 + 0 \cdot x_2 + 1 \cdot x_3 = 0 \
0 \cdot x_1 + 1 \cdot x_2 + 4 \cdot x_3 = 0 \
0 \cdot x_1 + 0 \cdot x_2 - 1 \cdot x_3 = 0
\end{cases}
$$

From the first and third equations:

$$
x_3 = 0
$$

From the second equation:

$$
x_2 = 0
$$

$$ x_1 $$ is free.

**Eigenvectors corresponding to $$ \lambda = 2 $$:**

$$
\mathbf{x} = \begin{pmatrix}
t \
0 \
0
\end{pmatrix}, \quad t \in \mathbb{R}
$$

#### **ii. Eigenvalue $$ \lambda = 3 $$**

Solve $$ (E - 3I)\mathbf{x} = 0 $$:

$$
E - 3I = \begin{pmatrix}
-1 & 0 & 1 \
0 & 0 & 4 \
0 & 0 & -2
\end{pmatrix}
$$

Set up the system:

$$
\begin{cases}
-1 \cdot x_1 + 0 \cdot x_2 + 1 \cdot x_3 = 0 \
0 \cdot x_1 + 0 \cdot x_2 + 4 \cdot x_3 = 0 \
0 \cdot x_1 + 0 \cdot x_2 - 2 \cdot x_3 = 0
\end{cases}
$$

From the second and third equations:

$$
x_3 = 0
$$

From the first equation:

$$
x_1 = x_3 = 0
$$

$$ x_2 $$ is free.

**Eigenvectors corresponding to $$ \lambda = 3 $$:**

$$
\mathbf{x} = \begin{pmatrix}
0 \
t \
0
\end{pmatrix}, \quad t \in \mathbb{R}
$$

#### **iii. Eigenvalue $$ \lambda = 1 $$**

Solve $$ (E - I)\mathbf{x} = 0 $$:

$$
E - I = \begin{pmatrix}
1 & 0 & 1 \
0 & 2 & 4 \
0 & 0 & 0
\end{pmatrix}
$$

Set up the system:

$$
\begin{cases}
1 \cdot x_1 + 0 \cdot x_2 + 1 \cdot x_3 = 0 \
0 \cdot x_1 + 2 \cdot x_2 + 4 \cdot x_3 = 0 \
0 \cdot x_1 + 0 \cdot x_2 + 0 \cdot x_3 = 0
\end{cases}
$$

From the first equation:

$$
x_1 = -x_3
$$

From the second equation:

$$
2x_2 = -4x_3 \implies x_2 = -2x_3
$$

Let $$ x_3 = t $$.

**Eigenvectors corresponding to $$ \lambda = 1 $$:**

$$
\mathbf{x} = \begin{pmatrix}
-t \
-2t \
t
\end{pmatrix} = t \begin{pmatrix}
-1 \
-2 \
1
\end{pmatrix}, \quad t \in \mathbb{R}
$$

## **Summary**

- **Matrix $$ A $$:**
  - **Eigenvalue:** $$ \lambda = -1 $$ (with multiplicity 2)
  - **Eigenvectors:** Any scalar multiple of $$ \begin{pmatrix} 0 \ 1 \end{pmatrix} $$

- **Matrix $$ E $$:**
  - **Eigenvalues and Eigenvectors:**
    - $$ \lambda = 2 $$: Eigenvectors are scalar multiples of $$ \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix} $$
    - $$ \lambda = 3 $$: Eigenvectors are scalar multiples of $$ \begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix} $$
    - $$ \lambda = 1 $$: Eigenvectors are scalar multiples of $$ \begin{pmatrix} -1 \ -2 \ 1 \end{pmatrix} $$
**Eigenvalues and Eigenvectors of Matrices $$ A $$ and $$ E $$** - **Matrix $$ A $$:** - **Eigenvalue:** $$ \lambda = -1 $$ (with multiplicity 2) - **Eigenvectors:** Any scalar multiple of $$ \begin{pmatrix} 0 \ 1 \end{pmatrix} $$ - **Matrix $$ E $$:** - **Eigenvalues:** $$ \lambda = 2, 3, 1 $$ - **Eigenvectors:** - For $$ \lambda = 2 $$: $$ \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix} $$ - For $$ \lambda = 3 $$: $$ \begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix} $$ - For $$ \lambda = 1 $$: $$ \begin{pmatrix} -1 \ -2 \ 1 \end{pmatrix} $$

1. Find eigenvalues and eigenvectors of the following matrices: \[ A=\left(\begin{array}{rr}-1 & 0 \\ 1 & -1\end{array}\right), \quad E=\left(\begin{array}{lll}2 & 0 & 1 \\ 0 & 3 & 4 \\ 0 & 0 & 1\end{array}\right) \]

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