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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UpStudy AI · Accepted Answer

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 $$ \lambda = 1 $$: $$ \begin{pmatrix} -1 \ -2 \ 1 \end{pmatrix} $$

Find eigenvalues and eigenvectors of the following matrices:

Upstudy AI Solution

Answer

Solution

1. Matrix

a. Finding Eigenvalues

b. Finding Eigenvectors

2. Matrix

a. Finding Eigenvalues

b. Finding Eigenvectors

i. Eigenvalue

ii. Eigenvalue

iii. Eigenvalue

Summary

The Deep Dive

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