Math / Algebra

QuestionLet $A$ be an $n \times n$ matrix and suppose that $\vec{v}$ is a $\lambda$ -eigenvector for $A$ .
Select all of the following statements which are true.
From the 5 options, select all that apply $A \vec{v}=\lambda \vec{v}$ $(\lambda I-A) \vec{v}=\overrightarrow{0}$ $\lambda \neq 0$ $\vec{v} \neq 0$ $\vec{v}$ is a solution to the system with augmented matrix $(A \mid 0)$
3

Studdy Solution

STEP 1

1. $A$ is an $n \times n$ matrix.
2. $\vec{v}$ is a non-zero vector such that $A\vec{v} = \lambda\vec{v}$ .
3. $\lambda$ is a scalar, known as the eigenvalue corresponding to the eigenvector $\vec{v}$ .

STEP 2

1. Verify the eigenvector equation.
2. Check the equation $(\lambda I - A) \vec{v} = \overrightarrow{0}$ .
3. Analyze the condition $\lambda \neq 0$ .
4. Confirm that $\vec{v} \neq 0$ .
5. Evaluate if $\vec{v}$ is a solution to the system with augmented matrix $(A \mid 0)$ .

STEP 3

Verify the eigenvector equation:
The definition of an eigenvector $\vec{v}$ corresponding to an eigenvalue $\lambda$ for the matrix $A$ is given by:
$A \vec{v} = \lambda \vec{v}$
This statement is true by the definition of eigenvectors and eigenvalues.

STEP 4

Check the equation $(\lambda I - A) \vec{v} = \overrightarrow{0}$ :
By rearranging the eigenvector equation $A \vec{v} = \lambda \vec{v}$ , we can write:
$A \vec{v} - \lambda \vec{v} = \overrightarrow{0}$ $(\lambda I - A) \vec{v} = \overrightarrow{0}$
This statement is true because it is equivalent to the eigenvector equation.

STEP 5

Analyze the condition $\lambda \neq 0$ :
The condition $\lambda \neq 0$ is not necessarily true. Eigenvalues can be zero or non-zero. Therefore, this statement is not always true.

STEP 6

Confirm that $\vec{v} \neq 0$ :
By definition, an eigenvector $\vec{v}$ must be non-zero. Therefore, this statement is true.

STEP 7

Evaluate if $\vec{v}$ is a solution to the system with augmented matrix $(A \mid 0)$ :
The system with augmented matrix $(A \mid 0)$ corresponds to the homogeneous system $A \vec{x} = \overrightarrow{0}$ . Since $A \vec{v} = \lambda \vec{v}$ and $\lambda \neq 0$ is not guaranteed, $\vec{v}$ is not necessarily a solution to $A \vec{x} = \overrightarrow{0}$ . Therefore, this statement is not true.
The true statements are: - $A \vec{v} = \lambda \vec{v}$ - $(\lambda I - A) \vec{v} = \overrightarrow{0}$ - $\vec{v} \neq 0$

Was this helpful?

Studdy Solution

STEP 1

1. $A$ is an $n \times n$ matrix.
2. $\vec{v}$ is a non-zero vector such that $A\vec{v} = \lambda\vec{v}$ .
3. $\lambda$ is a scalar, known as the eigenvalue corresponding to the eigenvector $\vec{v}$ .

STEP 2

1. Verify the eigenvector equation.
2. Check the equation $(\lambda I - A) \vec{v} = \overrightarrow{0}$ .
3. Analyze the condition $\lambda \neq 0$ .
4. Confirm that $\vec{v} \neq 0$ .
5. Evaluate if $\vec{v}$ is a solution to the system with augmented matrix $(A \mid 0)$ .

STEP 3

Verify the eigenvector equation:
The definition of an eigenvector $\vec{v}$ corresponding to an eigenvalue $\lambda$ for the matrix $A$ is given by:
$A \vec{v} = \lambda \vec{v}$
This statement is true by the definition of eigenvectors and eigenvalues.

STEP 4

STEP 5

Analyze the condition $\lambda \neq 0$ :
The condition $\lambda \neq 0$ is not necessarily true. Eigenvalues can be zero or non-zero. Therefore, this statement is not always true.

STEP 6

Confirm that $\vec{v} \neq 0$ :
By definition, an eigenvector $\vec{v}$ must be non-zero. Therefore, this statement is true.

STEP 7

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Studdy Solution

STEP 1

1. A A A is an n×n n \times n n×n matrix.2. v⃗ \vec{v} v is a non-zero vector such that Av⃗=λv⃗ A\vec{v} = \lambda\vec{v} Av=λv.3. λ \lambda λ is a scalar, known as the eigenvalue corresponding to the eigenvector v⃗ \vec{v} v.

STEP 2

STEP 3

Verify the eigenvector equation:The definition of an eigenvector v⃗\vec{v}v corresponding to an eigenvalue λ\lambdaλ for the matrix AAA is given by:Av⃗=λv⃗ A \vec{v} = \lambda \vec{v} Av=λvThis statement is true by the definition of eigenvectors and eigenvalues.

STEP 4

STEP 5

Analyze the condition λ≠0\lambda \neq 0λ=0:The condition λ≠0\lambda \neq 0λ=0 is not necessarily true. Eigenvalues can be zero or non-zero. Therefore, this statement is not always true.

STEP 6

Confirm that v⃗≠0\vec{v} \neq 0v=0:By definition, an eigenvector v⃗\vec{v}v must be non-zero. Therefore, this statement is true.

STEP 7

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. $A$ is an $n \times n$ matrix.
2. $\vec{v}$ is a non-zero vector such that $A\vec{v} = \lambda\vec{v}$ .
3. $\lambda$ is a scalar, known as the eigenvalue corresponding to the eigenvector $\vec{v}$ .

Verify the eigenvector equation:
The definition of an eigenvector $\vec{v}$ corresponding to an eigenvalue $\lambda$ for the matrix $A$ is given by:
$A \vec{v} = \lambda \vec{v}$
This statement is true by the definition of eigenvectors and eigenvalues.

Analyze the condition $\lambda \neq 0$ :
The condition $\lambda \neq 0$ is not necessarily true. Eigenvalues can be zero or non-zero. Therefore, this statement is not always true.

Confirm that $\vec{v} \neq 0$ :
By definition, an eigenvector $\vec{v}$ must be non-zero. Therefore, this statement is true.