Math Snap

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.

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:
- $A \vec{v} = \lambda \vec{v}$
- $(\lambda I - A) \vec{v} = \overrightarrow{0}$
- $\vec{v} \neq 0$