QuestionShow that $\mathbf{v}$ is an eigenvector of $A$ and find the corresponding eigenvalue, $\lambda$ . $A=\left[\begin{array}{rrr} 0 & 1 & -1 \\ 1 & 1 & 1 \\ 1 & 2 & 0 \end{array}\right], v=\left[\begin{array}{r} -2 \\ 1 \\ 1 \end{array}\right]$

Studdy Solution

STEP 1

1. We are given a matrix $A$ and a vector $\mathbf{v}$ .
2. We need to verify if $\mathbf{v}$ is an eigenvector of $A$ .
3. If $\mathbf{v}$ is an eigenvector, we need to find the corresponding eigenvalue $\lambda$ .

STEP 2

1. Calculate the product $A \mathbf{v}$ .
2. Check if $A \mathbf{v} = \lambda \mathbf{v}$ for some scalar $\lambda$ .
3. Solve for the eigenvalue $\lambda$ .

STEP 3

Calculate the product $A \mathbf{v}$ :
$A \mathbf{v} = \left[\begin{array}{rrr} 0 & 1 & -1 \\ 1 & 1 & 1 \\ 1 & 2 & 0 \end{array}\right] \left[\begin{array}{r} -2 \\ 1 \\ 1 \end{array}\right]$
Perform the matrix multiplication:
- First row: $0(-2) + 1(1) + (-1)(1) = 0 + 1 - 1 = 0$ - Second row: $1(-2) + 1(1) + 1(1) = -2 + 1 + 1 = 0$ - Third row: $1(-2) + 2(1) + 0(1) = -2 + 2 + 0 = 0$
So, $A \mathbf{v} = \left[\begin{array}{r} 0 \\ 0 \\ 0 \end{array}\right]$

STEP 4

Check if $A \mathbf{v} = \lambda \mathbf{v}$ :
Since $A \mathbf{v} = \left[\begin{array}{r} 0 \\ 0 \\ 0 \end{array}\right]$ and $\mathbf{v} = \left[\begin{array}{r} -2 \\ 1 \\ 1 \end{array}\right]$ , we can express this as:
$A \mathbf{v} = 0 \cdot \mathbf{v}$
Thus, $\lambda = 0$ .

STEP 5

The eigenvalue corresponding to the eigenvector $\mathbf{v}$ is:
$\lambda = 0$

Was this helpful?

Get Studdy

Scan QR code with your device to get Studdy on iOS or Android

QR code

Studdy solves anything!

Download the app

Start learning now

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

Parents Influencer program Contact Policy Terms