Math / Algebra

QuestionWithout calculation, find one eigenvalue and two linearly independent eigenvectors of $A=\left[\begin{array}{lll}2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2\end{array}\right]$ . Justify your answer.
One eigenvalue of $A$ is $\lambda=0$ because the columns of $A$ are linearly dependent. Two linearly independent eigenvectors of $A$ are $\square$ because the entries of each vector sum to 0 . (Use a comma to separate answers as needed.)

Studdy Solution

STEP 1

What is this asking? Find an eigenvalue and two eigenvectors of a matrix without any calculations and justify it! Watch out! Eigenvectors must be linearly independent, and we need to explain why our choices are valid.

STEP 2

1. Find an Eigenvalue
2. Find the Eigenvectors
3. Justify the Solution

STEP 3

The matrix $A$ has identical rows, which means its determinant is zero.
A matrix with a determinant of zero must have an eigenvalue of zero.
Why? Because the determinant of a matrix is the product of its eigenvalues!
If the product is zero, at least one of the eigenvalues has gotta be zero!
So, $\lambda = 0$ is an eigenvalue.

STEP 4

Remember, an eigenvector $v$ satisfies the equation $Av = \lambda v$ .
Since our eigenvalue is $\lambda = 0$ , we're looking for vectors $v$ such that $Av = 0 \cdot v = 0$ , the zero vector.

STEP 5

Let's think about what $A$ does to a vector.
Multiplying $A$ by any vector $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ gives us $2x + 2y + 2z$ for every entry in the resulting vector.
We want this result to be the zero vector, $\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$ , so we need $2x + 2y + 2z = 0$ , which simplifies to $x + y + z = 0$ .

STEP 6

Now, we need two linearly independent vectors that satisfy this equation.
Let's pick $x = 1$ and $y = -1$ .
Then, $1 + (-1) + z = 0$ , so $z = 0$ .
This gives us our first eigenvector: $\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}$ .

STEP 7

For our second eigenvector, let's choose $x = 1$ and $z = -1$ .
Then, $1 + y + (-1) = 0$ , so $y = 0$ .
This gives us our second eigenvector: $\begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}$ .
These two eigenvectors are linearly independent because one is not a scalar multiple of the other!

STEP 8

We found that $\lambda = 0$ is an eigenvalue because the determinant of $A$ is zero, and the determinant is the product of the eigenvalues.

STEP 9

We found two eigenvectors by solving $Av = 0$ .
Since the entries of both $\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}$ sum to zero, multiplying $A$ by these vectors results in the zero vector, satisfying the eigenvector equation for $\lambda = 0$ .

STEP 10

An eigenvalue is $\lambda = 0$ .
Two linearly independent eigenvectors are $\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}$ .

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Studdy Solution

STEP 1

What is this asking? Find an eigenvalue and two eigenvectors of a matrix without any calculations and justify it! Watch out! Eigenvectors must be linearly independent, and we need to explain why our choices are valid.

STEP 2

1. Find an Eigenvalue
2. Find the Eigenvectors
3. Justify the Solution

STEP 3

STEP 4

Remember, an eigenvector $v$ satisfies the equation $Av = \lambda v$ .
Since our eigenvalue is $\lambda = 0$ , we're looking for vectors $v$ such that $Av = 0 \cdot v = 0$ , the zero vector.

STEP 5

STEP 6

Now, we need two linearly independent vectors that satisfy this equation.
Let's pick $x = 1$ and $y = -1$ .
Then, $1 + (-1) + z = 0$ , so $z = 0$ .
This gives us our first eigenvector: $\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}$ .

STEP 7

For our second eigenvector, let's choose $x = 1$ and $z = -1$ .
Then, $1 + y + (-1) = 0$ , so $y = 0$ .
This gives us our second eigenvector: $\begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}$ .
These two eigenvectors are linearly independent because one is not a scalar multiple of the other!

STEP 8

We found that $\lambda = 0$ is an eigenvalue because the determinant of $A$ is zero, and the determinant is the product of the eigenvalues.

STEP 9

We found two eigenvectors by solving $Av = 0$ .
Since the entries of both $\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}$ sum to zero, multiplying $A$ by these vectors results in the zero vector, satisfying the eigenvector equation for $\lambda = 0$ .

STEP 10

An eigenvalue is $\lambda = 0$ .
Two linearly independent eigenvectors are $\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}$ .

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Studdy Solution

STEP 1

What is this asking? Find an eigenvalue and two eigenvectors of a matrix *without any calculations* and justify it! Watch out! Eigenvectors must be linearly independent, and we need to explain *why* our choices are valid.

STEP 2

1. Find an Eigenvalue2. Find the Eigenvectors3. Justify the Solution

STEP 3

STEP 4

Remember, an eigenvector vvv satisfies the equation Av=λvAv = \lambda vAv=λv.Since our eigenvalue is λ=0\lambda = 0λ=0, we're looking for vectors vvv such that Av=0⋅v=0Av = 0 \cdot v = 0Av=0⋅v=0, the zero vector.

STEP 5

STEP 6

Now, we need *two* linearly independent vectors that satisfy this equation.Let's pick x=1x = 1x=1 and y=−1y = -1y=−1.Then, 1+(−1)+z=01 + (-1) + z = 01+(−1)+z=0, so z=0z = 0z=0.This gives us our first eigenvector: [1−10]\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}⎣⎡​1−10​⎦⎤​.

STEP 7

STEP 8

We found that λ=0\lambda = 0λ=0 is an eigenvalue because the determinant of AAA is zero, and the determinant is the product of the eigenvalues.

STEP 9

STEP 10

An eigenvalue is λ=0\lambda = 0λ=0.Two linearly independent eigenvectors are [1−10]\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}⎣⎡​1−10​⎦⎤​ and [10−1]\begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}⎣⎡​10−1​⎦⎤​.

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What is this asking? Find an eigenvalue and two eigenvectors of a matrix without any calculations and justify it! Watch out! Eigenvectors must be linearly independent, and we need to explain why our choices are valid.

1. Find an Eigenvalue
2. Find the Eigenvectors
3. Justify the Solution

Remember, an eigenvector $v$ satisfies the equation $Av = \lambda v$ .
Since our eigenvalue is $\lambda = 0$ , we're looking for vectors $v$ such that $Av = 0 \cdot v = 0$ , the zero vector.

Now, we need two linearly independent vectors that satisfy this equation.
Let's pick $x = 1$ and $y = -1$ .
Then, $1 + (-1) + z = 0$ , so $z = 0$ .
This gives us our first eigenvector: $\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}$ .

We found that $\lambda = 0$ is an eigenvalue because the determinant of $A$ is zero, and the determinant is the product of the eigenvalues.

An eigenvalue is $\lambda = 0$ .
Two linearly independent eigenvectors are $\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}$ .