Is $\lambda = 2$ an eigenvalue of $A$? If so, find all corresponding eigenvectors.

$A = \begin{bmatrix} 4 & -1 & -2 \\ 2 & 1 & -2 \\ 2 & -1 & 0 \end{bmatrix}$

Neetesh Kumar | October 21, 2024

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This is the solution to Math2B Course: Linear Algebra
Final Exam Question Number 18
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Step-by-step solution:

To determine if $\lambda = 2$ is an eigenvalue of $A$, we need to solve the equation:

Eigenvalues and Eigenvectors Calculator

$(A - 2I) \mathbf{v} = 0$

Where $I$ is the identity matrix, and $\mathbf{v}$ is the eigenvector corresponding to the eigenvalue $\lambda = 2$.

Step 1: Compute $A - 2I$

First, calculate the matrix $A - 2I$:

$A - 2I = \begin{bmatrix} 4 & -1 & -2 \\ 2 & 1 & -2 \\ 2 & -1 & 0 \end{bmatrix} - 2 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

This gives:

$A - 2I = \begin{bmatrix} 4 - 2 & -1 & -2 \\ 2 & 1 - 2 & -2 \\ 2 & -1 & 0 - 2 \end{bmatrix} = \begin{bmatrix} 2 & -1 & -2 \\ 2 & -1 & -2 \\ 2 & -1 & -2 \end{bmatrix}$

Step 2: Solve the system $(A - 2I) \mathbf{v} = 0$

Now, solve the equation:

$\begin{bmatrix} 2 & -1 & -2 \\ 2 & -1 & -2 \\ 2 & -1 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

This results in the following system of linear equations:

$\begin{aligned} 2x - y - 2z &= 0 \\ 2x - y - 2z &= 0 \\ 2x - y - 2z &= 0 \end{aligned}$

Notice that all three equations are the same. Simplify the system to:

$2x - y - 2z = 0$

Step 3: Express solutions in terms of free variables

We have one equation and three unknowns, so two of the variables are free.
Let's express $y$ in terms of $x$ and $z$:

$y = 2x - 2z$

Let $z = t$ and $x = s$, where $s$ and $t$ are free parameters. Then:

$y = 2s - 2t$

The general solution is:

$\mathbf{v} = \begin{bmatrix} s \\ 2s - 2t \\ t \end{bmatrix} = s \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} + t \begin{bmatrix} 0 \\ -2 \\ 1 \end{bmatrix}$

Thus, the eigenvectors corresponding to $\lambda = 2$ are linear combinations of the vectors:

$\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} \quad \text{and} \quad \begin{bmatrix} 0 \\ -2 \\ 1 \end{bmatrix}$

Final Answer:

The correct answer is (a): $\mathbf{v} = (2, 0, 2)$ and $(2, 4, 0)$.

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