Find a basis for the eigenspace corresponding to the eigenvalue $\lambda = 3$ for the matrix:

$A = \begin{bmatrix} 4 & 3 & 2 \\ 2 & 9 & 4 \\ -1 & -3 & 1 \end{bmatrix}$

A basis for the eigenspace corresponding to $\lambda = 3$ is

Neetesh Kumar | October 16, 2024

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This is the solution to Math2B Course: Linear Algebra
Assignment: Ch5 Section 1 Question Number 6
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Step-by-step solution:

Eigenvalue and Eigenvector calculator

To find the eigenspace for $\lambda = 3$, we need to solve the equation:

$(A - 3I)v = 0$

where $I$ is the identity matrix.

Step 1: Compute $A - 3I$

We subtract $3I$ from matrix $A$:

$A - 3I = \begin{bmatrix} 4 & 3 & 2 \\ 2 & 9 & 4 \\ -1 & -3 & 1 \end{bmatrix} - 3 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 3 & 2 \\ 2 & 6 & 4 \\ -1 & -3 & -2 \end{bmatrix}$

Step 2: Solve the system of equations

We now solve the equation:

$\begin{bmatrix} 1 & 3 & 2 \\ 2 & 6 & 4 \\ -1 & -3 & -2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

Row reducing the augmented matrix:

$\begin{bmatrix} 1 & 3 & 2 & 0 \\ 2 & 6 & 4 & 0 \\ -1 & -3 & -2 & 0 \end{bmatrix}$

After performing Gaussian elimination , we get the following reduced matrix:

$\begin{bmatrix} 1 & 3 & 2 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$

This gives us the following system of equations:

$x_1 + 3x_2 + 2x_3 = 0$

From this equation, we can express $x_1$ in terms of $x_2$ and $x_3$:

$x_1 = -3x_2 - 2x_3$

Thus, the general solution for the eigenvector is:

$v = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -3x_2 - 2x_3 \\ x_2 \\ x_3 \end{bmatrix}$

We can break this into two independent vectors by setting $x_2 = 1, x_3 = 0$ for one vector and $x_2 = 0, x_3 = 1$ for the other:

$v_1 = \begin{bmatrix} -3 \\ 1 \\ 0 \end{bmatrix}, \quad v_2 = \begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix}$

Final Answer:

A basis for the eigenspace corresponding to $\lambda = 3$ is:

$\boxed{\begin{bmatrix} -3 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix}}$

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