To find the eigenvalues and the corresponding eigenspaces for the matrix $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 2 \\ 0 & -1 & -1 \end{bmatrix}$, we follow these steps:

Step 1: Find the characteristic equation

The eigenvalues $\lambda$ of a matrix $A$ are found by solving the characteristic equation:

$\text{det}(A - \lambda I) = 0$

Where $I$ is the identity matrix and $\lambda$ is the eigenvalue. The matrix $A - \lambda I$ is:

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

Now, compute the determinant:

$\text{det}(A - \lambda I) = \text{det} \begin{bmatrix} 1 - \lambda & 0 & 0 \\ 0 & 2 - \lambda & 2 \\ 0 & -1 & -1 - \lambda \end{bmatrix}$

This is a block triangular matrix, so the determinant is the product of the diagonal blocks:

$\text{det}(A - \lambda I) = (1 - \lambda) \cdot \text{det} \begin{bmatrix} 2 - \lambda & 2 \\ -1 & -1 - \lambda \end{bmatrix}$

Now, compute the determinant of the 2x2 matrix:

$\text{det} \begin{bmatrix} 2 - \lambda & 2 \\ -1 & -1 - \lambda \end{bmatrix} = (2 - \lambda)(-1 - \lambda) - (-2) = \lambda^2 + \lambda$

So the characteristic equation becomes:

$(1 - \lambda)(\lambda^2 + \lambda) = 0$

Step 2: Solve for the eigenvalues

We can now solve this equation:

$(1 - \lambda)(\lambda^2 + \lambda) = 0$

This gives two factors:

1. $1 - \lambda = 0$ → $\lambda = 1$
2. $\lambda(\lambda + 1) = 0$ → $\lambda = 0$ or $\lambda = -1$

Thus, the eigenvalues are $\lambda_1 = 1$, $\lambda_2 = 0$, and $\lambda_3 = -1$. The eigenvalue $\lambda = 1$ has multiplicity 1, and the eigenvalue $\lambda = -1$ has multiplicity 2.

Step 3: Find the eigenspaces

Now, let's find a basis for each eigenspace by solving $(A - \lambda I)\mathbf{v} = 0$ for each eigenvalue.

Eigenvalue $\lambda = 1$

For $\lambda = 1$, solve $(A - I)\mathbf{v} = 0$:

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

Solving this system, we have the equations:

$x_2 + 2x_3 = 0 \quad \text{and} \quad -x_2 - 2x_3 = 0$

This gives $x_2 = -2x_3$, so the solution is of the form:

$\mathbf{v} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$

Thus, the basis for the eigenspace corresponding to $\lambda = 1$ is $\left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \right\}$.

Eigenvalue $\lambda = -1$

For $\lambda = -1$, solve $(A + I)\mathbf{v} = 0$:

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

This gives the system of equations:

$3x_2 + 2x_3 = 0 \quad \text{and} \quad -x_2 = 0$

From the second equation, we get $x_2 = 0$. Plugging this into the first equation gives $2x_3 = 0$, so $x_3 = 0$. Thus, the solution is:

$\mathbf{v} = \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix}$

Therefore, the basis for the eigenspace corresponding to $\lambda = -1$ is $\left\{ \begin{bmatrix} 0 \\ 2 \\ -1 \end{bmatrix} \right\}$.

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