Question

Find (a) the characteristic equationand (b) the eigenvalues (and corresponding eigenvectors)of the matrix.

[2 3 1]
A = [0 -1 2]
{0 0 3]

Answer 1

The characteristic equation is
`\((\lambda - 2)(\lambda + 1)(\lambda - 3) = 0\)`. The eigenvalues are
`\(\lambda_1 = 2\), \(\lambda_2 = -1\), and \(\lambda_3 = 3\)`, with corresponding eigenvectors.

Let's go through the calculations.

(a) Characteristic Equation:

The characteristic equation is given by
`\(\det(A - \lambda I) = 0\)`, where A is the matrix,
`\(\lambda\)` is the eigenvalue, and I is the identity matrix.

`\[ A = \begin{bmatrix} 2 & 3 & 1 \\ 0 & -1 & 2 \\ 0 & 0 & 3 \end{bmatrix} \]\[ \det\left(\begin{bmatrix} 2-\lambda & 3 & 1 \\ 0 & -1-\lambda & 2 \\ 0 & 0 & 3-\lambda \end{bmatrix}\right) = 0 \]`

Calculate the determinant:

`\[ (2-\lambda) \cdot ((-1-\lambda) \cdot (3-\lambda) - 2 \cdot 0) = 0 \]`

Solve for
`\(\lambda\)` to find the characteristic equation.

(b) Eigenvalues (and Corresponding Eigenvectors):

Solve the characteristic equation to find the eigenvalues
`\(\lambda\)`. Once you have the eigenvalues, substitute them back into the equation
`\((A - \lambda I)\mathbf{x} = \mathbf{0}\)` to find the corresponding eigenvectors
`(\(\mathbf{x}\))`.

Let's proceed with the calculations:

`\[ (2-\lambda)((-1-\lambda)(3-\lambda)) = 0 \]`

Expand and simplify:

`\[ (2-\lambda)(\lambda^2 - 2\lambda - 3) = 0 \]`

Now, solve for
`\(\lambda\)`:

1.
`\(2 - \lambda = 0 \Rightarrow \lambda_1 = 2\)`

2.
`\(\lambda^2 - 2\lambda - 3 = 0\) (factor or use quadratic formula to find \(\lambda_2\) and \(\lambda_3\))`

Now, for each eigenvalue, solve the system of equations:

1. For
`\(\lambda_1 = 2\)`:

`\[ (A - 2I)\mathbf{x}_1 = \mathbf{0} \]`

2. For
`\(\lambda_2\) and \(\lambda_3\)`:

`\[ (A - \lambda_i I)\mathbf{x}_i = \mathbf{0} \] (where \(i = 2, 3\))`

(a) Characteristic Equation:

`\[ (2-\lambda)(\lambda^2 - 2\lambda - 3) = 0 \]`

Expanding and simplifying:

`\[ (2-\lambda)(\lambda^2 - 2\lambda - 3) = 0 \]\[ (2-\lambda)(\lambda+1)(\lambda-3) = 0 \]`

So, the characteristic equation is:

`\[ (\lambda - 2)(\lambda + 1)(\lambda - 3) = 0 \]`

(b) Eigenvalues and Corresponding Eigenvectors:

1. For
`\(\lambda_1 = 2\)`:

`\[ (A - 2I)\mathbf{x}_1 = \mathbf{0} \]\[ \begin{bmatrix} 0 & 3 & 1 \\ 0 & -3 & 2 \\ 0 & 0 & 1 \end{bmatrix} \mathbf{x}_1 = \mathbf{0} \]\[ \begin{bmatrix} 0 & 3 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \mathbf{x}_1 = \mathbf{0} \]`

The solution is
`\(x_(11) = 1\), \(x_(12) = 0\), \(x_(13) = -3\)`.

2. For
`\(\lambda_2 = -1\)`:

`\[ (A + I)\mathbf{x}_2 = \mathbf{0} \]\[ \begin{bmatrix} 3 & 3 & 1 \\ 0 & 0 & 2 \\ 0 & 0 & 4 \end{bmatrix} \mathbf{x}_2 = \mathbf{0} \]\[ \begin{bmatrix} 3 & 3 & 1 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{bmatrix} \mathbf{x}_2 = \mathbf{0} \]`

The solution is
`\(x_(21) = -1\), \(x_(22) = 1\), \(x_(23) = 0\)`.

3. For
`\(\lambda_3 = 3\)`:

`\[ (A - 3I)\mathbf{x}_3 = \mathbf{0} \]\[ \begin{bmatrix} -1 & 3 & 1 \\ 0 & -4 & 2 \\ 0 & 0 & 0 \end{bmatrix} \mathbf{x}_3 = \mathbf{0} \]`

The solution is
`\(x_(31) = 1\), \(x_(32) = (3)/(4)\), \(x_(33) = (1)/(2)\)`.

So, the eigenvalues are
`\(\lambda_1 = 2\), \(\lambda_2 = -1\), and \(\lambda_3 = 3\)`, and the corresponding eigenvectors are
`\(\mathbf{x}_1 = \begin{bmatrix} 1 \\ 0 \\ -3 \end{bmatrix}\), \(\mathbf{x}_2 = \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix}\), and \(\mathbf{x}_3 = \begin{bmatrix} 1 \\ (3)/(4) \\ (1)/(2) \end{bmatrix}\)`.