Isn't the equation from the determinant *λ^*2+4*λ*+4?

You're correct; the characteristic equation derived from the determinant of the matrix should indeed be \( \lambda^2 + 4\lambda + 4 = 0 \). Let's go through the determinant calculation again to clarify:

Given the matrix \( A = \begin{bmatrix} 2 & 16 \\ -1 & -6 \end{bmatrix} \), the characteristic polynomial is found by computing the determinant of \( A - \lambda I \), where \( I \) is the identity matrix and \( \lambda \) is the eigenvalue:

\[
A - \lambda I = \begin{bmatrix} 2 & 16 \\ -1 & -6 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 - \lambda & 16 \\ -1 & -6 - \lambda \end{bmatrix}
\]

Now, calculate the determinant of this matrix:

\[
\text{det}(A - \lambda I) = (2 - \lambda)(-6 - \lambda) - (16)(-1)
\]

Expanding this:

\[
= (2 - \lambda)(-6 - \lambda) + 16
\]

\[
= [2(-6 - \lambda) - \lambda(-6 - \lambda)] + 16
\]

\[
= [-12 - 2\lambda + 6\lambda + \lambda^2] + 16
\]

\[
= \lambda^2 + 4\lambda - 12 + 16
\]

\[
= \lambda^2 + 4\lambda + 4
\]

So the correct characteristic equation is:

\[
\lambda^2 + 4\lambda + 4 = 0
\]

Now, solving this quadratic equation:

\[
\lambda = \frac{-4 \pm \sqrt{(4)^2 - 4(1)(4)}}{2(1)} = \frac{-4 \pm \sqrt{16 - 16}}{2} = \frac{-4 \pm 0}{2} = \frac{-4}{2} = -2
\]

The eigenvalue is \( \lambda = -2 \) (with algebraic multiplicity 2).

### Is the Matrix Nondefective?

To check if the matrix is nondefective, we need to examine if there are enough linearly independent eigenvectors corresponding to the eigenvalue \( \lambda = -2 \).

For this, we find the eigenvector by solving \( (A - \lambda I)\mathbf{v} = 0 \):

\[
(A + 2I)\mathbf{v} = \begin{bmatrix} 2 + 2 & 16 \\ -1 & -6 + 2 \end{bmatrix} \mathbf{v} = \begin{bmatrix} 4 & 16 \\ -1 & -4 \end{bmatrix} \mathbf{v} = 0
\]

Row reducing this matrix:

\[
\text{Row reduce } \begin{bmatrix} 4 & 16 \\ -1 & -4 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 4 \\ 0 & 0 \end{bmatrix}
\]

This gives us the equation \( v_1 + 4v_2 = 0 \), or \( v_1 = -4v_2 \). Hence, the eigenvector corresponding to \( \lambda = -2 \) is \( \mathbf{v} = \begin{bmatrix} -4 \\ 1 \end{bmatrix} \).

Since we only have one linearly independent eigenvector and we needed two (due to the eigenvalue's algebraic multiplicity of 2), the matrix is **defective**, not nondefective.

Would you like further details or have any additional questions?

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Here are 8 questions that expand on this information:

1. What is the significance of a defective matrix in applications like differential equations?
2. Can a defective matrix be diagonalized?
3. How do you compute the Jordan Canonical Form for a defective matrix?
4. What role does geometric multiplicity play in determining whether a matrix is defective?
5. How does the presence of a defective matrix affect the stability of systems in control theory?
6. Can you explain how to find generalized eigenvectors for a defective matrix?
7. What is the relationship between a matrix's eigenvalues and its invertibility?
8. How would you apply this concept of defectiveness to complex eigenvalues?

**Tip:** A matrix with a repeated eigenvalue is defective if the number of linearly independent eigenvectors associated with that eigenvalue is less than its algebraic multiplicity.

Explore the solution to a defective matrix problem involving eigenvalues and eigenvectors. Learn how to compute the characteristic polynomial, solve the associated quadratic equation, and determine the defectiveness of a matrix. Suitable for college-level students and enthusiasts of linear algebra.

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