Let's analyze and solve the two parts of the problem given in the image.

Problem Statement:

Let $A$ be an $n \times n$ matrix. Show that:

Part (a)

If $\lambda$ is an eigenvalue of $A$, then $\frac{1}{\lambda}$ is an eigenvalue of $A^{-1}$, and vice versa.

Solution for Part (a):

1. By definition, if $\lambda$ is an eigenvalue of $A$, there exists a nonzero vector $\mathbf{v}$ such that: $A \mathbf{v} = \lambda \mathbf{v}.$

2. Assuming $\lambda \neq 0$, we can multiply both sides by $A^{-1}$ (since $A$ is invertible): $A^{-1} A \mathbf{v} = A^{-1} (\lambda \mathbf{v}).$ This simplifies to: $\mathbf{v} = \lambda A^{-1} \mathbf{v}.$

3. Rearranging this equation gives: $A^{-1} \mathbf{v} = \frac{1}{\lambda} \mathbf{v}.$

4. This shows that $\frac{1}{\lambda}$ is an eigenvalue of $A^{-1}$ corresponding to the same eigenvector $\mathbf{v}$.

5. Conversely, if $\mu$ is an eigenvalue of $A^{-1}$, then there exists a nonzero vector $\mathbf{w}$ such that: $A^{-1} \mathbf{w} = \mu \mathbf{w}.$ Multiplying both sides by $A$ gives: $A A^{-1} \mathbf{w} = A (\mu \mathbf{w}),$ which simplifies to: $\mathbf{w} = \mu A \mathbf{w}.$ Rearranging, we get: $A \mathbf{w} = \frac{1}{\mu} \mathbf{w}.$ Hence, $\frac{1}{\mu}$ is an eigenvalue of $A$.

Thus, we have shown that if $\lambda$ is an eigenvalue of $A$, then $\frac{1}{\lambda}$ is an eigenvalue of $A^{-1}$, and vice versa.

Part (b)

Show that the product of the eigenvalues of $A$ is equal to $\det(A)$.

Solution for Part (b):

1. The eigenvalues $\lambda_1, \lambda_2, \ldots, \lambda_n$ of $A$ are the roots of the characteristic polynomial of $A$, given by: $p(\lambda) = \det(A - \lambda I).$

2. For an $n \times n$ matrix $A$, the characteristic polynomial $p(\lambda)$ is a polynomial of degree $n$ with the form: $p(\lambda) = (-1)^n \lambda^n + c_{n-1} \lambda^{n-1} + \cdots + c_1 \lambda + c_0.$ where $c_0 = \det(A)$ is the constant term of the polynomial, which is also the product of the eigenvalues of $A$ (assuming they are non-zero).

3. By the Fundamental Theorem of Algebra, the characteristic polynomial can be factored as: $p(\lambda) = (-1)^n (\lambda - \lambda_1)(\lambda - \lambda_2) \cdots (\lambda - \lambda_n).$

4. Setting $\lambda = 0$ in this polynomial gives: $p(0) = (-1)^n (-\lambda_1)(-\lambda_2) \cdots (-\lambda_n) = \lambda_1 \lambda_2 \cdots \lambda_n = \det(A).$

Therefore, the product of the eigenvalues of $A$ is equal to $\det(A)$.

Let me know if you would like further details on any step, or if you have additional questions!

Further Related Questions

1. How does the eigenvalue relationship between $A$ and $A^{-1}$ change if $A$ is not invertible?
2. Can we use the trace of $A$ to find any specific sum of its eigenvalues?
3. How would the results change if $A$ is a non-square matrix?
4. What is the relationship between the eigenvalues of $A$ and those of $A^T$ (the transpose of $A$)?
5. How does the product of the eigenvalues relate to the determinant for other special types of matrices, like diagonal matrices?

Tip:

The determinant of a matrix is an important property that provides insights into its invertibility, eigenvalues, and overall behavior in transformations.