Eigenvalue Relationship Between A and its Inverse A^-1

This solution explains the relationship between the eigenvalues of a non-singular matrix A and the eigenvalues of its inverse matrix, A^-1.

Understanding Eigenvalues and Eigenvectors

In linear algebra, if A is a square matrix, a non-zero vector x is called an eigenvector of A when multiplying A by x results in a scaled version of x. The scaling factor is a scalar value called the eigenvalue, denoted by $\lambda$. The relationship is expressed mathematically as:

$ A \mathbf{x} = \lambda \mathbf{x} $

Here, $\lambda$ is the eigenvalue and x is the corresponding eigenvector.

Deriving the Eigenvalue of the Inverse Matrix

We are given that $\lambda$ is an eigenvalue of a non-singular matrix A. Key points to remember about a non-singular matrix are:

• It is invertible, meaning its inverse, denoted as A^-1, exists.
• None of its eigenvalues are zero. Therefore, if $\lambda$ is an eigenvalue of A, then $\lambda \neq 0$.

Let's start with the fundamental eigenvalue equation for matrix A:

$ A \mathbf{x} = \lambda \mathbf{x} $

Since A is non-singular, we know A^-1 exists. We can multiply both sides of the equation by A^-1:

$ \mathbf{A}^{-1} (A \mathbf{x}) = \mathbf{A}^{-1} (\lambda \mathbf{x}) $

Using the properties of matrix multiplication (associativity) and the definition of an inverse matrix ($\mathbf{A}^{-1} \mathbf{A} = \mathbf{I}$, where I is the identity matrix):

$ (\mathbf{A}^{-1} \mathbf{A}) \mathbf{x} = \lambda (\mathbf{A}^{-1} \mathbf{x}) $

$ \mathbf{I} \mathbf{x} = \lambda (\mathbf{A}^{-1} \mathbf{x}) $

Since multiplying any vector by the identity matrix results in the same vector ($\mathbf{I} \mathbf{x} = \mathbf{x}$):

$ \mathbf{x} = \lambda (\mathbf{A}^{-1} \mathbf{x}) $

As established earlier, because A is non-singular, its eigenvalue $\lambda$ cannot be zero ($\lambda \neq 0$). Thus, we can divide both sides of the equation by $\lambda$:

$ \frac{1}{\lambda} \mathbf{x} = \mathbf{A}^{-1} \mathbf{x} $

This resulting equation, $\mathbf{A}^{-1} \mathbf{x} = \frac{1}{\lambda} \mathbf{x}$, fits the definition of an eigenvalue and eigenvector. It shows that $\frac{1}{\lambda}$ is the eigenvalue of the inverse matrix A^-1, corresponding to the same eigenvector x.

Considering the Provided Options

The question asks for the eigenvalue of A^-1, given $\lambda$ is an eigenvalue of the non-singular matrix A. The options presented are $\frac{1}{\lambda}$, $\lambda$, $-\lambda$, and $-\frac{1}{\lambda}$. Based on our derivation, the correct eigenvalue for A^-1 is $\frac{1}{\lambda}$.

The provided correct answer text indicates that $\lambda$ is the correct option for this question.