Problem 32 - Differential Equations and Linear Algebra

Question

Let \(A\) be a nondefective matrix and let \(S\) be a matrix such that \(S^{-1} A S=\operatorname{diag}\left(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n}\right),\) where all \(\lambda_{i}\) are nonzero. (a) Prove that \(A\) is invertible. (b) Prove that $$S^{-1} A^{-1} S=\operatorname{diag}\left(\frac{1}{\lambda_{1}}, \frac{1}{\lambda_{2}}, \ldots, \frac{1}{\lambda_{n}}\right)$$

Step-by-Step Solution

Answer

In summary, since matrix \(A\) is nondefective, it has a complete set of linearly independent eigenvectors, and there exists a matrix \(S\) that can diagonalize it. We proved that \(A\) is invertible as its determinant is nonzero, and subsequently found the relationship \(S^{-1}A^{-1}S = \operatorname{diag}(\frac{1}{\lambda_1}, \frac{1}{\lambda_2}, \ldots, \frac{1}{\lambda_n})\).

Step 1: (Step 1: Prove that A is invertible)

Matrix \(A\) is nondefective, which means it has a complete set of linearly independent eigenvectors, and there exists a matrix \(S\) that can diagonalize it. The provided diagonal matrix is composed of eigenvalues as its diagonal entries. Since all \(\lambda_i\) are nonzero, we can conclude that the determinant of the diagonal matrix is nonzero as well. By the properties of determinant, we have \(\operatorname{det}(S^{-1}AS) = \operatorname{det}(S^{-1})\operatorname{det}(A)\operatorname{det}(S)\). This can be simplified to \(\operatorname{det}(A) = \frac{\operatorname{det}(S^{-1}AS)}{\operatorname{det}(S^{-1})\operatorname{det}(S)}\). Since the determinant of the diagonal matrix is nonzero and \(\operatorname{det}(S)\) and \(\operatorname{det}(S^{-1})\) exist, we can conclude that \(\operatorname{det}(A) \neq 0\), which proves that matrix \(A\) is invertible.

Step 2: (Step 2: Prove that \(S^{-1}A^{-1}S = \operatorname{diag}(\frac{1}{\lambda_1}, \frac{1}{\lambda_2}, \ldots, \frac{1}{\lambda_n})\))

Now that we know \(A\) is invertible, we can find \(A^{-1}\). Let's begin by multiplying both sides of the equation \(S^{-1}AS = \operatorname{diag}(\lambda_1, \lambda_2, \ldots, \lambda_n)\) with \(S\): \(A = S(\operatorname{diag}(\lambda_1, \lambda_2, \ldots, \lambda_n)S^{-1})\). Now, we will multiply both sides by \(A^{-1}\): \(\operatorname{I} = A^{-1}S(\operatorname{diag}(\lambda_1, \lambda_2, \ldots, \lambda_n)S^{-1})\) Since we have the identity matrix on the left-hand side of the equation, we can easily find the inverse of the right-hand side by multiplying the respective inverses in an opposite order: \(\operatorname{I} = S(\operatorname{diag}(\lambda_1, \lambda_2, \ldots, \lambda_n)S^{-1})A^{-1}\) \(\operatorname{I} = SO^{-1}A^{-1}\) Now we multiply both sides by the inverse of \(SO^{-1}\) from the left, which is \((SO^{-1})^{-1}\), to get: \((SO^{-1})^{-1} = A^{-1}\) Now, we can use the property of diagonal matrices that the inverse of a diagonal matrix is obtained by reciprocating its diagonal elements. So, the inverse of \(\operatorname{diag}(\lambda_1, \lambda_2, \ldots, \lambda_n)\) is \(\operatorname{diag}(\frac{1}{\lambda_1}, \frac{1}{\lambda_2}, \ldots, \frac{1}{\lambda_n})\). Thus, replacing with \(O^{-1} = \operatorname{diag}(\frac{1}{\lambda_1}, \frac{1}{\lambda_2}, \ldots, \frac{1}{\lambda_n})\), we get: \(S^{-1}A^{-1}S = \operatorname{diag}(\frac{1}{\lambda_1}, \frac{1}{\lambda_i}, \ldots, \frac{1}{\lambda_n})\), which is the result we need to prove.

Key Concepts

Matrix Diagonalization

Matrix diagonalization is a powerful method in linear algebra where a given square matrix is expressed in terms of its eigenvalues and eigenvectors. This technique allows us to convert the matrix into a diagonal form using a particular transformation. The diagonalized form represents the original matrix using the eigenvalues as the entries on the diagonal of a new matrix. If we have a square matrix \(A\), it can be diagonalized if there exists an invertible matrix \(S\) such that:

• The product \(S^{-1}AS\) is a diagonal matrix.
• The diagonal entries of this matrix are the eigenvalues of \(A\).

This process is extremely useful as working with diagonal matrices is much simpler, especially for computations involving powers of matrices and their exponentials. In this exercise, \(A\) is transformed into its diagonal form with \(S\), aligning with its eigenvalues, which simplifies further mathematical manipulations greatly.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are fundamental concepts in the field of linear algebra. When a square matrix \(A\) acts on an eigenvector \(\mathbf{v}\), it stretches or compresses it by a scalar factor called an eigenvalue, \(\lambda\), such that:

• \(A\mathbf{v} = \lambda\mathbf{v}\).

The eigenvalues are the diagonal elements in the diagonalized matrix form of \(A\), which means they reflect how \(A\) scales the directions given by its eigenvectors. For a matrix to be diagonalizable, each eigenvalue must have a sufficient number of linearly independent eigenvectors corresponding to it. Understanding the relationship between eigenvalues and eigenvectors is crucial. They provide simpler forms for matrix operations and unveil more properties of linear transformations that are otherwise not immediately obvious. Essentially, these help to completely describe the behavior of the matrix \(A\) in a geometrical and algebraic manner.

Invertible Matrix

A matrix is said to be invertible if there exists another matrix that can "undo" the action of \(A\) through multiplication, resulting in the identity matrix. In other words, a matrix \(A\) is invertible if there exists a matrix \(A^{-1}\) such that:

• \(AA^{-1} = A^{-1}A = I\),

where \(I\) is the identity matrix of the same dimension. One important condition for a matrix to be invertible is that its determinant must be nonzero. This exercise proves the matrix \(A\) is invertible by showing its determinant, derived from its nonzero eigenvalues, is non-zero. Having an invertible matrix is crucial in solving systems of linear equations and ensuring certain matrix operations can be performed. It confirms that each linear transformation represented by \(A\) can be reversed, leading to robustness in various applications of matrices.

Determinant Properties

Determinants play a critical role in understanding matrix properties, particularly in identifying whether a matrix is invertible and its behavior in linear transformations. For a diagonal matrix containing the eigenvalues of \(A\), its determinant is simply the product of its diagonal elements:

• \(\det(S^{-1}AS) = \lambda_1 \lambda_2 \cdots \lambda_n\).

Due to the properties of determinants, if this product (determinant) is non-zero, it confirms the invertibility of both the diagonal matrix and \(A\). Additionally, the transformation \(S^{-1}AS\) into its diagonal form is a determinant-preserving operation given that:

• \(\det(S^{-1})\det(A)\det(S) = \det(S^{-1}AS)\).

This means that the transformations used for diagonalization maintain determinant values, which simplifies confirming invertibility and understanding the scaling effects caused by \(A\) on vector spaces. Hence, determinant properties reveal essential insights about a matrix's capability to reverse its transformations.