Question

Let \(A\) be a diagonalizable \(n \times n\) matrix with eigenvalues \(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n}\) (including multiplicities). Show that: a. \(\operatorname{det} A=\lambda_{1} \lambda_{2} \cdots \lambda_{n}\) b. \(\operatorname{tr} A=\lambda_{1}+\lambda_{2}+\cdots+\lambda_{n}\)

Short answer

The determinant of \(A\) is the product of its eigenvalues, and the trace of \(A\) is the sum of its eigenvalues.

Step-by-step solution

Understand Diagonalizable Matrices

A diagonalizable matrix \(A\) can be expressed as \(A = PDP^{-1}\), where \(D\) is a diagonal matrix of eigenvalues, and \(P\) is the matrix of corresponding eigenvectors.

Recall Properties of Determinants

For any matrices \(A\), \(B\), and \(C\) of appropriate size, the determinant satisfies \( \operatorname{det}(ABC) = \operatorname{det}(A) \operatorname{det}(B) \operatorname{det}(C) \). Also, \( \operatorname{det}(I) = 1 \), where \(I\) is the identity matrix.

Compute Determinant Using Diagonalization

Given \(A = PDP^{-1}\), we have \(\operatorname{det}(A) = \operatorname{det}(PDP^{-1}) = \operatorname{det}(P)\operatorname{det}(D)\operatorname{det}(P^{-1})\). Since \( \operatorname{det}(P)\operatorname{det}(P^{-1}) = 1 \), it follows that \( \operatorname{det}(A) = \operatorname{det}(D)\).

Calculate Determinant of a Diagonal Matrix

The determinant of a diagonal matrix \(D\), where the main diagonal contains the eigenvalues \(\lambda_1, \lambda_2, \ldots, \lambda_n\), is given by the product of its diagonal elements. Thus, \( \operatorname{det}(D) = \lambda_1 \lambda_2 \cdots \lambda_n \). This proves \( \operatorname{det}(A) = \lambda_1 \lambda_2 \cdots \lambda_n \).

Recall Properties of Trace

The trace of a matrix is the sum of its diagonal elements. For any matrix, the trace satisfies \( \operatorname{tr}(ABC) = \operatorname{tr}(BCA) = \operatorname{tr}(CAB) \).

Compute Trace Using Diagonalization

For \(A = PDP^{-1}\), where \(D\) is diagonal with eigenvalues on its diagonal, \( \operatorname{tr}(A) = \operatorname{tr}(PDP^{-1}) = \operatorname{tr}(D) \).

Key concepts

Eigenvalues

Eigenvalues are special numbers associated with a square matrix. These values tell us about the behavior of the matrix, especially when it is applied to vectors. Imagine stretching or squashing vectors in a vector space; eigenvalues determine how much scaling occurs along particular directions. Given a matrix \(A\), if there is a non-zero vector \(v\) such that \(Av = \lambda v\), then \(\lambda\) is an eigenvalue of \(A\). The vector \(v\) is referred to as an eigenvector corresponding to that eigenvalue.

Eigenvalues are crucial for understanding if a matrix can be diagonalized, which simplifies many matrix operations. When a matrix is diagonalizable, it means that there are enough eigenvectors to form a basis for the space, making it possible to represent the matrix in a simpler diagonal form. In a diagonal matrix, the eigenvalues appear on the main diagonal, and the rest of the elements are zero. This property makes calculations, like finding the determinant, much easier.

Determinant

The determinant is a scalar value that is derived from a square matrix. It provides a variety of information about the matrix, such as whether the matrix is invertible. For any square matrix \(A\), the determinant is denoted by \(\operatorname{det}(A)\).

To relate determinants to diagonalizable matrices, consider matrices \(P\) and \(D\) such that \(A = PDP^{-1}\). The determinant of \(A\) is then equal to the product of the determinants of \(D\) (a diagonal matrix with eigenvalues as its entries) and the identity transformation from \(P\) and \(P^{-1}\), i.e., \(\operatorname{det}(P)\operatorname{det}(D)\operatorname{det}(P^{-1})\). However, \(\operatorname{det}(P)\operatorname{det}(P^{-1}) = 1\), so the determinant simplifies to the product of the eigenvalues: \(\lambda_1 \lambda_2 \cdots \lambda_n\). This relationship makes the calculation of determinants straightforward when dealing with diagonal matrices.

Trace

The trace of a matrix is the sum of its diagonal elements and is denoted by \(\operatorname{tr}(A)\). It is particularly useful because it is invariant under cyclic permutations, which means for any matrices \(A\), \(B\), and \(C\), the trace satisfies \(\operatorname{tr}(ABC) = \operatorname{tr}(BCA)\).

For a diagonalizable matrix \(A = PDP^{-1}\), the trace can be easily computed because \(P\) and \(P^{-1}\) in the product do not affect the trace due to this cyclic property. Therefore, \(\operatorname{tr}(A) = \operatorname{tr}(D)\). Since \(D\) is a diagonal matrix created from the eigenvalues, the trace of \(A\) becomes simply the sum of its eigenvalues: \(\lambda_1 + \lambda_2 + \cdots + \lambda_n\). This makes the trace an important tool for quickly evaluating key properties of linear transformations.

Eigenvectors

Eigenvectors are a fundamental part of understanding matrices, especially in diagonalization. If \(v\) is an eigenvector of a matrix \(A\), it means that applying \(A\) to \(v\) stretches or compresses \(v\) by the associated eigenvalue \(\lambda\), expressed as \(Av = \lambda v\).

Eigenvectors are critical because they allow matrices to be expressed in their simplest form. If a matrix has enough independent eigenvectors to span the space, it can be diagonalized. This means the matrix can be rewritten using its eigenvectors and eigenvalues, which simplifies complex computations.

Easily highlighting key directions and magnitudes of transformation, eigenvectors provide insight into the geometric nature of matrix operations. They are foundational in applications such as stability analysis, vibration analysis, and even in algorithms like Principal Component Analysis (PCA) for data dimensionality reduction.