Question

Prove that the trace of a matrix equals the sum of all \(n\) of its eigenvalues and the determinant is their product.

Short answer

Question: Prove that for a given square matrix A, the sum of its eigenvalues is equal to its trace, and the product of its eigenvalues is equal to its determinant. Answer: To prove this, we first begin by finding the eigenvalues of the matrix A using its characteristic equation. After obtaining the eigenvalues, we show that their sum corresponds to the trace of the matrix (sum of the main diagonal elements) by analyzing the terms of the expanded characteristic polynomial and using Vieta's formulas. Similarly, the product of eigenvalues can be shown to be equal to the determinant of the matrix A by equating the constant term in the characteristic polynomial to the product of the roots (eigenvalues).

Step-by-step solution

Refresher on Trace and Determinant of a Matrix

Before diving into the problem, let's remember what trace and determinant of a square matrix are. The trace of a square matrix \(A\) is the sum of its main diagonal elements (top-left to bottom-right), denoted by \(tr(A)\). The determinant of a square matrix \(A\), denoted by \(|A|\), is a scalar value that represents a special property of the matrix, and it can be found using various methods such as cofactor expansion or Gaussian elimination.

Finding the eigenvalues using the characteristic equation

Given a square matrix \(A\), we can find its eigenvalues by solving the characteristic equation: \(|A - \lambda I| = 0\), where \(\lambda\) represents the eigenvalues and \(I\) is the identity matrix of the same size as \(A\). This equation will result in a polynomial of degree \(n\) (for an \(n \times n\) matrix) called the characteristic polynomial, which we can solve to find the \(n\) eigenvalues.

Proving the sum of eigenvalues is equal to the trace of the matrix

Now that we have the eigenvalues of the matrix \(A\), let's denote them by \(\lambda_1, \lambda_2, ..., \lambda_n\). To prove that their sum is equal to the trace of \(A\), we can use the fact that the trace of the matrix is the sum of the main diagonal elements: \(tr(A)=a_{11}+a_{22}+...+a_{nn}\). Notice that the characteristic polynomial, in its expanded form, will have a term \(-\lambda (a_{11}+a_{22}+...+a_{nn})\) which, by Vieta's formulas, corresponds to the sum of the roots (eigenvalues). Therefore, the sum of eigenvalues is \(tr(A)\).

Proving the product of eigenvalues is equal to the determinant of the matrix

Using the same set of eigenvalues \(\lambda_1, \lambda_2, ..., \lambda_n\) found in Step 2, we want to prove that their product is equal to the determinant of the matrix. We again look at the characteristic polynomial. The constant term of the polynomial is in fact the determinant \(|A|\); specifically it is \((-1)^n|A|\). Furthermore, the constant term of the polynomial is also the product of its roots (eigenvalues) due to Vieta's formulas. Thus we can write $$ (-1)^n|A| = \lambda_1 \lambda_2 ... \lambda_n. $$ For an even value of \(n\), we obtain $$ |A| = \lambda_1 \lambda_2 ... \lambda_n. $$ For an odd value of \(n\), as the product, \(\lambda_1\lambda_2...\lambda_n\) is equal to the determinant \(A\) multiplied by \(-1\), the equation is still valid.

Key concepts

Matrix Trace

In linear algebra, the trace of a matrix is a very simple concept but holds significant importance.
Imagine a square matrix, which is a matrix with equal number of rows and columns.
The trace of such a matrix is the sum of the elements on its main diagonal. For example, if you have a 3x3 matrix:

• Let's say your matrix is: \[ A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33} \end{pmatrix} \]
• The trace is simply \( a_{11} + a_{22} + a_{33} \), the elements top-left to bottom-right.

The trace is often used because it is invariant under a change of basis.This means that its value does not change if you transform the matrix using different coordinates.

Matrix Determinant

The determinant of a square matrix gives us a scalar representation of the matrix.
This scalar indicates some intrinsic characteristics of the matrix itself.
One of the simplest properties of determinants is that they can help us understand the invertibility of a matrix:

• If the determinant is zero, the matrix does not have an inverse—it's singular.
• If the determinant is not zero, you can find an inverse for the matrix.

To calculate the determinant of a matrix, you can use various methods like cofactor expansion or row reduction.For a 2x2 matrix, it's straightforward: \[ \text{If } A = \begin{pmatrix} a & b \ c & d \end{pmatrix}, \text{ then its determinant } |A| = ad - bc.\]When you deal with larger matrices, you'd usually employ computational tools to help simplify the process.The determinant also plays a crucial role in the characteristic polynomial, specifically as the product of eigenvalues.

Characteristic Polynomial

The characteristic polynomial is a fascinating way to bridge between matrices and their eigenvalues.
For an \(n \times n\) matrix A, the characteristic polynomial is derived from the characteristic equation:

• We start with the equation \[ |A - \lambda I| = 0, \]where \(A\) is your matrix, \(\lambda\) represents its eigenvalues, and \(I\) is the identity matrix.
• Remember, when we mention \( |A - \lambda I| \), we're referring to the determinant of \(A - \lambda I\).

The result is a polynomial of degree \(n\), where the solutions \(\lambda\) are the eigenvalues of the matrix.These eigenvalues tell us about important properties of the system that the matrix represents, like stability and oscillatory patterns.A crucial aspect of the characteristic polynomial is how its coefficients relate to eigenvalues through tools like Vieta's formulas.

Vieta's Formulas

Vieta's formulas provide an insightful connection between the roots of a polynomial and its coefficients.
These formulas are applicable to any polynomial, including the characteristic polynomial of a matrix.
In the context of eigenvalues, Vieta's formulas tell us:

• The sum of all the roots (eigenvalues), without regarding sign, is equal to the negative coefficient of the \(x^{n-1}\) term in a polynomial of degree \(n\).
• The product of all the roots (eigenvalues) corresponds to the constant term of the polynomial, specifically the determinant of the matrix adjusted by \[ (-1)^n. \]

When solving a matrix's characteristic polynomial, Vieta's formulas allow us to easily deduce the sum and product of the eigenvalues.Thus, they link directly to the trace and determinant, showing their value lies in understanding these internal properties of the matrix in question.