Problem 98 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(A=\left[a_{i j}\right.\) be an \(n \times n\) matrix. The matrix \(A-\lambda I\) is called the characteristics matrix of \(A\), where \(\lambda\) is a scalar and \(I\) is the identity matrix. The determinant \(|A-\lambda I|\) is a non-null polynomial of degree \(n\) in \(\lambda\) and is called the characteristic polynomial of \(A\). The equation \(|A-\lambda I|=0\) is called the characteristic equation of \(A\) and its roots are called the characteristic roots or latent roots or eigen values of \(A\). The set of all eigenvalues of the matrix \(A\) is called the spectrum of A. The product of the eigenvalues of a matrix \(A\) is equal to the determinant \(A\). Which of the following statements are true? If \(A\) is any \(n \times n\) matrix and \(\lambda\) is a characteristic root of \(A\), then (A) \(A\) and \(A^{\prime}\) have the same characteristic roots (B) \(k \lambda\) is a characteristic root of \(k A\) ( \(k\) being scalar) (C) \(\lambda^{n}\) is a characteristic root of \(A^{n}\) ( \(n\) being positive integer) (D) \(\frac{1}{\lambda}\) is a characteristic root of \(A^{-1}\)

Step-by-Step Solution

Answer

Statements (A), (B), and (D) are true, (C) is false.

Step 1: Understanding Eigenvalues

Eigenvalues of a matrix \( A \) are the solutions \( \lambda \) of the equation \( |A - \lambda I| = 0 \). These are also known as characteristic roots or latent roots.

Step 1: Analyze Statement A

Statement A claims that a matrix \( A \) and its transpose \( A' \) have the same characteristic roots. This is true because the determinant is invariant under matrix transposition, i.e., \( |A - \lambda I| = |A' - \lambda I| \). Therefore, both have the same characteristic polynomial and hence, the same eigenvalues.

Step 2: Analyze Statement B

This statement suggests if \( \lambda \) is a characteristic root of \( A \), then \( k \lambda \) is a characteristic root of \( kA \). When \( A \) is multiplied by a scalar \( k \), the characteristic polynomial changes to \( |kA - \lambda I| = k^n|A - \frac{\lambda}{k}I| = 0 \), implying the roots scale by \( k \), making this statement true for each eigenvalue to scale \( \lambda \) as \( k\lambda \). Thus, statement (B) is true.

Step 3: Analyze Statement C

If \( \lambda \) is an eigenvalue of \( A \), it does not necessarily mean \( \lambda^n \) is an eigenvalue of \( A^n \). Eigenvalues of \( A^n \) are \( \lambda^n \) where \( \lambda \) are the eigenvalues of \( A \), but not all \( \lambda \) make it through this transition individually. Thus, statement C is false since \( \lambda^n \) is not automatically a characteristic root for any eigenvalue.

Step 4: Analyze Statement D

This statement claims \( \frac{1}{\lambda} \) is an eigenvalue of \( A^{-1} \) if \( \lambda \) is an eigenvalue of \( A \). Invertible matrices \( A \) with eigenvalue \( \lambda \) transform to \( A^{-1} \) with eigenvalue \( \frac{1}{\lambda} \), reflecting the theorem that if \( Av = \lambda v \), then \( A^{-1}v = \frac{1}{\lambda}v \). Therefore, statement D is true.

Key Concepts

Characteristic Polynomial

The characteristic polynomial of a square matrix is a crucial concept in linear algebra. It is derived from the matrix using a specific process. Given an \(n \times n\) matrix \(A\) and a scalar \(\lambda\), the matrix \(A - \lambda I\) is formed, where \(I\) is the identity matrix of the same size as \(A\). The determinant of this new matrix \(|A - \lambda I|\) is calculated, producing a polynomial in \(\lambda\). This polynomial, which is of degree \(n\), is known as the characteristic polynomial of the matrix \(A\).

In simpler terms, the characteristic polynomial is a way to encode important properties of the matrix, such as its eigenvalues, which are the roots of this polynomial.

• To find it, compute the determinant \(|A - \lambda I|\).
• It provides a polynomial equation where the major goal is to solve for \(\lambda\).
• The coefficients of the polynomial are functions of the entries of \(A\).

Characteristic Equation

The characteristic equation is closely related to the characteristic polynomial, as it is the equation you get when you set the characteristic polynomial equal to zero: \(|A - \lambda I| = 0\). This equation is fundamental for analyzing a matrix because its solutions, termed eigenvalues or characteristic roots, reveal significant properties about the matrix.

These eigenvalues can tell us if the matrix is invertible, as a non-zero determinant (derived from multiplying eigenvalues) implies invertibility. They also play a role in understanding the matrix's behavior in terms of transformations it represents, such as scaling and rotating vectors in a vector space.

• Solving \(|A - \lambda I| = 0\) provides the eigenvalues.
• The roots demonstrate key insights into the matrix's properties.
• Understanding these roots is critical for various applications in scientific computing and engineering.

Matrix Transposition

Matrix transposition is a simple yet important operation in linear algebra. It involves flipping a matrix over its diagonal, which means converting its rows into columns and columns into rows. If \(A\) is a matrix, the transpose of \(A\) is denoted by \(A'\) or sometimes \(A^T\).

This operation does not affect the eigenvalues of the matrix. Therefore, if \(\lambda\) is an eigenvalue of \(A\), it is also an eigenvalue of \(A'\). This invariance is due to the property that the determinants of a matrix and its transpose are equal, which in turn means the characteristic polynomials of \(A\) and \(A'\) are identical.

• Switch rows with columns to transpose a matrix.
• The transpose maintains the determinant, ensuring same eigenvalues as the original matrix.
• This property is vital in simplifying many problems in linear algebra.