Problem 41
Question
Let \(A\) be an \(n \times n\) matrix such that \(A^{n}=\alpha A\), where \(\alpha\) is a real number different from 1 and \(-1\). Then, the matrix \(A+I_{n}\) is (A) singular (B) non-singular, i.e., invertible (C) scalar matrix (D) None of these
Step-by-Step Solution
Verified Answer
The matrix \(A+I_n\) is non-singular.
1Step 1: Understanding the Given Condition
We are given that \(A^{n} = \alpha A\) for an \(n \times n\) matrix \(A\), where \(\alpha\) is a real number different from 1 and -1. This implies that \(A\) satisfies the polynomial \(x^n - \alpha x = 0\).
2Step 2: Analyzing the Polynomial Equation
Since \(x(x^{n-1} - \alpha) = 0\), the possible eigenvalues of \(A\) are 0 or the roots of \(x^{n-1} = \alpha\). These possible eigenvalues affect the behavior of \(A + I_n\).
3Step 3: Analyzing the Matrix \(A + I_n\)
The eigenvalues of \(A + I_n\) are \(\lambda + 1\), where \(\lambda\) are the eigenvalues of \(A\). If 0 is a root, the corresponding eigenvalue to \(A + I_n\) would be 1. However, if \(A\) has no zero eigenvalue, \(A + I_n\) is invertible.
4Step 4: Considering the Non-zero Eigenvalues
As from the polynomial \(x^n - \alpha x = 0\), if \(\alpha = 0\), then \(A\) has an eigenvalue 0 (not possible since \(\alpha eq 0\)). If the eigenvalues are non-zero, adding identity never makes \(A + I_n\) singular.
5Step 5: Conclusion Based on Eigenvalues
Since \(\alpha eq 0\), and \(\alpha eq 1\) or \(-1\), the eigenvalues of \(A\) (that would make \(A + I_n\) singular) are not allowed, confirming \(A + I_n\) is non-singular.
Key Concepts
Eigenvalues of MatricesPolynomial Equations in Linear AlgebraMatrix Theory
Eigenvalues of Matrices
Eigenvalues are crucial to understanding many properties of matrices. They are the scalars, \( \lambda \), associated with a matrix \( A \) that satisfy the equation \( A\mathbf{v} = \lambda\mathbf{v} \) for some non-zero vector \( \mathbf{v} \). These vectors, \( \mathbf{v} \), are known as eigenvectors.
Eigenvalues can provide insights into the matrix behavior, such as stability and response to linear transformations.
Eigenvalues can provide insights into the matrix behavior, such as stability and response to linear transformations.
- Zero Eigenvalues: If a matrix has zero as an eigenvalue, it indicates that the matrix is singular (non-invertible).
- Non-zero Eigenvalues: Non-zero eigenvalues suggest that the matrix could be invertible, depending on the specific nature of these eigenvalues.
Polynomial Equations in Linear Algebra
Polynomial equations play a central role in determining the characteristics and behaviors of matrices.
In the given problem, the polynomial equation \( x^n - \alpha x = 0 \) arises from the matrix equation \( A^n = \alpha A \), and serves to identify potential eigenvalues of the matrix \( A \).
Additionally, the polynomial nature allows us to explore higher order behaviors of matrices, uncovering deeper insights into their long-term transformations.
In the given problem, the polynomial equation \( x^n - \alpha x = 0 \) arises from the matrix equation \( A^n = \alpha A \), and serves to identify potential eigenvalues of the matrix \( A \).
- The roots derived from the polynomial, such as 0 or those satisfying \( x^{n-1} = \alpha \), are candidate eigenvalues.
Additionally, the polynomial nature allows us to explore higher order behaviors of matrices, uncovering deeper insights into their long-term transformations.
Matrix Theory
Matrix theory is a branch of mathematics focusing on the study of arrays of numbers and their transformations.
The exercise in question taps into several key concepts within this theory:
The exercise in question taps into several key concepts within this theory:
- Invertibility of Matrices: A matrix is invertible if there exists another matrix that when multiplied with yields the identity matrix.
- Identity Matrix: The matrix \( I_n \) plays an important role in ensuring changes in eigenvalues, as seen in the sum \( A + I_n \).
- Singular vs Non-Singular: Matrices are singular when they lack an inverse. The eigenvalues guide this distinction.”
Other exercises in this chapter
Problem 38
\(A\) and \(B\) are two non-singular matrices of the same order such that \(A^{n}=I\) for some positive integer \(n>1\). Then, \(B A^{n-1} B^{-1}-B A^{-1} B^{-1
View solution Problem 40
A skew-symmetric matrix \(A\) satisfies the relation \(A^{2}+\) \(I=0\), where \(I\) is a unit matrix. Then, \(A\) is (A) Idempotent matrix (B) Orthogonal matri
View solution Problem 42
If adj \(B=A\) and \(P, Q\) are two unimodular matrices, i.e., \(|P|=1=|Q|\), then \(\left(Q^{-1} B P^{-1}\right)^{-1}\) is equal to (A) \(P A Q\) (B) \(P B Q\)
View solution Problem 43
If \(A=\left[\begin{array}{cc}1 & \frac{\alpha}{n} \\ -\frac{\alpha}{n} & 1\end{array}\right]\), then (A) \(\lim _{n \rightarrow \infty} A^{n}=0\) (B) \(\lim _{
View solution