Eigenvalues of a Matrix and Its Squared Matrix

Linear algebra problems and solutions

Problem 183

Let $A$ be an $n \times n$ matrix. Suppose that the matrix $A^2$ has a real eigenvalue $\lambda>0$. Then show that either $\sqrt{\lambda}$ or $-\sqrt{\lambda}$ is an eigenvalue of the matrix $A$.

 
LoadingAdd to solve later

Sponsored Links


Hint.

Use the following fact: a scalar $\lambda$ is an eigenvalue of a matrix $A$ if and only if
\[\det(A-\lambda I)=0.\]

Proof.

Since $\lambda$ is an eigenvalue of $A^2$, the determinant of the matrix $A^2-\lambda I$ is zero, where $I$ is the $n \times n$ identity matrix:
\[\det(A^2-\lambda I)=0.\]

Now we have the following factorization.
\begin{align*}
A^2-\lambda I=(A-\sqrt{\lambda} I)(A+\sqrt{\lambda} I).
\end{align*}
Taking the determinant of both sides, we obtain
\begin{align*}
0&=\det(A^2-\lambda I)=\det \left((A-\sqrt{\lambda} I)(A+\sqrt{\lambda} I)\right)\\
&=\det(A-\sqrt{\lambda} I)\det(A+\sqrt{\lambda} I)
\end{align*}
by the multiplicative property of the determinant.

Therefore we have either
\[\det(A-\sqrt{\lambda} I)=0 \text{ or } \det(A+\sqrt{\lambda} I)=0\] Thus we conclude that either $\sqrt{\lambda}$ or $-\sqrt{\lambda}$ is an eigenvalue of the matrix $A$.


LoadingAdd to solve later

Sponsored Links

More from my site

  • Determinant/Trace and Eigenvalues of a MatrixDeterminant/Trace and Eigenvalues of a Matrix Let $A$ be an $n\times n$ matrix and let $\lambda_1, \dots, \lambda_n$ be its eigenvalues. Show that (1) $$\det(A)=\prod_{i=1}^n \lambda_i$$ (2) $$\tr(A)=\sum_{i=1}^n \lambda_i$$ Here $\det(A)$ is the determinant of the matrix $A$ and $\tr(A)$ is the trace of the matrix […]
  • Nilpotent Matrices and Non-Singularity of Such MatricesNilpotent Matrices and Non-Singularity of Such Matrices Let $A$ be an $n \times n$ nilpotent matrix, that is, $A^m=O$ for some positive integer $m$, where $O$ is the $n \times n$ zero matrix. Prove that $A$ is a singular matrix and also prove that $I-A, I+A$ are both nonsingular matrices, where $I$ is the $n\times n$ identity […]
  • If Eigenvalues of a Matrix $A$ are Less than $1$, then Determinant of $I-A$ is PositiveIf Eigenvalues of a Matrix $A$ are Less than $1$, then Determinant of $I-A$ is Positive Let $A$ be an $n \times n$ matrix. Suppose that all the eigenvalues $\lambda$ of $A$ are real and satisfy $\lambda <1$. Then show that the determinant \[ \det(I-A) >0,\] where $I$ is the $n \times n$ identity matrix. We give two solutions. Solution 1. Let […]
  • How to Find Eigenvalues of a Specific Matrix.How to Find Eigenvalues of a Specific Matrix. Find all eigenvalues of the following $n \times n$ matrix. \[ A=\begin{bmatrix} 0 & 0 & \cdots & 0 &1 \\ 1 & 0 & \cdots & 0 & 0\\ 0 & 1 & \cdots & 0 &0\\ \vdots & \vdots & \ddots & \ddots & \vdots \\ 0 & […]
  • Rotation Matrix in Space and its Determinant and EigenvaluesRotation Matrix in Space and its Determinant and Eigenvalues For a real number $0\leq \theta \leq \pi$, we define the real $3\times 3$ matrix $A$ by \[A=\begin{bmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta &\cos\theta &0 \\ 0 & 0 & 1 \end{bmatrix}.\] (a) Find the determinant of the matrix $A$. (b) Show that $A$ is an […]
  • Determine a Matrix From Its EigenvalueDetermine a Matrix From Its Eigenvalue Let \[A=\begin{bmatrix} a & -1\\ 1& 4 \end{bmatrix}\] be a $2\times 2$ matrix, where $a$ is some real number. Suppose that the matrix $A$ has an eigenvalue $3$. (a) Determine the value of $a$. (b) Does the matrix $A$ have eigenvalues other than […]
  • Characteristic Polynomial, Eigenvalues, Diagonalization Problem (Princeton University Exam)Characteristic Polynomial, Eigenvalues, Diagonalization Problem (Princeton University Exam) Let \[\begin{bmatrix} 0 & 0 & 1 \\ 1 &0 &0 \\ 0 & 1 & 0 \end{bmatrix}.\] (a) Find the characteristic polynomial and all the eigenvalues (real and complex) of $A$. Is $A$ diagonalizable over the complex numbers? (b) Calculate $A^{2009}$. (Princeton University, […]
  • Similar Matrices Have the Same EigenvaluesSimilar Matrices Have the Same Eigenvalues Show that if $A$ and $B$ are similar matrices, then they have the same eigenvalues and their algebraic multiplicities are the same. Proof. We prove that $A$ and $B$ have the same characteristic polynomial. Then the result follows immediately since eigenvalues and algebraic […]

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

More in Linear Algebra
Ohio State University exam problems and solutions in mathematics
Linear Transformation and a Basis of the Vector Space $\R^3$

Let $T$ be a linear transformation from the vector space $\R^3$ to $\R^3$. Suppose that $k=3$ is the smallest positive...

Close