Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$A$ is an $n\times n$ matrix of complex numbers. Prove that if $\lambda$ is an eigenvalue of $A^2,$ then $\sqrt{\lambda}$ or $-\sqrt{\lambda}$ is an eigenvalue of $A.$


If $\lambda$ is an eigenvalue of $A^2,$ we have $\lambda\alpha=A^2\alpha$ for some $\alpha.$ Then how can we find a $\beta$ s.t. $\sqrt{\lambda}\beta=A\beta?$

share|improve this question
    
If $\lambda$ is a complex eigenvalue then what is $\sqrt{\lambda}$ ? It is not well defined. –  jibounet Jul 27 at 9:56
    
$\lambda^{1/2}$ is a unique number even for $\lambda$ complex. I don't see any problem. @jibounet –  lovelesswang Jul 27 at 9:59
3  
@lovelesswang Not, it is not. The problem can still make sense, though. –  Git Gud Jul 27 at 10:04

4 Answers 4

up vote 7 down vote accepted

First note that: $$A^2 - \lambda I = (A-\sqrt\lambda I)(A+\sqrt\lambda I)$$ Let $v$ be an eigenvector of $A^2$ with eigenvalue $\lambda$. We can use $v$ to find an explicit eigenvector of $A$ with eigenvalue that is either $\sqrt\lambda$ or $-\sqrt\lambda$.

Since $(A^2-\lambda I)v = 0$, we must have either $(A+\sqrt\lambda I)v = 0$, in which case $v$ is also an eigenvector of $A$ with eigenvalue $-\sqrt\lambda$, or $(A+\sqrt\lambda I)v \neq 0$, in which case we set $ u:=(A+\sqrt\lambda I)v$, and note that $u$ is an eigenvector of $A$ with eigenvalue $\sqrt\lambda$, since $(A-\sqrt\lambda I)u=0$.

share|improve this answer

Since $\lambda$ is an eigenvalue of $A^2$, we know that $$\det (A^2 - \lambda I) = 0$$ From here we conclude that $$\det (A^2 - \lambda I) = \det((A - \sqrt{\lambda}I)(A + \sqrt{\lambda}I)) = \det(A - \sqrt{\lambda}I) \times\det ( A + \sqrt{\lambda}I)= 0$$ Hence $\sqrt{\lambda}$ or $-\sqrt{\lambda}$ is an eigenvalue of $A$.

share|improve this answer

Hint: use Jordan canonical form of A.

share|improve this answer

It follows from the Jordan decomposition that if $v_{\lambda^2}$ is an eigenvector of $A^2$ with eigenvalue $\lambda^2$, then it is either an eigenvector of $A$ corresponding to the eigenvalue $\lambda$ resp. $-\lambda$, or a combination $v_{\lambda}+v_{-\lambda}$. To see this, just square the Jordan block $$ \begin{pmatrix} \lambda & 1 & 0 & \ldots & 0\\ 0 & \lambda & 1 & \ldots & 0\\ & \ldots & & & & \\ 0 & \ldots & & 0 & \lambda \end{pmatrix} $$ to get a matrix with eigenvalue $\lambda^2$ and only one eigenvector $(1,0,\ldots, 0)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.