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Let $A$ be real square matrix. If $\det (A^2 - I) < 0$, then $A$ has an eigenvalue $\lambda \in (-1,1)$.

How to prove this?

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  • $\begingroup$ Have you tried anything? $\endgroup$ – DanZimm Jun 23 '14 at 12:02
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    $\begingroup$ Use the intermediate value theorem. And you should probably state that $A$ is a real matrix. $\endgroup$ – Daniel Fischer Jun 23 '14 at 12:03
  • $\begingroup$ @DanielFischer, well said, please consider making it into an answer. $\endgroup$ – Andreas Caranti Jun 23 '14 at 12:09
  • $\begingroup$ Can we assume $A$ to be symmetric? $\endgroup$ – user87543 Jun 23 '14 at 13:14
  • $\begingroup$ We can, but the problem applies to any real square matrix. $\endgroup$ – Yal dc Jun 23 '14 at 13:20
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Hint :

$\det(A^2-I)<0\Rightarrow \det(A+I)\cdot \det(A-I)<0$

Consider $f(x)=\det(A+xI)$ and consider case of $\det(A+I)>0$ and $\det(A-I)<0$

  • What is $f(1)$
  • What is $f(-1)$

Use Intermediate value theorem for this $f(x)$ and complete the rest...

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    $\begingroup$ This is really the best possible answer. $\endgroup$ – Rene Schipperus Jun 23 '14 at 12:11
  • $\begingroup$ @ReneSchipperus : Thank you so much... I feel happy only if OP understand this and complete the rest... $\endgroup$ – user87543 Jun 23 '14 at 12:12
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    $\begingroup$ The only thing I see from you decomposition is that eigenvalue cannot be $\pm 1$. $\endgroup$ – Yal dc Jun 23 '14 at 12:28
  • $\begingroup$ The same: 1 isn't eigenvalue of A. Sorry but nothing more. $\endgroup$ – Yal dc Jun 23 '14 at 12:42
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    $\begingroup$ @monalisa : Thank you :) $\endgroup$ – user87543 Jun 24 '14 at 8:18
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Hint: We have, if $\lambda_i$, $1 \le i \le n$ denote the eigenvalues of $A$, $$ \det(A^2 - 1) = \prod_{i=1}^n (\lambda_i^2 - 1) $$

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  • $\begingroup$ So what if some $\lambda_i$ are purely imaginary? (The characteristic polynomial of a real matrix need not split into real linear factors.) $\endgroup$ – Daniel Fischer Jun 23 '14 at 12:05
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    $\begingroup$ If $\lambda =\mu + i\nu$ is an eigenvalue of $A$, $\bar\lambda$ also is and the corresponding factors give $|\lambda_i^2 - 1|^2 \ge 0$ $\endgroup$ – martini Jun 23 '14 at 12:13
  • $\begingroup$ Let $n = 2$. $(\lambda^2_1 -1)(\lambda^2_2 -1) < 0 \Rightarrow$ one multiplier must be negative, the other must be positive $\Rightarrow \lambda^2_1 -1 < 0 \wedge \lambda^2_2 -1 > 0$. From the 1st inequality we get $|\lambda_1| < 1$ which is exactly what I need, but from the 2nd inequality: $\lambda_2 >1$ which is wrong. Did I make a mistake somewhere? $\endgroup$ – Yal dc Jun 23 '14 at 13:28

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