-2
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can you explain the question to me? thanks

Question : Give a 2 x 2 matrix matrix A such that A has no real eigenvalues $A^2$ has an eigenvalue of -1 with algebric and geometric multiplicity of 2

i dont know where to start

thanks

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  • $\begingroup$ is the eigenvalue of A (your example ) not equal to 0? $\endgroup$ – matheu96 Jul 13 '16 at 20:02
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Try

$$A=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$$

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  • $\begingroup$ the eigen value of A will be 0 and 0 is a real number $\endgroup$ – matheu96 Jul 13 '16 at 19:58
  • $\begingroup$ Are the eigenvalues of $A$ 0? What is the null space of $A-0I = A$? Or - Consider $A- \lambda I$ and solve for what $\lambda$ makes the null space nontrivial. You shouldn't get 0. $\endgroup$ – Christian Jul 13 '16 at 20:04
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    $\begingroup$ @matheu96 Check carefully before downvoting: the characteristic polynomial of $\;A\;$ is $\;x^2+1\;$ . Not only this, but zero is an eigenvalue of a matrix iff the matrix is singular, and this one is not singular. $\endgroup$ – DonAntonio Jul 13 '16 at 21:06
  • $\begingroup$ can you please explain how you got the matrix? thanks $\endgroup$ – matheu96 Jul 13 '16 at 22:44
  • $\begingroup$ @matheu96 Thinking of the easiest example of non-real complex number ($i$) which squared ($-1$) is already real. Figuring out a $\;2\times2\;$ matrix whose char. polynomial is $\;x^2+1\;$ was pretty easy. $\endgroup$ – DonAntonio Jul 14 '16 at 9:55
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If $A^2$ has eigenvalue $-1$ with geometric multiplicity $2$, then we must have $A^2 = -I$ (where $I$ is the identity matrix).

That is, $A^2$ yields the rotation by $180^\circ$. Can you think of a linear transformation that, when applied twice, results in a rotation by $180^\circ$?

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  • $\begingroup$ i got this transformation: $\endgroup$ – matheu96 Jul 14 '16 at 0:19
  • $\begingroup$ What transformation? $\endgroup$ – Omnomnomnom Jul 14 '16 at 0:48
  • $\begingroup$ i got A = \begin{bmatrix}cosx&-sinx\\sinx&cosx\end{bmatrix} is that correct? $\endgroup$ – matheu96 Jul 14 '16 at 1:00
  • $\begingroup$ But for what value of $x$? Not every $x$ will work here. $\endgroup$ – Omnomnomnom Jul 14 '16 at 3:45
  • $\begingroup$ for x = 90 degree is that correct? $\endgroup$ – matheu96 Jul 14 '16 at 5:12

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