matrices and eigenvalue

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

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

Try

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

• the eigen value of A will be 0 and 0 is a real number – matheu96 Jul 13 '16 at 19:58
• 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. – Christian Jul 13 '16 at 20:04
• @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. – DonAntonio Jul 13 '16 at 21:06
• can you please explain how you got the matrix? thanks – matheu96 Jul 13 '16 at 22:44
• @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. – DonAntonio Jul 14 '16 at 9:55

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 $180^\circ$?

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