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Let $A_n$ be square matrix where $n \geq 2$ and $A^2 +2A=0$. Then

  1. A is singular
  2. A is nonsingular
  3. 0 and -2 are eigenvalues of A
  4. either 0, or -2 is not an eigenvalue of A
  5. (1)-(4) are false.

My attempt: thanks to Cayley-Hamilton theorem, we know that characteristic polynomial of A equal zero. Replacing A with $\lambda$ gives $\lambda^2 + 2\lambda = 0$. Hence, 0 and -2 are eigenvalues of A, i.e. (3).

There are 2 problems: 1st, the answer should be (5), and 2nd problem is that assuming that (3) is true, (1) is also must be true due to 0 eigenvalue. So, I made mistake somewhere.

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The fact that $A^2 + 2A = 0$ implies that the minimal polynomial of $A$ divides $\lambda^2 + 2\lambda$. So the options for the minimal polynomial are $p_1(\lambda) = \lambda^2 + 2\lambda$, $p_2(\lambda) = \lambda$ and $p_3(\lambda) = \lambda + 2$. If the minimal polynomial of $A$ is $p_2$ then $A = 0$ (and so the only eigenvalue of $A$ is $0$) and if the minimal polynomial is $p_3$ then $A = -2I$ (and so the only eigenvalue of $A$ is $-2$).

If the minimal polynomial of $A$ is $p_1$ then since the minimal polynomial and the characteristic polynomial have the same roots it implies that the eigenvalues of $A$ are $0$ and $2$.

Using $A = -2I$, $A = 0$ and $A = \mathrm{diag}(0,-2)$, you see that the only possible answer is $5$.

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  • $\begingroup$ Aha, so Cayley-Hamilton theorem works only in one direction, not in the direction I used it? $\endgroup$ – Yal dc Jul 14 '14 at 7:53
  • $\begingroup$ Indeed. The Cayley-Hamilton theorem says that if you plug $A$ into the characteristic polynomial, you get $0$ (the zero matrix). If you have some polynomial $p$ such that $p(A) = 0$, it doesn't mean that $p$ is the characteristic polynomial. $\endgroup$ – levap Jul 14 '14 at 7:55
  • $\begingroup$ Thank you for such elaborate answer. $\endgroup$ – Yal dc Jul 14 '14 at 7:57
  • $\begingroup$ You're welcome. $\endgroup$ – levap Jul 14 '14 at 7:58
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$A=0$ rules out a couple of the answers. Can you find a non-singular matrix that satisfies $A^2+2A=0$?

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  • $\begingroup$ For example, $diag(-2,-2)$ $\endgroup$ – Yal dc Jul 14 '14 at 7:47
  • $\begingroup$ But what went wrong in my solution? $\endgroup$ – Yal dc Jul 14 '14 at 7:47

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