0568 #50

Forum for the GRE subject test in mathematics.
jayre
Posts: 11
Joined: Wed Nov 04, 2009 10:30 pm

0568 #50

Postby jayre » Fri Nov 06, 2009 2:24 pm

50. Let A be a real 2x2 matrix. Which of the following statements must be true?
I. All of the entries of A^2 are nonnegative.
II. The determinant of A^2 is nonnegative.
III. If A has two distinct eigenvalues, then A^2 has two distinct eigenvalues.
A) I only B) II only C) III only D) II and III only E) I, II, III

I guessed D, but the answer is B, only II must be true.

I wasn't able to prove or disprove any of them to myself except I, which is easy to think of a counterexample for. II seemed to be true, and since I couldn't disprove III I guessed that both were true. Is there any way short of thinking up a counterexample to show that III is false? And is there a way to show that II is true? I guessed that it was, but couldn't show it.

Thanks.

joey
Posts: 32
Joined: Fri Oct 16, 2009 3:53 pm

Re: 0568 #50

Postby joey » Fri Nov 06, 2009 3:28 pm

Determinants multiply nicely, so
\mbox{det}A^2=\mbox{det}A\cdot\mbox{det}A\ge 0
This proves II.

To see that III may be false, use the definition of an eigenvalue. If \lambda is an eigenvalue of A, then its corresponding eigenvector \bar{x} must satisfy
A\bar{x}=\lambda\bar{x}
A^2\bar{x}=A\left(\lambda\bar{x}\right)=\lambda^2\bar{x}
Hence, \lambda^2 is an eigenvalue of A^2. This gives us a hint: If the distinct eigenvalues of A are opposites, we might only get one eigenvalue for A^2. Let's try constructing a 2x2 matrix with eigenvalues 1 and -1. We'll just put those values on the diagonal of a triangular matrix:
A=\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)
Then A^2=I, which has only 1 as an eigenvalue.

jayre
Posts: 11
Joined: Wed Nov 04, 2009 10:30 pm

Re: 0568 #50

Postby jayre » Fri Nov 06, 2009 3:42 pm

Thanks, that's exactly what I was looking for. I feel a little dumb about II, that's so obvious... but I did feel like it should be true. Anyway, thanks for the explanation.

joey
Posts: 32
Joined: Fri Oct 16, 2009 3:53 pm

Re: 0568 #50

Postby joey » Fri Nov 06, 2009 4:06 pm

No problem. Good luck tomorrow! (Is it tomorrow for you? It is for me.)

jayre
Posts: 11
Joined: Wed Nov 04, 2009 10:30 pm

Re: 0568 #50

Postby jayre » Fri Nov 06, 2009 4:38 pm

yep, tomorrow it is.




Return to “Mathematics GRE Forum: The GRE Subject Test in Mathematics”



Who is online

Users browsing this forum: pumpkinator and 3 guests