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.

## 0568 #50

### Re: 0568 #50

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.

$$\mbox{det}A^2=\mbox{det}A\cdot\mbox{det}A\ge 0$$

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.

### Re: 0568 #50

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.

### Re: 0568 #50

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