# Let $A$ be a $2 × 2$ matrix with real entries which is not a diagonal matrix and which satisﬁes $A^3 = I$. Pick out the true statements:

Let $A$ be a $2 \times 2$ matrix with real entries which is not a diagonal matrix and which satisﬁes $A^3 = \mathcal{I}_2$. Pick out the true statements:

• $\operatorname{tr}(A) = −1$
• $A$ is diagonalizable over $\mathbb{R}$
• $λ = 1$ is an eigenvalue of $A$

The characteristic polynomial of $A$ will divide the equation $x^3-1$ and it has degree $2$. So the characteristic polynomial will be $x^2+x+1$ which has two non real complex roots and sum of them is $-1$.
so only (a) is true.

Am I right?

• @ chopak : Can you please explain why The characteristic polynomial of A will divide the equation $x^3-1$? – Topology Jun 11 '14 at 11:19

## 1 Answer

I would say you're 90% right.

The only detail I would point out to you is about the fact the characteristic polynomial dividing $x^3 -1$. Perhaps you already know this, and decided it's not worth mentioning it, but I would explain it more slowly.

It's not true, in general, that the characteristic polynomial divides no matter what annhilating polynomial, but the minimal one does. So the minimal polinomial of $A$ could be either

$$\ x - 1 , \ x^2+x+1 , \ \text{or}\ x^3 - 1 \ .$$

Since it's a $2\times 2$ matrix, the last one is excluded. Also the first one, because they are telling us that $A$ it's not a diagonal matrix.

Hence the only possibility for the minimal polynomial is $x^2+x+1$. Since it divides the characteristic polynomial too and the later must have degree two, we conclude that indeed $x^2 + x + 1$ is the characteristic polynomial.

• Good answer +1. But there is no such thing as a percentage of being right. – Marc van Leeuwen Jan 7 '14 at 8:21