Does there exists a non-diagonal $2 \times 2$ matrix $A$, such that $A^3 =I$

Question

Does there exist a $2 \times 2$ non-diagonal matrix $A$ such that $A^3 = I$ ?

Suppose it exists then $\lambda^3 = 1$ where $\lambda$ is the eigenvalue of $A$

Now a $2 \times 2$ matrix can have $2$ eigenvalues.

Those look like cube roots of unity $1,\omega , \omega ^ 2$.

Now, can I generate a $2 \times 2$ matrix whose eigen values are 1,1 and it's non diagonal?

I think yes $A = \begin{bmatrix} 1 & 1\\ 0 & 1\\ \end{bmatrix}$

But I see that cubing this does not give me $I$, even though it satisfies the determinant and trace conditions.

Is this approach correct?

Any interesting problem can be formulated from this?

Do you require matrix to have real entries, or do you allow complex entries? — Martin Sleziak, Oct 03 '17 at 09:17
Here are older questions about real matrices: How to construct a $2\times 2$ real matrix $A$ not equal to Identity such that $A^3=I$? and Is there any matrix $2\times 2$ such that $A\neq I$ but $ A^3=I$. Found using Approach0. — Martin Sleziak, Oct 03 '17 at 09:20
@MartinSleziak it would be interesting the cases of both real and complex entries,also still a question inside lurking around is "can some kind of interesting problem be thought from this problem?" :),Also can this be extended to higher order matrices?. — BAYMAX, Oct 03 '17 at 09:23
Can we find a nondiagonal but diagonalizable matrix of this kind? — Dimension Entangled, Aug 15 '21 at 15:04

score 8 · Accepted Answer · answered Oct 02 '17 at 06:08

8

Note that $(x-\omega)(x-\omega^2)=x^2+x+1$ so look for a matrix with characteristic function $x^2+x+1$, that is with trace $-1$ and determinant $1$, for instance $$\pmatrix{-1&1\\-1&0}.$$

answered Oct 02 '17 at 06:08

Angina Seng

158,341

But would that matrix cube become the unit matrix? I don't think so... – skyking Oct 02 '17 at 06:20
@skyking suck it and see $\ddot\smile$ – Angina Seng Oct 02 '17 at 06:22

score 5 · Answer 2 · answered Oct 02 '17 at 06:17

Of course, a rotation by $2\pi/3$ would have the property that $A^3=I$. The matrix for that is:

$$A=\begin{pmatrix} -1/2 & -\sqrt{3}/2 \\ \sqrt{3/2} & -1/2 \end{pmatrix}$$

As for eigenvalues this one has only complex eigenvalues that happen to cube up to $1$.

score 0 · Answer 3 · 2017-10-08T04:25:05.400

Note that $x^3-1=(x-1)(x^2+x+1)=(x-1)(x-j)(x-j^2)$ where $j=\exp(2i\pi/3)$; since $A$ is real, $spectrum(A)=\{1,1\}$ or $spectrum(A)=\{j,j^2\}$. Since $A$ is not diagonal, the first case does not work because $A$ should be similar to $B=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ which does not satisfy $B^3=I$.

Thus $(*)$ $spectrum(A)=\{j,j^2\}$ and, consequently $A^2+A+I=0$ -that implies $A^3=I$-. We conclude with the following gneralization.

EDIT 2. Considering the @Lord Shark the Unknown's comment , I rewrite my two propositions into one.

Proposition. Let $P(x)$ be a REAL polynomial of degree $p\geq 2$ and $q$ be an even integer $\geq 2$ (resp. of odd degree $p\geq 2$ and $q$ be an odd integer $\geq 2$); then there is a non-diagonal real $q\times q$ matrix $A$ s.t. $P(A)=0$.

Proof. We may write $P(x)=(x^2+ax+b)Q(x)$ where $x^2+ax+b$ and $Q$ are real polynomials. Let $U\in M_2(\mathbb{R})$ be the companion matrix of $x^2+ax+b$.

Case 1. $q=2r$, we choose $A=diag(A_1,\cdots,A_r)$ where $A_i=U$.

Note that when $P(x)=x^2+1$ and $q$ is odd, such a matrix $A\in M_q$ does not exist.

Case 2. $p$ and $q=2r+1$ are odd. Then $P$ admits a real root $u$ and we choose $A=diag(U_1,\cdots,U_r,u)$.

Another way of getting a non-diagonal matrix annihilated by a given polynomial is to take the companion matrix. — Angina Seng, Oct 02 '17 at 16:01
@ Lord Shark the Unknown. I wanted to explicitly use the eigenvalues but it is not a good idea; the method using the companion matrix is better because there is a sole case. — , Oct 02 '17 at 16:39

Does there exists a non-diagonal $2 \times 2$ matrix $A$, such that $A^3 =I$

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