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Suppose that the matrix $A$ is as follows: $$A=\begin{bmatrix} -3&4\\ 2&3\end{bmatrix}$$

We need to prove that $A^{2n + 1}=A$.

The way I tackled this problem is as follows:

  1. If $A^{2n + 1} =A$, then $A^{2n}$ must be the same as the identity matrix $I$.
  2. Thus $(A^2)^n$ must be the same as $I$.
  3. By calculation $A^2=17I$.
  4. Thus the statement can't be proved.

I'm not sure if I'm correct. I believe we have to make use of Cayley Hamilton's Theorem and diagonalization to solve the problem but I can't seem to wrap my head around it. Any help will be appreciated.

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I suspect that a sign is wrong in one of the entries of $A$ (changing any of the four signs results in $\det A=\pm1$ instead of $\det A = -17$) – Hagen von Eitzen Sep 2 '12 at 9:43
I think Hagen is right. Change for example $2$ to $-2$. You will see that your conclusions are indeed correct and it is provable. – Seyhmus Güngören Sep 2 '12 at 9:49
Are you sure you did't mistype the signs of one of the entries? As stated the matrix indeed does not have the stated property. But your reasoning is flawed: the matrix $\begin{pmatrix}-1&0\\0&0\end{pmatrix}$ does have the stated property, but fails your point 1. On the other hand, you may check that the property for $n=1$ implies the property for all $n>1$, which simplifies checking it. – Marc van Leeuwen Sep 2 '12 at 9:53
up vote 4 down vote accepted

If one considers the matrix $$ A=\left[\begin{array}{cc}-3&4s_1\cr2s_2&3\end{array}\right], $$ where $s_1,s_2 \in \{-1,1\}$ with $s_1s_2=-1$, then the characteristic polynomial $p_A$ of $A$ is given by $$ p_A(\lambda)=\lambda^2-\text{trace}(A)\lambda+\det(A)=\lambda^2-1. $$ It follows from Cayley Hamilton's Theorem that $p_A(A)=0$, i.e. $$ A^2=I. $$ Hence $A^{2n}=I$ and $A^{2n+1}=A$ for every $n \in \mathbb{Z}$.

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EDIT: As @Hagen states in the comment on the question, that perhaps one of the signs is wrong in the matrix $A$. This working follows if that is true. If not, then this answer can be discarded.

The characteristic polynomial of $A$ is $$f(\lambda)=\lambda^2-1=(\lambda+1)(\lambda-1).$$ You can show that this matrix is diagonisable (you ahould check this), so that we can write $A=VDV^{-1}$ and $g(A)=Vg(D)V^{-1}$, where here $g(x)=x^{2n+1}$.

Since $D$ is a diagonal matrix with entries of $1$ and $-1$, then $D^{2n+1}=D$ with $n\in\mathbb{Z}$.

Thus, $$A^{2n+1}=VD^{2n+1}V^{-1}=VDV^{-1}=A$$ when $n$ is an integer.

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The entries of the matrix as given are false. Change the following $(A)_{2,1}$=-2 instead of 2 and it works. Solve simply by induction on $n$, it is very easy then.

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Show $A^3=A$ by direct computation.

Then proceed by induction on $n$: $$A^{2n+1}= A^{2(n-1)} A^3 = A^{2(n-1)} A = A^{2(n-1)+1} = A$$

No need for Cayley-Hamilton.

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