Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
4  
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
1  
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

4 Answers 4

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 $n \in \mathbb{Z}$.

share|improve this answer

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.

share|improve this answer

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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.