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Let $A=\begin{bmatrix}0&1\\1&0\end{bmatrix}$. Find the formula for $A^n$ and prove it.

The way I tried to solve it is like this:

If we find $A^2$, $A^3$ and so on you will notice this patern: If $n$ is odd $A^n=\begin{bmatrix}0&1\\1&0\end{bmatrix}$. If $n$ is even then $A^n=I$.

To prove this I used mathematical induction. We see that the base case is $A^1$ and $A^2$. Now we suppose the statement holds for a natural number $n$. If $n+1$ is odd then this means that n is even and we see that the matrix $A^{n+1}=A^nA=\begin{bmatrix}0&1\\1&0\end{bmatrix}$, so the statement holds in this case. If $n+1$ is even then this means that $n$ is odd and we see that the matrix $A^{n+1}=A^nA=I$, so the statement holds in this case also. As a consequence the statement holds for all natural numbers.

I want to know if this proof is right or not? Can someone help me?

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This looks absolutely correct. Another way to view it is that matrix multiplication corresponds to function composition. In particular, this is the map $T:\mathbb R^2 \to \mathbb R^2$ given by $(x,y) \mapsto (y,x)$, which is a reflection about the line $y=x$, so $T \circ T=Id$, which in turn implies that $T \circ \dots\circ T=T^{2n+1}=T \circ T^{2n}=T$, and $T^{2n}=Id$

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Yes. Another approach is by diagonalising the (symmetric) matrix $A$ $$ A = V D V^T $$ where $D$ is the diagonal matrix of eigenvalues $(\pm1)$ and $V$ is the matrix formed from eigenvectors of $A$

Then $$ A^n = V D^n V^T $$ so it depends on $$D^n=\left( \begin{array}{cc}1 &0 \\ 0& -1 \end{array} \right)^n$$

A key thing is that to compute a diagonal matrix to a power you raise the diagonal elements to that power.

  • If n is even then $D^n=I$, so $A^n=V I V^T = I $ due to orthogonality of $V$.
  • If n is odd then $D^n=D$, so $A^n=V D V^T = A $
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