If all eigenvalues are 1 or -1, is then $A^{12}=I$? True or false: If all the eigenvalues of A are either $\lambda=1$ or $\lambda = -1$ then $A^{12}$= I
If we have a matrix $$\mathbf A = \begin{pmatrix}1&0\\0&-1\end{pmatrix}$$ this has eigenvalues of 1 or -1 then $$\mathbf A = \begin{pmatrix}\lambda-1&0\\0&\lambda+1\end{pmatrix}$$ and if we plug in $\lambda=1 $ $$\mathbf A = \begin{pmatrix}0&0\\0&2\end{pmatrix}$$ So $A^{12}$ $\neq$ I  
Is this a correct way of proving this is false?
 A: Compute powers of the matrix $A$, where
$$A=\begin{pmatrix} 1 & 1\\0&1\end{pmatrix}.$$
You will find that 
$$A^n=\begin{pmatrix} 1 & n\\0&1\end{pmatrix}.$$
Remark: If a $2\times 2$ matrix $A$ has both $1$ and $-1$ as eigenvalues, then we will have $A^2=I$. But we can find a $3\times 3$ matrix that has both $1$ and $-1$ as eigenvalues such that $A^n\ne I$ for all positive integers $n$. 
A: The characteristic polynomial is $p(\lambda) = \lambda^2-1$. By Cayley-Hamilton theorem, we have $A^2-I = 0$. Hence, $A^{12} = I$.
A: In your example, $\mathbf A^2=\mathbf I$ (easy to check by hand) and therefore $\mathbf A^{12}=(\mathbf A^2)^6=\mathbf I^6=\mathbf I$. Andre Nicolas explains a correct reason that the answer is in fact "False".
Your method does not work because you were confusing two operations. In line 2 you replaced $\mathbf A$ with $\lambda \mathbf I - \mathbf A$ but this is what we do when trying to find the eigenvalues (we want the determinant of this matrix to be zero).
To compute large powers of a (diagonalizable) matrix, we instead diagonalize it $\mathbf A = \mathbf S \Lambda \mathbf S$ and then raise $\Lambda$ to the same power: $\mathbf A^n = \mathbf S \Lambda^n \mathbf S$.
A: EDIT: Use the decomposition in terms of eigenvalues. (Sorry i called this the "SVD" the first time, haven't used lin alg acronyms in a while). Since the eigenvalues are roots of 1, we know that $D^{12}=I$ because 12 is even. So $$A^{12}=SD^{12}S^{-1}=SIS^{-1}=SS^{-1}=I$$ as desired. 
However, you would have to check that the matrix is actually diagonalizable in all instances. I'm not sure if the question is assuming this or not.
