So I'm currently taking my first linear algebra course, and recently we've been working with inverses, determinants and cofactor expansion, and diagonalization.
I haven't had any trouble working out computational problems, but some of the conceptual exercises on my diagonalization homework have been giving me some trouble. For reference, I've included problems below:
3.3.13. If A is diagonalizable and 1 and -1 are the only eigenvalues, show that (A^-1) = A.
3.3.14. If A is diagonalizable and 0 and 1 are the only eigenvalues, show that (A^2) = A.
I was able to figure out 3.3.14 fairly easily after realizing: $$ D = diag(0,1) $$ $$ D^2 = diag(0^2, 1^2) = diag(0,1) = D $$ $$ A = PDP^{-1} \Rightarrow A^2 = PD^2P^{-1} = PDP^{-1} = A$$
However, I was unable to solve 3.3.13 without doing some really messy math paired with trial-and-error, which is obviously insufficient for proofs. My feeling is that there must be some implications for a matrix A based on its eigenvalues, namely for values like -1, 0, and 1.
Is this true? And if so, how could I use these implications to work out 3.3.13?
If A is a diagonalizable matrix, whose only eigenvalues are $\pm1$, then there is an invertible matrix $P$ such that $$A=PDP^{-1}$$ where $$D=\text{diag}(d_1,d_2\dots,d_n)$$ and $d_k=\pm1,\ k=1,2,\dots,n$. Then $$D^2=\text{diag}(d_1^2,d_2^2\dots,d_n^2)=I$$ so that $$A^2 =PD^2P^{-1}=PIP^{-1}=I$$ so $$A^{-1}=A$$