Show that if A is diagonalizable, then sin^2(A) + cos^2(A) = I. Does this identity also hold for nondiagonalizable matrices?

Question

Show that if A is diagonalizable, then $\sin^2(A)+\cos^2(A)=I$. Does this identity also hold for nondiagonalizable matrices? This is what I got so far: $$ e^{iA}= \cos A +i\sin A \\ \cos A= \frac{e^{iA}+e^{-iA}}{2} \\ \sin A= \frac{ie^{-iA}-ie^{iA}}{2} \\ \cos^2(A) + \sin^2(A)= I $$ What should I write to complete this proof? How do I show that this identity hold in general(i.e nondiagonalizable matrices) ? Please help

Answer 1

You are essentially there (and without using diagonalisability), provided you know the expressions for $\cos A$ and $\sin A$ are indeed generally valid for square matrices $A$ (for instance they might be the definitions). I would avoid writing down the conclusion immediately, as it may raise the suspicion that you did so without having an argument. To show you do, just write $$ \cos(A)^2=\frac{e^{2iA}+2I+e^{-2iA}}4 \quad\text{and}\quad \sin(A)^2=\frac{i^2e^{-2iA}-2i^2I+(-i)^2e^{2iA}}4 $$ so that $\cos(A)^2+\sin(A)^2=\frac{2I--2I}4=I$.