Can anyone please help me with the following questions:
Let $K $ be a commutative field and $ M_n(K)$ is the vector space on $ K$ of square matrices of order n with coefficients in $K.$
1: prove that any matrix $ M$ in $M_n(K)$ not having $ 0$ as eigenvalue is invertible and give the expression of it's inverse.
2: prove that a nilpotent matrix $ M$ in $ M_n(K) $ is diagonalizable if and only if it is equal to $ 0$ .
3: let $ M $ in $ M_n(K)$ such that $M^3 = I$ where $ I $ is the identity matrix in $ M_n(K)$ , prove that $ M $ is diagonalized
1: maybe it’s useful to know that the characteristc polynomial (I will call it $p_M$) evaluated in the given matrix gives back the zero matrix. So, let $p_M=x^n+a_{n-1}x^{n-1}+\cdots+ a_1 x + a_0$. Now,since $0$ isn’t an eigenvalue we know that $x$ does not divide this polynomial, so we can ensure that $a_0\ne0$. Now $\mathbf{O}=M^n+a_{n-1}M^{n-1}+\cdots+a_1 M + a_0 \mathbf{I}.$ We now have $\mathbf{I}={a_0}^{-1}(M^{n-1}+a_{n-1}M^{n-2}+\cdots+a_1\mathbf{I})M.$ So this point is proved.
2: a nilpotent matrix has only $0$ as eigenvalue (why?) so a diagonal matrix with all zeroes in the diagonal is the zero matrix.
3: as already pointed in the comments this point is not true if the field is not algebraically closed. For example you can’t diagonalize a generic 2x2 rotation matrix Over the reals, because it’s eigenvalues are complex numbers (check it).