Once again, i can't even start to put ideas together to solve this problem, any hint or advise would be very helpful. NOTE: I don't ask necessarily for full solution, instead I ask for hints or advice to put my brain to work the problem. The problem says:
Let $A \in M_n \mathbb(C)$ be a normal matrix. Use the spectral decomposition $$A=E_1 \lambda_1 + E_2 \lambda_2 + \cdots + E_k \lambda_k$$ to prove:
a) If $q$ is a polynomial then $q(A)=\sum_{i=1}^{k} q(\lambda_i) E_i$.
b) If $A^n\equiv 0$ for some $n$, then $A \equiv 0$.
c) A matrix commute with $A$ iff commutes with each $E_i, i=1,2,...,k$.
d) $A$ has a normal square root, i.e., there exists $B \in M_n(\mathbb{C})$ normal such that $B^2=A$.
It is all straightforward.
For a), the spectral decomposition has the projections $E_j$ with the property that $E_j^*=E_j=E_j^2$ and $E_jE_h=0$ if $h\ne j$. Then $$\tag{1} A^2=\lambda_1^2E_1+\cdots+\lambda_n^2E_k $$ (all the products $E_jE_h$ are zero). Similarly, $$\tag{2}A^m=\lambda_1^mE_1+\cdots+\lambda_n^mE_k.$$ Taking linear combinations, we get $$\tag{3} q(A)=q(\lambda_1)E_1+\cdots+q(\lambda_n)E_k. $$
If $A^m=0$, then $0=\lambda_1^mE_1+\cdots+\lambda_n^mE_k$. Multiplying the equality by $E_1$, we get $0=\lambda_1^mE_1$, and so $\lambda_1=0$. Similarly, $\lambda_h=0$ for all $h$, and so $A=0$.
For c), it is useful to note that the decomposition assumes that $\lambda_j\ne\lambda_h$ if $h\ne k$. Then we can choose a polynomial $q$ with $q(\lambda_j)=1$ and $q(\lambda_h)=0$ for all $h\ne j$. Then $q(A)=E_j$. Now, if $BA=AB$, then $BA^2=A^2B$, and similarly $BA^m=A^mB$; taking linear combinations, $Bp(A)=p(A)B$ for all polynomials $p$. In particular, $BE_j=Bq(A)=q(A)B=E_jB$.
For d), just choose complex numbers $\mu_1,\ldots,\mu_k$ with $\mu_j^2=\lambda_j$.