If $A^3 =0$, what is the determinant of $(A-3I)$? From $A^3 = 0$, I can conclude that A must be singular. But how to proceed from there?
 A: Since $A^3=0$, $A$ is a nilpotent matrix and all its eigenvalues are $0$. Now we can see that eigenvalues of $A-3I$ are $-3,-3,\ldots, -3$ ($n-$times). For, let $X$ be an eigenvector of $A$ such that $AX=0$. Then $(A-3I)X=AX-3IX=-3X$. Therefore $-3$ is an eigenvalue of $A-3I$. Since Det$(A-3I)=$product of eigenvalues, Det$(A-3I)=(-3)^n$.
Here one thing that should be noticed that the algebraic multiplicity of the eigenvalue $-3$ is $n$. Using the Jordan canonical form you can see this.   
A: You do not need to assume you are working over $\mathbb{C}$; you also do not need the Jordan canonical form (though such a matrix would have one over any field, since the minimal polynomial splits).
Since $A^3=0$, that means that $A$ satisfies the polynomial $t^3$. This means the minimal polynomial of $A$ divides $t^3$. Thus, the only irreducible factor of the minimal polynomial, regardless of the field you are working over, is $t$.
Since every irreducible factor of the characteristic polynomial is an irreducible factor of the minimal polynomial (over any field), then the characteristic polynomial of $A$ is necessarily equal to $(-1)^nt^n$, where $n$ is the size of $A$ (that is, $A$ is an $n\times n$ matrix). Thus, the eigenvalue $0$ has algebraic multiplicity $n$.
Now, if a matrix $B$ has eigenvalues $\lambda_1,\ldots,\lambda_n$, repeated with algebraic multiplicity, then $\alpha B + \beta I$ has eigenvalues $\alpha\lambda_1+\beta$, $\alpha\lambda_2+\beta,\ldots,\alpha\lambda_n+\beta$. 
Thus, the eigenvalue of $A-3I$ is $-3$, with algebraic multiplicity $n$.
Now, the determinant of a matrix is the constant term of its characteristic polynomial (which is $\det(A-tI)$). This would be the product of all its eigenvalues (with algebraic multiplicity) over an algebraic closure of the field of definition; but since here we already have that the characteristic polynomial is $(-1)^n(t+3)^n$, it follows that the determinant of $A-3I$ is $(-1)^n3^n$. 
No Jordan forms needed. No complex field assumption needed.
