I know it might sound dumb, but specifically, Why NULL matrices are not the only NULPOTENT matrices? I am thinking that as all eigen values of NILPOTENT matrices are 0, then $\lambda = 0$, and as per Cayley-Hamilton theorem, when $\lambda =0$ then $A =0$ will also satisfy?
So instead of $A^n = 0$, because all eigen values are 0, thus, $A = 0$, also satisfies?
I know its wrong, but please point out where?
On
Nilpotent matrices aren't necessarily zero because matrices like $$ \begin{bmatrix}0&1\\0&0\end{bmatrix} $$ exist. Upper triangular matrices with zeroes on the diagonal are always nilpotent.
It is true that a diagonalizable nilpotent matrix must be the zero matrix, for reasons very similar to the eigen-argument you demonstrate. Such matrices are entirely determined by their eigenvectors and eigenvalues. However, non-diagonalizable matrices, like the one above, have too few eigenvectors to do this.
On
Suppose we have a $n \times n$ nilpotent matrix $A.$ Then its characteristics equation is $ \lambda ^n.$ Now, the minimal polynomial of the matrix $A$ could be $ \lambda ^k$ for any $ 1 \leq k \leq n$ ( the equation of the minimal polynomial depends on the matrix). The matrix is $0$ if $k =1$, and for the rest of the values of $k$ we will have a non-zero nilpotent matrix. For reference, you can take a look at Jordan normal form of a matrix.
As an example of non-zero nilpotent matrix, take any strictly upper triangular matrix. One such matrix is mentioned in some of the answers.
Cayley-Hamilton says that the characteristic polynomial $\chi_A$ of $A$ satistfies $\chi(A) = 0$. Since all eigenvalues of a nilpotent matrix are zero, we have : $$\chi_A (X) = \prod_{i} (X - \lambda_i) = X^n$$ Therefore, $A^n = 0$. You cannot "move from $\lambda$ to $A$" if $A$ is not diagonalizable.
If $A$ is diagonalizable, there is an invertible matrix $ P$ such that : $$ A = P \begin{pmatrix} \lambda_1 \\ & \ddots \\ && \lambda_n \end{pmatrix}P^{-1}$$ If $A $ is nilpotent, we have $\lambda_1 = \ldots = \lambda_n = 0$. Therefore $A= 0$.