Which one of the following are correct? 
Let, $A=   \left[ {\begin{array}{cc}    -1 & 2 \\    0 & -1 \\  
 \end{array} } \right]$ , and $B = A + A^2 + A^3 +···+ A^{50}$. Then
$(A) B^2 = I $
$(B) B^2 = 0$ 
$(C) B^2 = A$ 
$(D) B^2 = B$

Eigenvalues of $A$ are $-1,-1$. So, Eigenvalues of $B$ are $0,0$. So, $\det(B)=0$. So, options (A) and (C) can be eliminated. How do I eliminate further?. By Cayley-Hamilton Theorem $A^2+2A+I=0$. I found $A^3,A^4,...$ using Cayley-Hamilton Theorem. It was time-consuming. Given matrix is not diagonalizable. Can you please help me?
 A: The sum and product of upper-triangular matrices is again upper triangular. Therefore, $B$ is upper triangular, and since its diagonal entries are $0$, we have $$B=\begin{bmatrix}0&c\\ 0&0\end{bmatrix}.$$ So $B^{2}=0$. 
A: Let us begin by finding the remainder when the polynomial $x + \cdots + x^{50}$ is divided by $(x+1)^2$.  So, suppose
$$x + \cdots + x^{50} = (x+1)^2 q(x) + (ax + b).$$
Then by substituting $x := -1$, we see that $-a + b = 0$.  Now, if we take the derivative on both sides of the above equation, we see
$$1 + 2x + \cdots + 50 x^{49} = 2 (x+1) q(x) + (x+1)^2 q'(x) + a.$$
Substituting $x := -1$ in this equation gives $a = 1 - 2 + \cdots - 50 = -25$.  Thus, $a = b = -25$.
Now, if we substitute $A$ for $x$ in the original equation, we conclude
$$B = A + \cdots + A^{50} = (A+I)^2 q(A) - 25 (A + I) = \begin{bmatrix} 0 & -50 \\ 0 & 0 \end{bmatrix}.$$
Here, we use that $(A + I)^2 = 0$ to eliminate the $(A + I)^2 q(A)$ term.  From here, it should be easy to answer the original question.

(Or, if you want a briefer way to eliminate the possibility that $B = 0$: essentially, what we are doing here is showing that $x + \cdots + x^{50}$ is not divisible by $(x+1)^2$ by showing that the polynomial's derivative at $x := -1$ is not zero.  Therefore, since $(\lambda+1)^2$ is the minimal polynomial of $A$, it follows that applying this polynomial to $A$ gives a nonzero matrix.  On the other hand, since the polynomial is divisible by $x+1$, its square is divisible by $(x+1)^2$, which implies that $B^2 = 0$.)
A: Since $A-I$ is triangular with nonzero diagonal entries, it follows that $A-I$ has nonzero determinant, so is invertible, hence $I-A$ is also invertible.

Note also that any two poynomials in $A$ commute.

\begin{align*}
\text{Then}\;\;(I-A)B&=(I-A)(A+A^2+A^3+\cdots+A^{50})\\[4pt]
&=(A + A^2 + A^3 +···+ A^{50})-(A^2 + A^2 + A^3 +···+ A^{51})\\[4pt]
&=A-A^{51}\\[4pt]
&=(I-A^{50})A\\[4pt]
&=(I+A^{25})(I-A^{25})A\\[4pt]
&=(I+A)(1-A+A^2-\cdots+A^{24})(I-A^{25})A\\[4pt]
&=(I+A)C,\;\text{where}\;C=(1-A+A^2-\cdots+A^{24})(I-A^{25})A\\[4pt]
\implies\;\bigl((A-I)B\bigr)^2&=\bigl((I+A)C\bigr)^2\\[4pt]
\implies\;(A-I)^2B^2&=(I+A)^2C^2\qquad\text{[since $B,C$ commute with all polynomials in $A$]}\\[4pt]
\implies\;(A-I)^2B^2&=0\qquad\text{[since $(I+A)^2=A^2+2A+I=0$]}\\[4pt]
\implies\;B^2&=0\qquad\text{[since $(I-A)$ is invertible]}\\[4pt]
\end{align*}
A: The Jordan normal form of $A$ is:
$$A = PJP^{-1}$$
where $J = \begin{bmatrix}-1 & 1 \\ 0 & -1\end{bmatrix}$ and $P = \begin{bmatrix}1 & 0 \\ 0 & \frac12\end{bmatrix}$.
For any entire function $f$ we have
$$f(A) = f(PJP^{-1}) = Pf(J)P^{-1} = P\begin{bmatrix}f(-1) & f'(-1) \\ 0 & f(-1)\end{bmatrix}P^{-1} $$
In particular, since $B = A + A^2 + \cdots + A^{50}$ we have
$$f(z)  = z + z^2 + \cdots + z^{50} = z \cdot \frac{x^{50}-1}{z-1}$$
$$f'(z) = \frac{51z^{50} - 1}{z-1} - \frac{z(x^{50}-1)}{(z-1)^2}$$
hence $f(-1) = 0$ and $f'(-1) = -25$.
$$B = f(A) = P\begin{bmatrix}0 & -25 \\ 0 & 0\end{bmatrix}P^{-1} = \begin{bmatrix}0 & -50 \\ 0 & 0\end{bmatrix}$$
Therefore $B^2 = 0$.
