Proof of : Show that for Any Upper-Diagonal Matrix $A$ n $\times$ n with its diagonal coefficients equal to 0, $A^n = 0$ 
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*I want to show the following: For any square  matrix $A$ n $\times$ n of the form: $A=\begin{pmatrix} 
0 & a_{1,2} & . & . & . & a_{1,n} \\ 
0 & 0 & a_{2,3} & . & . & a_{2,n} \\ 
0 & 0 & 0 & . & .  & .\\
. & . & . & . & . & . \\
. & . & . & . & . & a_{n-1,n} \\
. & . & . & . & . & 0  
\end{pmatrix}$,


$A^n=0$
In order to do that, I want to show that if you multiply such a matrix b times. All the coefficients $(a)_{ij}$ such that $i<i+b-1$is equal to 0


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*My attempt
I am trying to show that $\forall b, \leq n$ $(A^b)_{i,i+b-1}= 0$
I will try to show this by induction


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*basis Let's show that $(A^2)_{i,i+1}=0$
By definition, $(A^2)_{i,i+1} = \Sigma^n_{k=0}a_{i,k}a_{k,i+1}$


For $i \geq k$, $a_{i,k}=à \implies a_{i,k}a_{k,i+1} = 0$
For $i=k-1$, $i+1=k \implies i+1 \geq k \implies a_{k,i+1} = 0 \implies a_{i,k}a_{k,i+1} = 0$ And therefore, for $i < k, a_{i,k}a_{k,i+1}=0$
Therefore, $(A^2)_{i,i+}=0$
We now have a matrix of the form: $A^2 = \begin{pmatrix} 
0 & 0 & a_{1,3} & . & . & a_{1,n} \\ 
0 & 0 & 0 & a_{2,4} & . & a_{2,n} \\ 
0 & 0 & 0 & . & .  & .\\
. & . & . & . & . & a_{n,n-2} \\
. & . & . & . & . & 0 \\
. & . & . & . & . & 0  
\end{pmatrix}$


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*Inductive step And here is where the struggle starts (I cannot finish this step).


Let $(A^b)_{i,i+b-1}$ be TRUE, let's show that it implies that $(A^{b+1})_{i,i+b}$ is TRUE:
$(A^b)_{i,i+b-1}=0$
$\implies$ $(A)_{i,i+b-1}(A^b)_{i,i+b-1}=0$
$\implies \Sigma^n_{k=0} a_{i,i+b-1}(a^b)_{i,i+b-1}= 0$.
EDIT: I think I have solved my problem, can you guys confirm
We now have a multiplication of the form: $A \times A^b =\begin{pmatrix} 
0 & a_{1,2} & . & . & . & . & . & . & . & a_{1,n} \\ 
0 & 0 & a_{2,3} & . & . & . & . & . & . & a_{2,n} \\ 
0 & 0 & 0 & . & .  & . & . & . & . & .\\
. & . & . & . & . & . & . & . & . & . \\
. & . & . & . & . & . & . & . & . & a_{n-1,n} \\
. & . & . & . & . & . & . & . & . & . \\
. & . & . & . & . & . & . & . & . & . \\
. & . & . & . & . & . & . & . & . & . \\
. & . & . & . & . & . & . & . & . & . \\
. & . & . & . & . & . & . & . & . & 0 \\
\end{pmatrix} \times  \begin{pmatrix} 
0 & 0 & . & 0 & a_{1,b+1} & a_{1,b+2} & . & . & . & a_{1,n} \\ 
0 & 0 & 0 & . & 0 & a_{2,b+2} & a_{2,b+3} & . & . & a_{2,n} \\ 
0 & 0 & 0 & . & .  & 0 & . & . & . & .\\
. & . & . & . & . & . & . & . & . & . \\
. & . & . & . & . & . & . & . & . & a_{n-b,n} \\
. & . & . & . & . & . & . & . & . & 0 \\
. & . & . & . & . & . & . & . & . & . \\
. & . & . & . & . & . & . & . & . & . \\
. & . & . & . & . & . & . & . & . & . \\
0 & . & . & . & . & . & . & . & . & 0 \\ 
\end{pmatrix}$
Let's call the first matrix $N$ and the second matrix $M$.
Let's show that $(NM)_{i,i+b} = 0$:
$(NM)_{i,i+b} = \Sigma_{k=0}^n n_{i,k}m_{k,i+b} = (0 \times \Sigma_{k=0}^i n_{k,i+b}) + (\Sigma_{k=n-i}^n a_{i,k} \times 0) = 0$
Therefore $(NM)_{i,i+b} = (AA^b)_{i,i+b} = (A^{b+1})_{i,i+b} = 0$
I think I have succeeded to prove by induction that $(A^b)_{i,i+b-1} = 0$ But I am not certain
Can someone confirm that my proof by induction is right?
 A: Since $A$ is an upper triangular matrix, it's diagonal entries are its eigenvalues. However, all of the diagonal entries are zero thus all of its eigenvalues are zero as well. Hence the characteristic polynomial is:
$$p_A(\lambda)=\prod_{i=1}^{n}{(\lambda-\lambda_i)}=\lambda^n$$
But the Cayley-Hamilton Theorem states that any square matrix satisfies its characteristic polynomial, and so $A^n=0$.
EDIT (Alternative Method)
Let $\mathbf{e_1},\mathbf{e_2}, ..., \mathbf{e_n}$ be the standard basis on ${\mathbb{R}^n}$
Note that $A\mathbf{e_1}=0$ and that $A\mathbf{e_i}=0$ is simply a linear combination of the previous $i-1$ basis vectors for $i\in\{2,...,n\}$, and so it is now easy to show by induction on $i$ that $A^n\mathbf{e_i}=0$ for all $\mathbf{e_i}$and hence $A^n=0$.
A: Show that in the 3rd power, the next diagonal from below will vanish. This is much easier to prove inductively. Then $A^n$ will have all diagonals zero, so zero all over.
Raising to successive powers sweeps up the matrix vanishing it down to zero.
