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Let $A=(a_{ij})$ be the $10\times10$ matrix in which $a_{ij}=\begin{cases}1 &\text{if}\quad j=i+1 \\ 0 &\text{otherwise}\end{cases}$, then the least integer $k$ such that $A^k=0$ is ................

I tried to use the property that a maximum rank of a nilpotent matrix is $n/2$ where $n$ is the index. But can't get the answer. Please help. Please also note that I don't know anything about Sylvester inequality so please don't include that in your answer...

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    Have you tried for smaller matrices? Like 2x2 and 3x3? – 2017-01-17
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    Have you looked at smaller examples and tried multiplying them out to look at $A^2$ and $A^3$? Can you make a conjecture as to the pattern? Can you prove your conjecture? E.g. if this was a $4\times 4$ matrix instead, it would be $A=\left[\begin{smallmatrix}0&1&0&0\\0&0&1&0\\0&0&0&1\\0&0&0&0\end{smallmatrix}\right]$. What does $A^2$ look like? $A^3$? $A^4$? – 2017-01-17
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    Please do not use pictures for critical portions of your post. Pictures may not be legible, cannot be searched and are not view-able to some, such as those who use screen readers. I have [edit]ed your question to reflect this principle. – 2017-12-21

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