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The index of a matrix $A\in \mathbb{R}^{n\times n}$ is the smallest nonnegative integer $k$ such that ${\rm rank}\ (A^{k+1})={\rm rank}\ (A^k)$.

I am looking for a singular matrix with index 1. Is it possible to construct such type of matrix? Also, it should be non-symmetric.

HK Lee
  • 19,964
Kavita Bisht
  • 61

2 Answers2

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Take a $2\times2$ matrix: $$A=\begin{pmatrix}a&&b\\c&&d\end{pmatrix}$$ For $X$ to be singular, you need that $\det(A)=0$, that is $ad-bc=0$. In that case, $rank(A)=1$ (assuming there is at least one nonzero element, of course). Moreover, $\det(A^2)=(bc-ad)^2=0$, which again reduces to $ad-bc=0$ and $rank(A^2)=1$.

Now you just need to try several values and solve for the rest of them. For instane, take $a=d=1$ and follow.

AugSB
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You can take a projector. For the case $n=2$ for example $$\begin{pmatrix}1&-1\\0&0\end{pmatrix}.$$