# Find a example of $A$ be $4 \times 4$ matrix such that $A$ has rank $2$ but $A^2 =0$?

Find a example of $$A$$ be $$4 \times 4$$ matrix such that $$A$$ has rank $$2$$ but $$A^2 =0$$?

My attempt :

$$A=\begin{bmatrix} 0 & 0 & 1 &0\\0 & 0 & 1 & 0\\0 &0 &0 &0 \\0 &0 &0 &0 \\\end{bmatrix}$$

Is it correct ??

Any hints/solution will be apprecaited

thanks u

• Your example has rank $1$. You might be better off considering the Jordan canonical form. – EuYu Dec 8 '18 at 9:28
• Along the right lines, but looks like rank $1$ to me ... – Mark Bennet Dec 8 '18 at 9:28
• Ya ,that is my misunderstanding @ Mark and @EuYu – jasmine Dec 8 '18 at 9:29
• Your question is unclear. Do you mean to find an example (rather than a counterexample) of a 4-by-4 matrix $A$ such that $A$ has rank 2 but $A^2=0$? – user1551 Dec 8 '18 at 9:38
• One way of thinking about the rank is as the dimension of the image space of a linear map. The image here will be a subspace $W$ of four dimensional space $V$ - because you want $A^2=0$ you end up applying $A$ to $W$ and getting zero. (I'm using terms loosely here). Rank $2$ means that the rank-nullity theorem tells you that $W$ has dimension $2$. Pick a convenient two dimensional subspace $W$ to go to zero, and then make sure you map everything in $V$ to $W$ and that both basis vectors in $W$ appear in the image of $V$. It is this last bit which didn't work for your first attempt. – Mark Bennet Dec 8 '18 at 9:42

Your first 2 lines are not independent thus the rank is 1 and not 2. Look at $$A=\begin{bmatrix} 0 & 0 & 1 &0\\0 & 0 & 0 & 0\\0 &0 &0 &0 \\0 &1 &0 &0 \\\end{bmatrix}$$