I understand how to do it if the question changed $A^3$ to $A^2$, because then you can just use the rank–nullity theorem. $\text{Rank}+\text{Nullity}=6$, $\text{Rank}=4$ so $\text{Nullity}=2$ so of the $6$ column vectors $v_1,v_2,\dots,v_6,$ only two of them satisfy $Av_i=0$. I don't know what to do with $A^3$ however because I'm not sure what nullity says about $A^2v_i$.
2026-04-06 19:31:52.1775503912
True or False: There is a $6\times 6$ matrix $A$ with $\text{Rank}(A)=4$ and $A^3 =0$
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If $A = \begin{bmatrix} 0 && 0 && 1 && 0 && 0 && 0 \\ 0 && 0 && 0 && 1 && 0 && 0 \\ 0 && 0 && 0 && 0 && 1 && 0 \\ 0 && 0 && 0 && 0 && 0 && 1 \\ 0 && 0 && 0 && 0 && 0 && 0 \\ 0 && 0 && 0 && 0 && 0 && 0 \end{bmatrix}$
Then
$A^2 = \begin{bmatrix} 0 && 0 && 0 && 0 && 1 && 0 \\ 0 && 0 && 0 && 0 && 0 && 1 \\ 0 && 0 && 0 && 0 && 0 && 0 \\ 0 && 0 && 0 && 0 && 0 && 0 \\ 0 && 0 && 0 && 0 && 0 && 0 \\ 0 && 0 && 0 && 0 && 0 && 0 \end{bmatrix}$
And
$A^3 = \begin{bmatrix} 0 && 0 && 0 && 0 && 0 && 0 \\ 0 && 0 && 0 && 0 && 0 && 0 \\ 0 && 0 && 0 && 0 && 0 && 0 \\ 0 && 0 && 0 && 0 && 0 && 0 \\ 0 && 0 && 0 && 0 && 0 && 0 \\ 0 && 0 && 0 && 0 && 0 && 0 \end{bmatrix} $