If $A=\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right)$, find the rank of $A+A^2$

Question

If $A=\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right)$, find the rank of $A+A^2$

Note: Ans is given as Rank=2.

$A(e_1) = e_3, A(e_2) = e_1, A(e_3) = e_2$. Thus $(A+A^2)(e_1) = e_3+e_2$, $(A+A^2)(e_2) = e_1+e_3$ and $(A+A^2)(e_3) = e_2+e_1$. Thus $A+A^2$ has rank .....??????.

Answer 1

Thus $A+A^2$ has rank $3$!

You have already noted that $A+A^2(e_1) = (e_3+e_2),A+A^2(e_2) = (e_3+e_1),A+A^2(e_3) = (e_1+e_2)$. To show that these three vectors are linearly independent, consider a linear combination of these vectors equal to zero: $$ c_1(e_3+e_2)+c_2(e_3+e_1) + c_3(e_1+e_2) = 0 \implies (c_3+c_2)e_1+(c_1+c_3)e_2+(c_1+c_2)e_3=0 $$ Hence, because $e_1,e_2,e_3$ are linearly independent, we get $c_1+c_2=c_2+c_3=c_3+c_1=0$, which on solving gives $c_1=c_2=c_3=0$. Hence the three image vectors are linearly independent, so the rank of the transformation $A+A^2$ is $3$.