If $A=\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right)$, find the rank of $A+A^2+A^3$

Question

If $A=\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right)$, find the rank of $A+A^2+A^3$

We have $\rho(A+A^2+A^3)\leq\rho(A)+\rho(A^2)+\rho(A^3) $ where $\rho(A)$ means rank of A.

Here $\rho(A)=3.$ Also $\rho(A^2)=3$ and $\rho(A^3)=3$ as $\det A\neq 0$ and $\det(A^2)=\det(A)\det(A)\neq 0$ and so on.

But how to get the desired result.

Answer 1

$A(e_1) = e_3, A(e_2) = e_1, A(e_3) = e_2$. Thus $(A+A^2+A^3)(e_1) = e_3+e_2+e_1$, $(A+A^2+A^3)(e_2) = e_1+e_3+e_2$ and $(A+A^2+A^3)(e_3) = e_2+e_1+e_3$. Thus $A+A^2+A^3$ has rank 1.

Answer 2

Don't worry about inequalities. Just compute $A+A^2+A^3$, then consider the rank of the resulting matrix.

Answer 3

$A$ is a permutation matrix, meaning $A^2$ and $A^3$ are also permutation matrices. This gives you an easy way to find $A+A^2+A^3$ and from there, it's easily seen that rank is $1$.

EDIT To elaborate, a permutation matrix $P$ is a matrix that acts upon another matrix by permuting it's rows. Permutation matrix is just an identity matrix with permuted rows (In your case, $A$ is an identity matrix with second row shifted up, first row shifted 2 positions down and third row in the "middle"). This means that when you multiply $B$ by $A$ from the left, it will just reorder rows of $B$ ( same order as they are replaced in $A$ with respect to identity matrix ).

In your case, you multiply $A$ by this permutation matrix, meaning you will just get $A$ with permuted rows. This means that:

$$A^2 = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} $$