linear algebra – Determining the sum of a left circulating and a nonsingular matrix on a 2-order field

For any matrix $ A = (a_ {ij}) in M_n ( mathbb {F} _2) $, denote $ bar {A} = ( bar {a_ {ij}}, $ or $ bar {a} = $ 0 if $ a = $ 1 and $ bar {a} = $ 1 if $ a = $ 0. assume $ n $ is a power of two. Let $ A $ (circulating left) and $ I $ be $ n times n $ matrices on $ mathbb {F} _2 $ Defined by,
$$ X = begin {pmatrix}
x_1 & x_2 & x_3 & ldots & x_ {n-1} & x_n \
x_2 & x_3 & x_4 & ldots & x_ {n} & x_1 \
x_3 & x_4 & x_5 & ldots & x_ {1} & x_2 \ & & & vdots \
x_ {n-1} & x_n & x_1 & ldots & x_ {n-3} & x_ {n-2} \
x_n & x_1 & x_2 & ldots & x_ {n-2} & x_ {n-1}
end {pmatrix}, I'm
begin {pmatrix}
0 & 0 & 0 & ldots & 0 & 0 \
0 & 0 & 0 & ldots & 0 & 1 \
0 & 0 & 0 & ldots & 1 & 1 \ & & vdots \
0 & 0 & 1 & ldots & 1 & 1 \
0 & 1 & 1 & ldots & 1 & 1
end {pmatrix}. $$

Then, it seems that neither $ X + I $ or $ X + bar {I} is invertible. But I can not prove it. Is there a simple way to prove it?