# 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?