Ytukyg

For k $ge2$ we recursively define $A_{2^k}$ as $begin{bmatrix} A_{2^{k-1}} & A_{2^{k-1}} \ A_{2^{k-1}} & -A_{2^{k-1}} end{bmatrix}$ and $A_1=[1]$

The problem is to show that $|det(A_n)|=n^{n/2}$

My attempt: we do an induction on $k$

$|det(A_2)|=2=2^{2/2}$. Induction hypothesis: $|det(A_{n})|=n^{n/2}$ and we want to show that $|det(A_{2n})|=(2n)^n$

using block matrix properties

$|det(A_{2n})|=|det(begin{bmatrix} A_{n} & A_{n} \ A_{n} & -A_{n} end{bmatrix})|=|det(-A)det(A+AA^{-1}A)|=|2^ndet(A_n)^2|=|2^nn^n|=(2n)^n$

Can somone confirm that there are no flaws in the reasoning please?

Your proof looks good.

Alternately, for the induction-step, notice that
$$
begin{vmatrix}
A_n & A_n \
A_n & -A_n
end{vmatrix}
=
begin{vmatrix}
A_n & A_n \
0 & -2A_n
end{vmatrix},
$$
since the matrices
$$
begin{bmatrix}
A_n & A_n \
A_n & -A_n
end{bmatrix}
text{and}
begin{bmatrix}
A_n & A_n \
0 & -2A_n
end{bmatrix}
$$
are row-equivalent.

Thus,
$$|A_{2n}| = |A_n|cdot |-2A_n| = n^{n/2}(-2)^n n^{n/2} = 2^n n^n = (2n)^{2n/2}.$$