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

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

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

  • 2
    $\begingroup$ Looks ok. Alternatively you can prove (by induction on $k$) that $A_{2^k}A_{2^k}^T=2^k I$. As the determinants of the transposes are equal, the claim follows up to sign from this. The matrices you get are known as Hadamard matrices. $\endgroup$ – Jyrki Lahtonen Mar 2 '18 at 17:57

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


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