Equivalence of Definitions of Hadamard Matrix

Definition 1
A Hadamard matrix $H$ is a square matrix such that:

Definition 2
A Hadamard matrix $H$ is a square matrix such that:

Proof
Let $H$ be a Hadamard matrix whose arbitrary element in the $j$th row and $k$th column is given by $\sqbrk a_{j k}$.

By either definition of Hadamard matrix:
 * $a_{j k} = \pm 1$

and so:
 * ${a_{j k} }^2 = 1$

By definition of transpose, column $j$ of $H_\intercal$ is ${r_j}^\intercal$, where $r_j$ is row $j$ of $H$.

$(1)$ implies $(2)$
Let $H = \sqbrk a_n$ be a Hadamard matrix by definition $1$.

We have that the rows of $H$ are pairwise orthogonal.

That is:
 * $r_j \cdot {r_k}^\intercal = 0$

when $j \ne k$.

Let $r_j$ be an arbitrary row of $H$:
 * $r_j = \begin {bmatrix} a_{j 1} & a_{j 2} & \cdots & a_{j n} \end {bmatrix}$

Then by definition of matrix product:
 * $h_{j j} = \ds \sum_{k \mathop = 1}^n {a_{j k} }^2$

where by definition of Hadamard matrix (either one) each of $a_{j k}$ is either $1$ or $-1$.

Hence:
 * $\forall k \in \set {1, 2, \ldots, n}: {a_{j k} }^2 = 1$

Then:

Similarly:

Thus $H H^\intercal$ is of the form:


 * $h_{j k} = n \delta_{j k}$

where $\delta_{j k}$ is the Kronecker delta.

So by definition of identity matrix:
 * $H H^\intercal = n \mathbf I_n$

Thus $H$ is a Hadamard matrix by definition $2$.

$(2)$ implies $(1)$
Let $H = \sqbrk a_n$ be a Hadamard matrix by definition $2$.

We have that:
 * $H H^\intercal = n \mathbf I_n$

From that last line:
 * $j \ne k \implies r_j \cdot {r_k}^\intercal = 0$

That is, the rows of $H$ are pairwise orthogonal.

Thus $H$ is a Hadamard matrix by definition $1$.