这是一次Johns Hopkins大学的Matrix Theory课程的作业代写成功案例
Hadamard’s inequality states that if $N$ is the matrix having columns ${ }^{[3]} v_j$, then
$$
|\operatorname{det}(N)| \leq \prod_{i=1}^n\left|v_i\right| .
$$
Prove it.
By using the rescaling method, we can perform a scaling transformation on a row of the matrix, and the change in the determinant value is equivalent to the change in the value on the right-hand side of the inequality. Therefore, we can repeatedly apply this transformation to every row of the matrix to make each row a unit vector, thereby reducing the problem.
Now we need to prove that for a reduced unit matrix, each of its rows is a unit vector and its determinant is less than or equal to 1. However, we know that the determinant of this matrix corresponds to the volume of the parallelotope formed by the matrix’s rows, but since these unit vectors are not orthogonal, the volume can be estimated by their edge lengths. This is similar to the simplest case where the area of a right-angled triangle formed by two sides is always maximal.
A Hadamard matrix is a square matrix in which every element is either 1 or -1, and it is an orthogonal matrix. The well-known Hadamard inequality tells us that the product of its diagonal elements can determine the determinant of this matrix.
In 1933 Paley stated that the order $n(n \geq 4)$ of any Hadamard matrix is divisible by 4 . This is easy to prove. The converse has been a long-standing conjecture：
For every positive integer $n$, there exists a Hadamard matrix of order $4 n$.
Now, if we want to find an n×n Hadamard matrix, essentially we are searching for a set of characters composed of 1s and -1s that are as perpendicular as possible. If we use a greedy algorithm, we can start by choosing a character and then know that half of the characters perpendicular to it have the same element at corresponding positions, while the other half has the opposite element at corresponding positions. Next, the third character we want to find will satisfy the same relationship with the first character and also with the second character. If we repeat this process continuously, we will get a structure similar to the Cantor set, which will break down each part into smaller and smaller fragments and distribute 1s and -1s as evenly as possible. Through this process, we can construct many Hadamard matrices with row numbers of 2 to the power of k.
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