Hadamard’s Inequality
Let \boldsymbol{A}=(a_{ij})_{n\times n} be a real matrix of order n. Its determinant satisfies Hadamard’s inequality: \left(\det \boldsymbol{A}\right)^2 \leq \prod\limits_{i=1}^{n}\left(a_{1i}^2+a_{2i}^2+\dots+a_{ni}^2\right)
This was first published by Hadamard in 1893. Based on the notion that volume is represented by the determinant, the above inequality has a very clear geometric meaning. When n=2, it states that the area of a parallelogram is no greater than the product of the lengths of its two sides; when n=3, it states that the volume of a parallelepiped is no greater than the product of the lengths of its three edges; higher dimensions can be analogized similarly. These conclusions are almost "obviously true" in geometry. Therefore, it is difficult to understand why this inequality was only discovered in 1893. Of course, algebra does not accept such general statements; it requires a rigorous proof.
However, whether in textbook hints or materials I have searched for on the Internet (such as the article in the attachment), they almost invariably use the concept of positive definite matrices. But for such a basic inequality with such obvious geometric meaning, requiring the foundation of positive definite matrix knowledge to prove it seems like putting the cart before the horse. This also reflects a common issue where many researchers lack independent mathematical insight and merely follow others. (Once the first person does it, many follow, and even if the methods differ slightly, the original path remains unchanged.) In fact, positive definite matrices are not required at all; this inequality can be proven through a purely geometric method.
Optimal Coordinate System
Consider a matrix as a set of n column vectors \boldsymbol{a}_1, \boldsymbol{a}_2, \dots, \boldsymbol{a}_n: \boldsymbol{A}=[\boldsymbol{a}_1, \boldsymbol{a}_2, \dots, \boldsymbol{a}_n]
To find the volume of the n-dimensional parallelotope formed by these n column vectors, we need to rotate the original coordinate system slightly. That is, we change to a different rectangular coordinate system so that the representation of these n vectors becomes simpler. In algebraic language, this means finding an orthogonal matrix \boldsymbol{U} such that \boldsymbol{A}=\boldsymbol{U}\boldsymbol{T}, where the form of \boldsymbol{T} is as simple as possible (making the determinant easy to calculate).
What kind of coordinate system is optimal? Let \boldsymbol{e}_1, \boldsymbol{e}_2, \dots, \boldsymbol{e}_n be the basis of the new coordinate system; they are, of course, mutually orthogonal unit vectors. (Only in this way is the volume preserved; in algebraic language, orthogonal transformations preserve the determinant.) Naturally, we select one of the vectors, for example, \boldsymbol{e}_1 = \frac{\boldsymbol{a}_1}{|\boldsymbol{a}_1|}, as a coordinate axis, so that \boldsymbol{a}_1 has the simplest expression. To make the expression of \boldsymbol{a}_2 as simple as possible, we should place \boldsymbol{a}_2 on a coordinate plane. From a construction perspective, this means constructing a vector from \boldsymbol{e}_1 and \boldsymbol{a}_2 such that it becomes the \boldsymbol{e}_2 axis; in this way, \boldsymbol{a}_2 is a linear combination of \boldsymbol{e}_1 and \boldsymbol{e}_2. Similarly, to make the expression of \boldsymbol{a}_3 as simple as possible, \boldsymbol{a}_3 should be expressed as a linear combination of \boldsymbol{e}_1, \boldsymbol{e}_2, \boldsymbol{e}_3, which is equivalent to constructing \boldsymbol{e}_3 from \boldsymbol{a}_1, \boldsymbol{a}_2, \boldsymbol{a}_3. Continuing this process, it is actually the orthonormalization process (Gram-Schmidt) of the vectors \{\boldsymbol{a}_1, \boldsymbol{a}_2, \dots, \boldsymbol{a}_n\}!
That is to say, the set of vectors \boldsymbol{e}_1, \boldsymbol{e}_2, \dots, \boldsymbol{e}_n obtained by orthonormalizing \boldsymbol{a}_1, \boldsymbol{a}_2, \dots, \boldsymbol{a}_n is one of the optimal rectangular coordinate systems for representing the original vectors. This set of vectors forms an orthogonal matrix \boldsymbol{U}: \boldsymbol{U}=[\boldsymbol{e}_1, \boldsymbol{e}_2, \dots, \boldsymbol{e}_n]
Under this coordinate system, we have: \boldsymbol{A}=[\boldsymbol{a}_1, \boldsymbol{a}_2, \dots, \boldsymbol{a}_n]=[\boldsymbol{e}_1, \boldsymbol{e}_2, \dots, \boldsymbol{e}_n]\boldsymbol{T}=\boldsymbol{U}\boldsymbol{T} where \boldsymbol{T}=\begin{bmatrix} \lambda_1 & \cdot & \dots & \cdot \\ 0 & \lambda_2 & \dots & \cdot \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & \lambda_n \end{bmatrix} is an upper triangular matrix.
Completing the Proof
With the above conclusions, the proof of Hadamard’s inequality becomes very simple. Since \boldsymbol{U} is an orthogonal matrix, we naturally have: \begin{array}{l} |\boldsymbol{a}_1| = |(\lambda_1)| \geq \lambda_1, \quad |\boldsymbol{a}_2| = \left| \begin{pmatrix} \cdot \\ \lambda_2 \end{pmatrix} \right| \geq \lambda_2 \\ \dots \\ |\boldsymbol{a}_n| = \left| \begin{pmatrix} \cdot \\ \cdot \\ \vdots \\ \lambda_n \end{pmatrix} \right| \geq \lambda_n \end{array}
Therefore: \begin{aligned} (\det\boldsymbol{A})^2 &= (\det\boldsymbol{T})^2 = (\lambda_1 \lambda_2 \dots \lambda_n)^2 \\ &\leq |\boldsymbol{a}_1|^2 |\boldsymbol{a}_2|^2 \dots |\boldsymbol{a}_n|^2 \\ &= \prod\limits_{i=1}^{n}\left(a_{1i}^2+a_{2i}^2+\dots+a_{ni}^2\right) \end{aligned} This completes the proof of Hadamard’s inequality. A geometric proof!
Conclusion
In studying Higher Algebra, especially matrix-related theories, it is very beneficial and necessary to firmly grasp their geometric meanings; I have a deep appreciation for this. Whether from the perspective of memory or reasoning, geometry provides us with the most intuitive ideas, preventing us from accepting a concept without any context. It cannot be overemphasized that rote memorization is useless in mathematics.
Attachment
Proof and Geometric Meaning of Hadamard’s Inequality.pdf
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