# Hadamard's inequality via Lagrange multiplier

I want to prove Hadamard's inequality which is formulated as follows:

Let $$X=(x_{ij})_{1\leq i,j\leq m}$$ be the $$m\times m$$ real matrix. Prove that $$|\det X|\leq \sqrt{\sum_{k=1}^m x_{1k}^2}\times \cdots \times \sqrt{\sum_{k=1}^m x_{mk}^2}$$

It suffices to prove the following result:

Let $$X=(x_{ij})_{1\leq i,j\leq m}$$ be the $$m\times m$$ real matrix such that $$\lVert x_i\rVert=1$$ for each $$i\in [m]$$, where $$x_i=(x_{i1},\dots,x_{im})$$. Prove that $$|\det X|\leq 1.$$

I'll show my attempts which are based on Lagrange's multipliers method.

Proof attempt: Let $$f:\mathbb{R}^{m^2}\to \mathbb{R}$$ be defined as follows: $$f(\vec{x})\equiv f(x_{11},\dots,x_{1m},\dots,x_{m1},\dots,x_{mm}):=\det \begin{pmatrix} x_{11}\dots x_{1m}\\ \cdots \cdots \\ x_{m1}\dots x_{mm} \end{pmatrix}.$$ Also define constraints functions $$F_1,\dots, F_m:\mathbb{R}^{m^2}\to \mathbb{R}$$ as follows: $$F_i(\vec{x})=(x_{i1})^2+\dots+(x_{im})^2-1 \quad \text{for each} \quad i\in [m].$$ Let $$S:=\{x\in \mathbb{R}^{m^2}: F_1(x)=\dots=F_m(x)=0\}$$. We notice that for each $$x\in S$$ the $$\text{rank}(F_1(x),\dots,F_m(x))=m$$ and hence $$S$$ defines $$(m^2-m)$$-dimensional smooth submanifold in $$\mathbb{R}^{m^2-m}$$.

Now we need to consider the following optimization problem: $$\begin{cases} f\to \text{extremum}\\ x\in S\end{cases}$$ or equivalently $$f|_S\to \text{extremum}$$ and my goal to show that $$\big|f|_S(x)\big|\leq 1$$.

Consider Lagrangian function $$\mathcal{L}(\vec{x},\lambda)=f(\vec{x})-\sum\limits_{i=1}^m \lambda_i F_i(\vec{x})$$. We know that if $$\vec{x}_0\in S$$ is local extremum of $$f|_S$$, then $$(\vec{x_0},\widehat{\lambda})$$ is a critical point of $$\mathcal{L}(x,\lambda)$$, where $$\widehat{\lambda}\in \mathbb{R}^m$$ follows from the necessary condition for extremum with constraints. Let's find critical point of $$\mathcal{L}(x,\lambda)$$, i.e. $$\begin{cases} \dfrac{\partial \mathcal{L}}{\partial x_{k\ell}}=\dfrac{\partial f}{\partial x_{k\ell}}(x)-\sum\limits_{i=1}^m \lambda_i \dfrac{\partial F_i}{\partial x_{k\ell}}(x)=0, \quad k,\ell\in [m] \\ \dfrac{\partial \mathcal{L}}{\partial \lambda_{i}}=-F_i(x)=0, \quad i\in [m] \end{cases}$$ In order to find partial derivative of $$f(x)$$ which is a determinant function, we can write $$f(x)=x_{i1}A_{i1}(x)+\dots+x_{im}A_{im}(x)$$ which is basically a cofactor expansion of $$f(x)=\det X$$. Now we see that $$\frac{\partial f}{\partial x_{ij}}(x)=A_{ij}(x)$$. Also notice that $$\frac{\partial F_i}{\partial x_{k\ell}}=\begin{cases} 0, \quad \text{if}\ i\neq k\\ 2x_{k\ell},\quad \text{if}\ i=k \end{cases}$$

This implies that $$\begin{cases} A_{k\ell}(x)=\lambda_k 2x_{k\ell}, \quad k,\ell\in [m]\\ F_i(x)=0, \quad i\in [m] \end{cases}$$

Multiplying the first equation by $$x_{k\ell}$$ and summing over $$1\leq\ell\leq m$$ and using the fact that $$F_k(x)=0$$ it follows that $$\lambda_k=f(x)/2$$. Thus we were able to show that $$\lambda_1=\dots=\lambda_k=f(x)/2$$.

In particular, this implies that $$A_{k\ell}(x)=x_{k\ell}f(x)$$. Assume that $$f(x)\neq 0$$, then the fact that $$A_{k\ell}(x)=x_{k\ell}f(x)$$ implies that $$X^{-1}=X^T$$, where $$X=(x_{ij})$$. Hence $$(\det X)^2=\det X \det X^T=\det XX^T=\det XX^{-1} =1$$. It implies that $$(f(\vec{x}))^2=1$$.

Now I am confused and I'd like to ask the following questions:

1. In the last paragraph I assume that $$f(x)\neq 0$$. But what happens if $$f(x)=0$$?

2. So this argument show that if $$x_0\in S$$ is a local extremum of $$f|_S$$, then $$f(x_0)^2=1$$. But does it imply that $$|f(x)|\leq 1$$ for all $$x\in S$$ because this is what I want to prove eventually.

3. All this argument makes sense if local extremum of $$f|_S$$ exists. But what if it does not exist?

I'd be grateful if someone can answer my questions and help me to finish the solution! Thank you!

Addendum: After reading @Youem's answer I was able to polish my proof. We notice that $$S$$ is a compact subset of $$\mathbb{R}^{m^2}$$ and hence the function $$f|_S$$ has a maximum and minimum at some points $$x_M\in S$$ and $$x_m\in S$$, respectively.

1. We notice that $$f(x_M)\geq 1$$ because the point $$\widehat{x}=(\underbrace{1,0,\dots,0}_{\text{m elements}}, \underbrace{0,1,\dots,0}_{\text{m elements}},\dots,\underbrace{0,0,\dots 1}_{\text{m elements}})\in S$$ and $$f(\widehat{x})=1$$ (I am doing this because I want to be sure that $$f(x_M)\neq 0$$). Applying what I've done above implies that $$(f(x_M))^2=1$$ and hence $$f(x_M)=1$$. Therefore, for all $$x\in S$$ we have $$f(x)\leq 1$$.

2. Also we notice that $$f(x_m)\leq -1$$ because the point $$\tilde{x}=(\underbrace{-1,0,\dots,0}_{\text{m elements}}, \underbrace{0,1,\dots,0}_{\text{m elements}},\dots,\underbrace{0,0,\dots 1}_{\text{m elements}})\in S$$ and $$f(\tilde{x})=-1$$ (I am doing this because I want to be sure that $$f(x_m)\neq 0$$). Applying what I've done above implies that $$(f(x_m))^2=1$$ and hence $$f(x_m)=-1$$. Therefore, for all $$x\in S$$ we have $$f(x)\geq -1$$.

Combining both conclusions we obtain that $$\forall x\in S$$, we have $$|f(x)|\leq 1$$.

1. $$f(x) = 0$$ gives you trivially the inequality.
2. Use the multilinearity of the determinant for $$x\not\in S$$.
• Thank you! I've edited my answer (see addendum). It turns out there is no need to use any multilinearity of the determinant for $x\notin S$.