If A positive definite then $\det(A) \leq \prod_{i=1}^n a_{i,i}$

Question

If $A=(a_{ij})_{ij=1,...n}\in \mathbb{C}^{n \times n}$ positive definite $ \Rightarrow \det(A) \leq \prod_{i=1}^n a_{i,i}$

I've argumented with the Cholesky-decomposition (which exists because A is pos. def.) that $A = CC^*$

with $C$ being an lower triangular matrix and $C^*$ being an upper triangular matrix.

Then $C$ and $C^*$ have the form...

$C =\left( \begin{array}{rrrr} x_{1,1} & 0 & 0 & 0 \\ x_{1,2} & x_{2,2} & 0 & 0 \\ ... & ... & ... & 0 \\ x_{1,n} & x_{2,n} & ... & x_{n,n} \\ \end{array}\right) $ and $C^*=\left( \begin{array}{rrrr} x_{1,1} & x_{1,2} & ... & x_{1,n} \\ 0 & x_{2,2} & ... & x_{2,n} \\ 0 & 0 & ... & ... \\ 0 & 0 & 0 & x_{n,n} \\ \end{array}\right) $

then $CC^* = \left( \begin{array}{rrrr} x_{1,1}^2 & & & * \\ & x_{2,2}^2 +x_{1,2}^2 & & \\ & & ... & \\ * & & & x_{n,n}^2 + x_{1,n}^2+x_{2,n}^2+...+x_{n-1,n}^2\\ \end{array}\right) $

Obviously, $\prod_{i=1}^n a_{i,i} = x_{1,1}^2*(x_{2,2}^2+x_{1,2}^2)*...*(x_{n,n}^2+x_{1,n}^2+...+x_{n-1,n}^2)$

And $\det(A) = \det(CC^*) = \det(C) * \det(C^*) = \det(C^*)*\det(C^*) = x_{1,1}^2*x_{2,2}^2 *...*x_{n,n}^2$

If what I've proven so far is correct, I'm very happy. But something bothers me. In $\mathbb{R}$ it is clear that $\det(A) \leq \prod_{i=1}^n a_{i,i}$ because the other sums are positive real numbers. Since I'm in the complex realm this formular doesn't hold true, does it?

1. I cannot define $"\leq"$ with complex numbers since it's not an order relation.
2. There is no guarantee that the other sums besides $x_{1,1}^2*x_{2,2}^2 *...*x_{n,n}^2$ are greater than 0 or even a real number, because $(a+bi)^2 = (a^2-b^2)+i(2ab)$ is still a complex number in most cases.

Can someone check my findings and elaborate?

Thanks in advance.

Answer 1

The problem as stated probably should have hermitian in place of positive definite, so that things work out, because in the resulting Cholesky decomposition, $C^*$ will be the conjugate transpose of $C$, which will ensure that each $x_{i,j}$ only gets multiplied with $\overline{x_{i,j}}$, hence resulting in a real number. The final expressions involve sums and products of these expressions only, so everything works out as you state it.

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Here's another proof, in the spirit of being elementary. Let $A$ be a positive definite Hermitian matrix, with $\tilde{ A}$ its cofactor matrix. We prove the following lemma , which provides a base for induction.

Let $B = \begin{pmatrix}A& x \\ y^T & 1\end{pmatrix}$ for some column vectors $x,y$, where $A$ is $n \times n$ and $x,y$ are $n \times 1$ (So $B$ is a block matrix, partitioned into last entry(which is $1$), last column, last row and the rest). Then $$\boxed{\det A - \det B = x^T\tilde{A}y}$$

How stunning is that? We've basically related the determinant of a matrix to the determinant of the matrix formed by deleting the last row and last column.

Proof : Check for yourself that $\det A - \det B$ is a bilinear map in $x,y$ (the determinant being multilinear in its columns and rows is the key). So it's enough to check what happens when $x = e_i,y=e_j$ for $e_i,e_j$ the standard basis elements.

When we set $x= e_i$ and $y=e_j$, however, it is clear from the Laplace expansion(or usual formula) along the last row $[y^T 1]$, that for the $1$ you get a $\det A$ which cancels out with the $\det A$ in the bilinear form, and the $y^T$ contains just a $1$ in position $j$,which along with the position $i$ of the last column gives the cofactor $\tilde{A}_{ij}$ as the determinant by definition. So in this case, the bilinear form does equal $x^TAy$.

The result follows from the fact that linear transformations coinciding on a basis are equal.

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Finally, we have the :

Corollary : If $A$ is a positive definite Hermitian matrix and $x$ is an $n\times 1$ vector, then $\det\begin{pmatrix}A& x \\ \bar{x}^T & 1\end{pmatrix} \leq \det A$.

Proof : RHS $-$ LHS is $\bar{x}^T\tilde{A}x = (\det A)\bar{x}^TA^{-1}x$. Recalling that the inverse is also positive definite gives the result. Note that equality holds iff $x=0$.

From here, it is an easy conclusion that for any such positive definite matrix $A$, $\det A \leq A_{nn}\tilde{A}_{nn}$, but then the matrix whose determinant is $\tilde{A}_{nn}$, is also positive definite (why? Any principal submatrix of a PSD Hermitian matrix is PSD Hermitian, use the definition), so the result follows by induction.

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More can be proven :

Let $\lambda_i$ , $1 \leq i \leq n$ denote the eigenvalues of $A$, a PSD Hermitian matrix, and let $A_{ii}$ be the diagonal entries. Let $\sigma_r$ denote the $r$th fundamental symmetric polynomial in $n$ variables for $1 \leq r \leq n$. Then $\sigma_r(\lambda_i) \leq \sigma_r(A_{ii})$. The determinant case is when $r = n$, so $\sigma_r$ is just the product of all inputs. Note that $r=1$ is the fact that the trace is at most the sum of the diagonal entries : we know this to be an equality.