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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.

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    $\begingroup$ Hint: $A$ is positive definite if and only if its Hermitian part $M = \dfrac{1}{2}(A+A^*)$ is positive definite. If you phrase your problem in temrs of $M,$ then you can have all the nice Hermitian properties. $\endgroup$ – dezdichado May 26 '20 at 20:39
  • $\begingroup$ @dezdichado thanks but I'm not sure how it can simplify my problem. Do you mean that when M is hermitian then I know that $a_{j,i} = \overline{a_{i,j}}$ (i've used the same index in my matrices, in this case both are $a_{i,j}$) and thus $a_{i,j}^2 = |z|^2 \in \mathbb{R}$ ? $\endgroup$ – Quotenbanane May 26 '20 at 20:48
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    $\begingroup$ also, $*$ is complex conjugation, so the entries of $C^*$ will be $\overline{x_{i,j}}$ $\endgroup$ – dezdichado May 26 '20 at 20:49
  • $\begingroup$ Oh yes, you're right!! Thanks a lot, that solves it! Does the rest of my proof seems OK to you? $\endgroup$ – Quotenbanane May 26 '20 at 20:51
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    $\begingroup$ yes, they look fine but Cholesky decomposition and Positive Definiteness are usually associated with Hermitian matrices, so the original problem might have typo. $\endgroup$ – dezdichado May 26 '20 at 20:52
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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.


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.


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.


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.

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