tanh transformation of correlation matrix

Question

Let a symmetric positive definite matrix $\boldsymbol{S}\in\mathbb{R}^{d\times d}$ be given with ones on the main diagonal. All other entries can assumed to be $\in (-1,1)$

Consider the following transformation: tanh is applied on the off-diagonal entries and the diagonal entries stay equal to $1$. Is the resulting matrix again positive definite?

If we do not require that the diagonal entries are equal $1$, it does not work, see tanh transformation of symmetric positive definite matrix

Now I am wondering if it works if the diagonal entries of the original matrix are equal to $1$

Answer 1

What follows is a partial answer with additional assumption that all off-diagonal entries of $S$ are nonpositive. In this case we have $$S=I-A,\qquad a_{ii}=0,\ a_{ij}\ge 0$$ We are interested in the nontrivial case when $A\neq 0.$ The condition $I-A\ge 0$ means that $$\langle Ax,x\rangle \le \langle x,x\rangle,\quad x\in \mathbb{R}^d $$ therefore it is equivalent to the property that the greatest eigenvalue $\lambda_\max$ of $A$ satisfies $\lambda_\max\le 1.$ We want to show that $\|A\|=\lambda_\max\le 1.$

For $x=(x_1,x_2,\ldots,x_d)^T$ let $|x|=(|x_1|,|x_2|,\ldots, |x_d|)^T.$ By well known formulas concerning symmetric matrices we have $$\|A\|=\max_{x\neq 0}{|\langle Ax,x\rangle|\over \langle x,x\rangle }\\ \lambda_\max =\max_{x\neq 0}{\langle Ax,x\rangle\over \langle x,x\rangle }=\max_{x\neq 0}{\langle A|x|,|x|\rangle\over \langle x,x\rangle }=\|A\|$$ (the second equality in the second line follows from $a_{ij}\ge 0$).

Let $B=\tanh A$ denote the matrix with entries $b_{ij}=\tanh(a_{ij}).$ As $0\le \tanh x\le x,$ for $x\ge 0,$ we get $$0\le b_{ij}\le a_{ij}$$ Thus $$\|B\|=\max_{x\neq 0}{\langle B|x|,|x|\rangle\over \langle x,x\rangle } \le \max_{x\neq 0}{\langle A|x|,|x|\rangle\over \langle x,x\rangle }=\|A\|\le 1$$ Therefore $-I\le B\le I.$ In particular we get $$0\le I-B=I-\tanh(A)=I+\tanh(-A)$$

Remark The proof covers also the case $S=I+A,$ $a_{ij}\ge 0,$ with assumption $I-A\ge 0$ (different from $S\ge 0$).

Answer 2

Partial Answer

We know that a $2\times 2$-correlation matrix $$\tag{1} \begin{pmatrix}1&\rho\\\rho&1\end{pmatrix} $$ is positive definite if and only if $\rho\in(-1,1)$. Since $\tanh(\rho)\in(-1,1)$ for all $\rho$ the answer to your question is yes in that case. Even better:

• If (1) is not even a correlation matrix because $|\rho|>1$ its $\tanh$-transform will always be a strictly positive definite one.

Answer 3

If a matrix is a symmetric and positive definite then all eigenvalues are positive. If a matrix has only positive eigenvalues the matrix is symmetric and positive definite.

The $tanh$ function is a contraction on the interval $[-1,1]$. The resulting interval under the picture of the $tanh$ is only approximate $[0.761594,0.761594]$. The $tanh$ is strictly monotone increasing on the reals.

Since the problem assumes that all values of the matrix under transformation are in $(-1,1)$. So the values get a little smaller but do not change signs.

So it is easy to understand that the given transformation preserves the symmetry attribute. It even keeps the order between the outer diagonal matrix entries. $a_{i,j}==a_{j,i}$ for $i \not = j$ and $-d<=i,j<=d$.

So there is equivalently to show that all eigenvalues remain positive under the transformation or the matrix remains indeed positive definite.

Since this transformation is designed like the step in the eigenvalue calculation where only from the diagonal elements the eigenvalues are subtracted and then the determinant is set to zero to calculate the eigenvalue there is a chance that the transformation does.

Starting in two dimension the transform appears to change the pair of eigenvalues from $\{1+b,1-b\}$ to $\{1+tanh(b),1-tanh(b)\}$. These can easily be proven to be both positive under the condition $-1<b<1$ alone and the $tanh$-contraction.

In general, the eigenvalue polynomial is a polynomial function of the outer diagonal matrix elements in exponents up to one lower the dimension and the highest exponent is the dimension and the second highest is proportional to the dimension so the contraction makes the constant and the factor of the lower exponents smaller. There are many attributes of the coefficients well-known and this is one. The eigenvalues remain all positive under the given transformation.

So the equivalence of the positive eigenvalue set and the symmetry and positive definiteness does the trick. All that is to do for the second task of your question is that this property passes away if the diagonal elements are not one anymore.

From the fundamental but not general result for two dimensions the stems a closer conditions together with the positiveness of the eigenvalues.

The problem is formulating it for general coefficients. As in the three-dimensional case, this only reduces the number of degrees of freedom to the dimension of the matrix. The rest is already free. So there is a need to do a choice. To avoid too much complexity only the simplest possible cases are considered. So most of the outer diagonal elements are zero and only one like in the two-dimensional case is symmetric and had the given attributes. Then the eigenvalue polynomial remains simple and the proof can be comfortably be done like in the two dimensions.

The increase in dimensions loses the complexity but the generality remains. If not all outer diagonal elements are non zero then transform the basis in a manner that it does. This should always be possible like calculating the eigenvalues and so on and so further.

I hope that helps and clarifies that for a suitable answer the complete settings have to be added to the question if it should be a fit for the progress in a course. If this is for fundamental maths then even the set of knowledge remains important to be given. Nobody starts from zero. Matrice are a generalization from numbers and much more delicate to tread.