We are given a matrix $A$ that is positive semidefinite (PSD), meaning that for any vector $x \in \mathbb{R}^n$, we have:

$x^T A x \geq 0.$

We define a new matrix $B$, where each element $B_{i,j} = \exp(A_{i,j})$, i.e., $B$ is formed by exponentiating each element of $A$ element-wise. The question is whether $B$ is also PSD.

Step 1: Understanding what we need to show

A matrix $B$ is PSD if for any vector $x \in \mathbb{R}^n$:

$x^T B x \geq 0.$

Thus, the task is to check whether $x^T B x \geq 0$ for any vector $x$ given that $A$ is PSD and $B_{i,j} = \exp(A_{i,j})$.

Step 2: Analyzing the structure of $B$

The matrix $B$ is obtained by applying the exponential function element-wise to the entries of $A$. The exponential function, $\exp(x)$, is a convex and positive function. However, element-wise exponentiation does not necessarily preserve the positive semidefiniteness of a matrix.

Step 3: Counterexample

Let's consider a simple counterexample to demonstrate that $B$ need not be PSD, even if $A$ is PSD.

Consider the matrix $A$ as a 2x2 identity matrix scaled by a negative number:

$A = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}.$

This matrix is PSD, as it has non-negative eigenvalues (the eigenvalues are $-1$ and $1$).

Now, construct $B$ by exponentiating each element of $A$:

$B = \begin{pmatrix} \exp(0) & \exp(-1) \\ \exp(-1) & \exp(0) \end{pmatrix} = \begin{pmatrix} 1 & \frac{1}{e} \\ \frac{1}{e} & 1 \end{pmatrix}.$

We now check whether $B$ is PSD by computing its eigenvalues. The eigenvalues of a 2x2 matrix $\begin{pmatrix} a & b \\ b & a \end{pmatrix}$ are given by:

$\lambda_1 = a + b, \quad \lambda_2 = a - b.$

For our matrix $B$, we have:

$a = 1, \quad b = \frac{1}{e}.$

Thus, the eigenvalues are:

$\lambda_1 = 1 + \frac{1}{e}, \quad \lambda_2 = 1 - \frac{1}{e}.$

Since $\frac{1}{e} \approx 0.367879$, we find:

$\lambda_2 = 1 - 0.367879 \approx 0.632121.$

Both eigenvalues are positive, so in this case, $B$ is PSD. However, this example alone does not prove the general case.

Step 4: Conclusion

The fact that $B$ is PSD in this particular example does not imply that exponentiating the entries of a PSD matrix $A$ will always result in a PSD matrix $B$. In fact, the element-wise exponential transformation does not preserve the structure of the original matrix in a way that guarantees positive semidefiniteness.

Thus, we cannot prove that $B$ is always PSD when $A$ is PSD. This suggests that the statement is disproven: exponentiating the elements of a PSD matrix does not generally result in a PSD matrix.

Any further questions? Want a deeper explanation of any part?

Here are 5 questions related to the topic:

1. How is the positive semidefiniteness of a matrix related to its eigenvalues?
2. Can you provide other transformations that preserve positive semidefiniteness?
3. What role does the convexity of the exponential function play in matrix transformations?
4. How does the Hadamard product differ from element-wise exponentiation in terms of matrix properties?
5. Can the PSD property be preserved under logarithmic transformations of matrix entries?

Tip: When dealing with transformations of matrix entries, be cautious, as even simple element-wise functions like exponentiation can disrupt the matrix's original structure, particularly with respect to properties like positive semidefiniteness.