STEM Problem Detail

Problem Statement

Prompt

Provide me stepwise solution of question 1.51.

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## Problem 1.51

**Problem:** Show that the equation
\[
\cosh(A) =
\begin{bmatrix}
1 & a & a & \cdots & a \\
1 & a & a & \cdots & a \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
a & a & \cdots & 1 & 1
\end{bmatrix} \in \mathbb{C}^{n \times n}
\]
has no solutions for \( a \neq 0 \) and \( n > 1 \).

### Step 1: Denote the right-hand side matrix

Let
\[
B =
\begin{bmatrix}
1 & a & a & \cdots & a \\
1 & a & a & \cdots & a \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
a & a & \cdots & 1 & 1
\end{bmatrix}.
\]
We assume \( a \neq 0 \) and \( n > 1 \).

### Step 2: Determine the Jordan structure of \( B \)

Consider \( B - I \):
\[
B - I =
\begin{bmatrix}
0 & a & a & \cdots & a \\
1 & a-1 & a & \cdots & a \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
a & a & \cdots & 0 & 0
\end{bmatrix}.
\]
It can be verified that \( \operatorname{rank}(B - I) = n - 1 \) (since the first row is a linear combination of the others, and the remaining \( n-1 \) rows are linearly independent for \( a \neq 0 \)).
A matrix with eigenvalue 1 and \( \operatorname{rank}(B - I) = n-1 \) has exactly one Jordan block for the eigenvalue 1. Thus \( B \) has a single \( n \times n \) Jordan block corresponding to eigenvalue 1.

### Step 3: Analyze the equation \( \cosh(A) = B \)

Let \( f(z) = \cosh z \). The equation is \( f(A) = B \).
The eigenvalues of \( B \) are all 1 (since it is a single Jordan block with eigenvalue 1). Therefore the eigenvalues \( \lambda \) of \( A \) must satisfy
\[
f(\lambda) = \cosh \lambda = 1.
\]
The solutions are \( \lambda = 0 \) (plus possibly multiples of \( 2\pi i \), but the principal branch gives \( \lambda = 0 \)).

Thus all eigenvalues of \( A \) are 0.

### Step 4: Apply Theorem 1.36

We examine the derivatives of \( f \) at \( \lambda = 0 \):
\[
f'(z) = \sinh z, \quad f'(0) = 0,
\]
\[
f''(z) = \cosh z, \quad f''(0) = 1 \neq 0.
\]
Theorem 1.36(b) (with \( \ell = 2 \)) states: if \( f'(\lambda) = 0 \) but \( f''(\lambda) \neq 0 \), then a Jordan block of size \( m \) in \( A \) for eigenvalue \( \lambda \) gives rise to two Jordan blocks in \( f(A) \) when \( m \ge 2 \). More precisely, a single Jordan block of size \( m \) in \( A \) splits into two Jordan blocks in \( f(A) \).

### Step 5: Derive a contradiction

Since \( A \) must have eigenvalue 0 only, and \( f(A) = B \) has a single Jordan block of size \( n \) (with eigenvalue 1), Theorem 1.36 forces \( A \) to have a single Jordan block of size \( n \) for eigenvalue 0.
But then, because \( f'(0) = 0 \) and \( f''(0) \neq 0 \), the image under \( f \) of a single \( n \times n \) Jordan block (with \( n \ge 2 \)) would split into at least two Jordan blocks. Thus \( f(A) \) would have more than one Jordan block, contradicting the fact that \( B \) has exactly one Jordan block.

### Step 6: Conclusion

Therefore no matrix \( A \) can satisfy \( \cosh(A) = B \) when \( a \neq 0 \) and \( n > 1 \).

\[
\boxed{\text{For } a \neq 0 \text{ and } n > 1, \text{ the equation } \cosh(A) = B \text{ has no solution.}}
\]

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Problem #6