STEM Problem Detail

Problem Statement

Prompt

Provide me stepwise solution of question 1.20.

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## Problem 1.20: Symmetry Property for Matrix Functions

**Problem Statement:** Let $$f$$ satisfy $$f(-z) = \pm f(z)$$. Show that $$f(-A) = \pm f(A)$$ whenever the primary matrix functions $$f(A)$$ and $$f(-A)$$ are defined. (Hint: Problem 1.2 can be used.)

**Given Hint/Proof Sketch:**

- Let $$Z^{-1}AZ = J = \text{diag}(J_k)$$ be a Jordan canonical form.
- Then $$Z^{-1}(-A)Z = -J$$ and so $$Z^{-1}f(-A)Z = f(-J) = \text{diag}(f(-J_k))$$.
- $$f(-z) = \pm f(z)$$ implies $$f^{(j)}(-z) = \pm (-1)^j f^{(j)}(z)$$.
- From (1.34) we see that $$f(-J_k) = \pm f(J_k)$$.
- Hence $$\text{diag}(f(-J_k)) = \text{diag}(\pm f(J_k)) = \pm f(J)$$, and the result follows.

### Step 1: Understand the property

We are given that for scalars $$z$$:
\[
f(-z) = \pm f(z) \quad \text{for all } z,
\]
where the $$\pm$$ sign is fixed (either always $$+$$ or always $$-$$).

Examples:

- Even functions: $$f(-z) = f(z)$$ (e.g., $$f(z) = \cos z$$, $$z^2$$)
- Odd functions: $$f(-z) = -f(z)$$ (e.g., $$f(z) = \sin z$$, $$z^3$$)

We need to show this property extends to matrices: $$f(-A) = \pm f(A)$$.

### Step 2: Jordan canonical form approach

Let $$A = Z J Z^{-1}$$ be the Jordan decomposition, where:
\[
J = \text{diag}(J_1, J_2, \ldots, J_m)
\]
and each $$J_k$$ is a Jordan block for eigenvalue $$\lambda_k$$.

#### Step 2.1: Transform $$-A$$

\[
-A = Z(-J)Z^{-1} = Z \text{diag}(-J_1, -J_2, \ldots, -J_m) Z^{-1}.
\]

#### Step 2.2: Apply function to $$-A$$

Since $$f(X) = Z f(J) Z^{-1}$$ when $$X = Z J Z^{-1}$$:
\[
f(-A) = Z f(-J) Z^{-1} = Z \text{diag}(f(-J_1), f(-J_2), \ldots, f(-J_m)) Z^{-1}.
\]

#### Step 2.3: Analyze $$f(-J_k)$$ using Problem 1.2

From Problem 1.2, we have formula (1.34) for $$f(-J_k)$$:
\[
f(-J_k) =
\begin{bmatrix}
f(-\lambda_k) & -f'(-\lambda_k) & \cdots & (-1)^{m_k-1} \frac{f^{(m_k-1)}(-\lambda_k)}{(m_k-1)!} \\
& f(-\lambda_k) & \ddots & \vdots \\
& & \ddots & -f'(-\lambda_k) \\
& & & f(-\lambda_k)
\end{bmatrix}.
\]

### Step 3: Use the scalar property $$f(-z) = \pm f(z)$$

#### Step 3.1: Derivative relations

If $$f(-z) = \pm f(z)$$, then differentiating both sides:
\[
-f'(-z) = \pm f'(z) \quad \Rightarrow \quad f'(-z) = \mp f'(z).
\]
Differentiating again:
\[
f''(-z) = \pm f''(z).
\]
In general:
\[
f^{(j)}(-z) = \pm (-1)^j f^{(j)}(z).
\]

#### Step 3.2: Apply to matrix entries

For the entries of $$f(-J_k)$$:

- Diagonal entries: $$f(-\lambda_k) = \pm f(\lambda_k)$$
- First superdiagonal: $$-f'(-\lambda_k) = -[\mp f'(\lambda_k)] = \pm f'(\lambda_k)$$
- Second superdiagonal: $$(-1)^2 \frac{f''(-\lambda_k)}{2!} = \frac{\pm f''(\lambda_k)}{2!} = \pm \frac{f''(\lambda_k)}{2!}$$
- In general: $$(-1)^{j} \frac{f^{(j)}(-\lambda_k)}{j!} = (-1)^j \frac{\pm (-1)^j f^{(j)}(\lambda_k)}{j!} = \pm \frac{f^{(j)}(\lambda_k)}{j!}$$

Thus each entry of $$f(-J_k)$$ is $$\pm$$ times the corresponding entry of $$f(J_k)$$.

Therefore:
\[
f(-J_k) = \pm f(J_k) \quad \text{for each Jordan block}.
\]

### Step 4: Combine for full matrix

Since this holds for each Jordan block:
\[
f(-J) = \text{diag}(f(-J_1), f(-J_2), \ldots, f(-J_m)) = \text{diag}(\pm f(J_1), \pm f(J_2), \ldots, \pm f(J_m)).
\]

Since the $$\pm$$ is the same for all blocks (same function $$f$$):
\[
f(-J) = \pm \text{diag}(f(J_1), f(J_2), \ldots, f(J_m)) = \pm f(J).
\]

### Step 5: Return to original basis

\[
f(-A) = Z f(-J) Z^{-1} = Z (\pm f(J)) Z^{-1} = \pm Z f(J) Z^{-1} = \pm f(A).
\]

Thus:
\[
\boxed{f(-A) = \pm f(A)}.
\]

### Step 6: Alternative proof using Cauchy integral (for analytic $$f$$)

For analytic $$f$$, we can use the Cauchy integral definition.

Let $$\Gamma$$ be a closed contour enclosing $$\Lambda(A)$$ (spectrum of $$A$$).

From (1.12):
\[
f(A) = \frac{1}{2\pi i} \int_{\Gamma} f(z)(zI - A)^{-1} \, dz.
\]

For $$f(-A)$$, we need a contour enclosing $$\Lambda(-A) = \{-\lambda : \lambda \in \Lambda(A)\}$$.

Let $$\Gamma_1$$ be the reflection of $$\Gamma$$ in the imaginary axis (encloses $$\Lambda(-A)$$).

Then:
\[
f(-A) = \frac{1}{2\pi i} \int_{\Gamma_1} f(z)(zI + A)^{-1} \, dz.
\]

Make substitution $$w = -z$$, $$dw = -dz$$:
\[
f(-A) = \frac{1}{2\pi i} \int_{\Gamma} f(-w)(wI - A)^{-1} \, dw.
\]

But $$f(-w) = \pm f(w)$$, so:
\[
f(-A) = \pm \frac{1}{2\pi i} \int_{\Gamma} f(w)(wI - A)^{-1} \, dw = \pm f(A).
\]

### Step 7: Examples

**Example 1: Even function $$f(z) = \cos z$$**
\[
\cos(-z) = \cos z \quad \Rightarrow \quad \cos(-A) = \cos A.
\]

**Example 2: Odd function $$f(z) = \sin z$$**
\[
\sin(-z) = -\sin z \quad \Rightarrow \quad \sin(-A) = -\sin A.
\]

**Example 3: Polynomial $$f(z) = z^3$$**
\[
(-z)^3 = -z^3 \quad \Rightarrow \quad (-A)^3 = -A^3.
\]

**Example 4: Exponential (neither even nor odd)**
\[
e^{-z} \neq \pm e^z \quad \text{so the theorem doesn't apply}.
\]

### Step 8: Summary

We have proven that if a function satisfies $$f(-z) = \pm f(z)$$ for scalars, then the same relation holds for matrices: $$f(-A) = \pm f(A)$$.

Two proofs were given:

1. Using Jordan form and the explicit formula from Problem 1.2
2. Using Cauchy integral formula (for analytic functions)

This is a useful symmetry property that extends scalar function properties to matrix functions.

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