STEM Problem Detail

Problem Statement

Prompt

Provide me stepwise solution of question 1.27

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## Problem 1.27

Let $$A \in \mathbb{C}^{n \times n}$$ have no eigenvalues on $$\mathbb{R}^-$$ except possibly for a semisimple zero eigenvalue. Show that:

1. There is a unique square root $$X$$ of $$A$$ that is a primary matrix function of $$A$$ and whose nonzero eigenvalues lie in the open right half-plane.
2. If $$A$$ is real then $$X$$ is real.

**Solution:**

1. **Given conditions:**

- $$A$$ has no eigenvalues on $$\mathbb{R}^- = (-\infty, 0]$$
- Exception: A semisimple zero eigenvalue is allowed
- "Semisimple" means diagonalizable (Jordan blocks are $$1 \times 1$$)
2. **Jordan canonical form:** Write $$A = Z J Z^{-1}$$ where:
\[
J = \begin{bmatrix} J_1 & 0 \\ 0 & J_0 \end{bmatrix}
\]

- $$J_1$$ contains all Jordan blocks with nonzero eigenvalues
- $$J_0$$ contains Jordan blocks for eigenvalue 0
- Since zero eigenvalue is semisimple, $$J_0 = 0$$ (diagonal)
3. **Define the square root function:** Let $$f(t) = t^{1/2}$$ be the principal square root, defined with branch cut along $$\mathbb{R}^-$$ and $$f(1) = 1$$.
4. **Primary square root of $$A$$:** By definition of primary matrix function:
\[
X = f(A) = Z f(J) Z^{-1} = Z \begin{bmatrix} f(J_1) & 0 \\ 0 & f(J_0) \end{bmatrix} Z^{-1}
\]
5. **Compute $$f(J_0)$$:** Since $$J_0 = 0$$ and zero is semisimple, $$f(0) = 0$$ is well-defined (no Jordan block structure issues).
6. **Compute $$f(J_1)$$:** All eigenvalues of $$J_1$$ are in $$\mathbb{C} \setminus \mathbb{R}^-$$, so by Theorem 1.29:

- $$f(J_1) = J_1^{1/2}$$ is the unique primary square root of $$J_1$$
- Its eigenvalues lie in the open right half-plane $$\Re(z) > 0$$
7. **Form of $$X$$:**
\[
X = Z \begin{bmatrix} J_1^{1/2} & 0 \\ 0 & 0 \end{bmatrix} Z^{-1}
\]
8. **Properties of $$X$$:**

1. **Uniqueness:**

- Any primary square root of $$A$$ must have form $$Z \begin{bmatrix} Y_1 & 0 \\ 0 & Y_0 \end{bmatrix} Z^{-1}$$
- $$Y_0^2 = 0$$ and $$Y_0$$ primary function of 0 $$\Rightarrow Y_0 = 0$$
- $$Y_1^2 = J_1$$ and $$Y_1$$ primary function of $$J_1$$ $$\Rightarrow Y_1 = J_1^{1/2}$$ by Theorem 1.29
- Hence $$X$$ is unique
2. **Part 2: Realness when $$A$$ is real**

**Method 1 (Using real Schur decomposition):**

- For real $$A$$, use real Schur decomposition: $$A = Q T Q^T$$ with $$Q$$ orthogonal, $$T$$ quasitriangular
- Zero eigenvalue blocks are $$1 \times 1$$ (semisimple)
- Apply principal square root to $$T$$ using block algorithm
- Result remains real

**Method 2 (Using Theorem 1.18):**

- Theorem 1.18: $$f(A)$$ is real if $$A$$ is real and $$f$$ satisfies $$f(\overline{z}) = \overline{f(z)}$$
- Principal square root satisfies this for $$\mathbb{C} \setminus \mathbb{R}^-$$
- For zero eigenvalue: $$f(0) = 0$$ is real

**Method 3 (Using Remark 1.9):**

- Primary matrix functions preserve realness when:

- Principal square root satisfies these conditions
3. **Conclusion:**

1. There exists a unique primary square root $$X$$ of $$A$$ with nonzero eigenvalues in open right half-plane
2. If $$A$$ is real, then $$X$$ is real

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Problem #2