STEM Problem Detail

Problem Statement

Prompt

Provide me stepwise solution of question 1.23.

Solution Rendering...

## Problem 1.23

Verify the Cauchy integral formula (1.12) in the case $$f(\lambda) = \lambda^j$$ and $$A = J_n(0)$$,
the Jordan block with zero eigenvalue.

The Cauchy integral formula (1.12) is:
\[
f(A) = \frac{1}{2\pi i} \int_{\Gamma} f(z)(zI - A)^{-1} dz
\]
where $$\Gamma$$ is a closed contour enclosing the eigenvalues of $$A$$.

**Solution:**

1. **Given:** $$A = J_n(0) = \begin{bmatrix}
0 & 1 & 0 & \cdots & 0 \\
0 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & 1 \\
0 & 0 & 0 & \cdots & 0
\end{bmatrix}$$, and $$f(\lambda) = \lambda^j$$.
2. **Eigenvalues:** $$A$$ has eigenvalue 0 with algebraic multiplicity $$n$$.
3. **Resolvent $$(zI - A)^{-1}$$:** For $$z \neq 0$$,
\[
zI - A = \begin{bmatrix}
z & -1 & 0 & \cdots & 0 \\
0 & z & -1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & -1 \\
0 & 0 & 0 & \cdots & z
\end{bmatrix}
\]

The inverse is:
\[
(zI - A)^{-1} = \begin{bmatrix}
z^{-1} & z^{-2} & z^{-3} & \cdots & z^{-n} \\
0 & z^{-1} & z^{-2} & \cdots & z^{-n+1} \\
0 & 0 & z^{-1} & \cdots & z^{-n+2} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & z^{-1}
\end{bmatrix}
\]
4. **Integrand:** $$f(z)(zI - A)^{-1} = z^j (zI - A)^{-1}$$:
\[
z^j (zI - A)^{-1} = \begin{bmatrix}
z^{j-1} & z^{j-2} & z^{j-3} & \cdots & z^{j-n} \\
0 & z^{j-1} & z^{j-2} & \cdots & z^{j-n+1} \\
0 & 0 & z^{j-1} & \cdots & z^{j-n+2} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & z^{j-1}
\end{bmatrix}
\]
5. **Contour integral:** Take $$\Gamma$$ to be a circle around 0. The $$(k,l)$$ entry of $$f(A)$$ is:
\[
[f(A)]_{kl} = \frac{1}{2\pi i} \int_{\Gamma} z^{j - (l-k+1)} dz
\]
for $$l \geq k$$, and 0 otherwise.
6. **Apply Cauchy residue theorem:** For integer $$m$$,
\[
\frac{1}{2\pi i} \int_{\Gamma} z^m dz =
\begin{cases}
1, & m = -1 \\
0, & \text{otherwise}
\end{cases}
\]
7. **Compute entries:**

- The integral is nonzero only when $$j - (l-k+1) = -1$$, i.e., $$l - k = j$$.
- For $$j \geq 0$$, this occurs when $$l = k + j$$.
- In that case, the entry is 1.
- All other entries are 0.
8. **Resulting matrix:**
\[
f(A) = \begin{cases}
\begin{bmatrix}
0 & 1 & 0 & \cdots & 0 \\
0 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & 1 \\
0 & 0 & 0 & \cdots & 0
\end{bmatrix}^j, & 0 \leq j < n \\
0, & j \geq n
\end{cases}
\]
9. **Compare with $$A^j$$:** For $$0 \leq j < n$$,
\[
A^j = \begin{bmatrix}
0 & 0 & \cdots & 1 & 0 & \cdots & 0 \\
0 & 0 & \cdots & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & 0 & 0 & \cdots & 0
\end{bmatrix}
\]
with 1's on the $$j$$-th superdiagonal. This matches the result from the Cauchy integral.
10. **For $$j \geq n$$:** $$A^j = 0$$ since $$A$$ is nilpotent with index $$n$$, and the integral gives 0.
11. **Conclusion:** The Cauchy integral formula gives $$f(A) = A^j$$, verifying (1.12) for $$f(\lambda) = \lambda^j$$ and $$A = J_n(0)$$.

Loading solution...

Problem #4