STEM Problem Detail

Problem Statement

Prompt

Provide me stepwise solution of question 4.4.

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## Problem 4.4

**Problem:** Let $$S, T \in \mathbb{C}^{n \times n}$$ be upper triangular and let
$$X = f(T^{-1}S)$$, $$Y = f(ST^{-1})$$. Show that
\[
SX - YS = 0, \quad TX - YT = 0,
\tag{4.47}
\]
and show how to compute $$X$$ and $$Y$$ together without forming $$T^{-1}S$$.

### Step 1: Express $$Y$$ in terms of $$X$$

Since $$ST^{-1} = T(T^{-1}S)T^{-1}$$, by Theorem 1.13(a):
\[
Y = f(ST^{-1}) = f(T(T^{-1}S)T^{-1}) = T f(T^{-1}S) T^{-1} = T X T^{-1}.
\]
So $$Y = TXT^{-1}$$.

### Step 2: Prove $$SX = YS$$

From $$Y = TXT^{-1}$$, we have $$YS = TXT^{-1}S$$.
Since $$X$$ commutes with $$T^{-1}S$$:
\[
X(T^{-1}S) = (T^{-1}S)X.
\]
Multiply by $$T$$ on left:
\[
T X T^{-1} S = S X.
\]
Thus $$YS = SX$$, so $$SX - YS = 0$$.

### Step 3: Prove $$TX = YT$$

From $$Y = TXT^{-1}$$, multiply by $$T$$ on right:
\[
YT = TX.
\]
Thus $$TX - YT = 0$$.

### Step 4: Diagonal entries

From $$Y = TXT^{-1}$$, diagonal entries satisfy:
\[
y_{ii} = t_{ii} x_{ii} t_{ii}^{-1} = x_{ii}.
\]
Also $$x_{ii} = f(\lambda_i)$$ where $$\lambda_i = s_{ii}/t_{ii}$$.
Thus:
\[
x_{ii} = y_{ii} = f(s_{ii}/t_{ii}), \quad i = 1,\dots,n.
\]

### Step 5: Off-diagonal recurrence

For $$i < j$$, from $$SX = YS$$:
\[
(SX)_{ij} = s_{ii} x_{ij} + \sum_{k=i+1}^{j-1} s_{ik} x_{kj} + s_{ij} x_{jj},
\]
\[
(YS)_{ij} = y_{ii} s_{ij} + \sum_{k=i+1}^{j-1} y_{ik} s_{kj} + y_{ij} s_{jj}.
\]
Equate:
\[
s_{ii} x_{ij} - s_{jj} y_{ij} = s_{ij}(y_{ii} - x_{jj}) + \sum_{k=i+1}^{j-1} (y_{ik} s_{kj} - s_{ik} x_{kj}).
\tag{1}
\]

From $$TX = YT$$:
\[
t_{ii} x_{ij} - t_{jj} y_{ij} = t_{ij}(y_{ii} - x_{jj}) + \sum_{k=i+1}^{j-1} (y_{ik} t_{kj} - t_{ik} x_{kj}).
\tag{2}
\]

### Step 6: Solve for $$x_{ij}$$ and $$y_{ij}$$

Equations (1) and (2) give:
\[
\begin{bmatrix}
s_{ii} & -s_{jj} \\
t_{ii} & -t_{jj}
\end{bmatrix}
\begin{bmatrix}
x_{ij} \\
y_{ij}
\end{bmatrix}
=
\begin{bmatrix}
s_{ij}(y_{ii} - x_{jj}) + \sum_{k=i+1}^{j-1} (y_{ik} s_{kj} - s_{ik} x_{kj}) \\
t_{ij}(y_{ii} - x_{jj}) + \sum_{k=i+1}^{j-1} (y_{ik} t_{kj} - t_{ik} x_{kj})
\end{bmatrix}.
\]

### Step 7: Recurrence procedure

1. Compute $$x_{ii} = y_{ii} = f(s_{ii}/t_{ii})$$ for all $$i$$.
2. For $$j=2,\dots,n$$ and $$i=j-1,\dots,1$$:
- Compute right-hand side using known entries.
- Solve $$2 \times 2$$ system for $$x_{ij}, y_{ij}$$.

### Step 8: Breakdown condition

The recurrence fails when the $$2 \times 2$$ matrix is singular:
\[
\det \begin{bmatrix} s_{ii} & -s_{jj} \\ t_{ii} & -t_{jj} \end{bmatrix} = -s_{ii}t_{jj} + s_{jj}t_{ii} = 0.
\]
That is, when $$s_{ii}t_{jj} = s_{jj}t_{ii}$$, or $$s_{ii}/t_{ii} = s_{jj}/t_{jj}$$ (equal eigenvalues).

### Step 9: Summary

We can compute $$X$$ and $$Y$$ without forming $$T^{-1}S$$, using a recurrence that breaks down when $$s_{ii}/t_{ii} = s_{jj}/t_{jj}$$ for some $$i \neq j$$.

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Problem #8