STEM Problem Detail

Problem Statement

Prompt

Provide me stepwise solution of question 4.3.

Solution Rendering...

## Problem 4.3

**Question:** Let
\[
T = \begin{bmatrix} T_{11} & T_{12} \\ 0 & T_{22} \end{bmatrix} \in \mathbb{C}^{n \times n}
\]
be block upper triangular with $$\Lambda(T_{11}) \cap \Lambda(T_{22}) = \emptyset$$. Let $$f$$ be defined on the spectrum of $$T$$ and let
\[
F = f(T) = \begin{bmatrix} f(T_{11}) & F_{12} \\ 0 & f(T_{22}) \end{bmatrix}.
\]

The block Parlett recurrence yields $$F_{12}$$ as the solution of the Sylvester equation
\[
T_{11}F_{12} - F_{12}T_{22} = f(T_{11})T_{12} - T_{12}f(T_{22}). \tag{4.44}
\]

But $$F_{12}$$ can also be obtained by block diagonalization. Show that if
\[
T_{11}X - XT_{22} = T_{12} \tag{4.45}
\]
then
\[
\begin{bmatrix} T_{11} & 0 \\ 0 & T_{22} \end{bmatrix} = \begin{bmatrix} I & -X \\ 0 & I \end{bmatrix}^{-1} \begin{bmatrix} T_{11} & T_{12} \\ 0 & T_{22} \end{bmatrix} \begin{bmatrix} I & -X \\ 0 & I \end{bmatrix},
\]
and deduce that
\[
F_{12} = f(T_{11})X - Xf(T_{22}). \tag{4.46}
\]

**Solution:**

1. **Step 1: Show the block diagonalization formula**

Given $$T_{11}X - XT_{22} = T_{12}$$ from (4.45).

Consider the transformation matrix:
\[
M = \begin{bmatrix} I & -X \\ 0 & I \end{bmatrix}.
\]

Its inverse is:
\[
M^{-1} = \begin{bmatrix} I & X \\ 0 & I \end{bmatrix}.
\]

Now compute $$M^{-1}TM$$:
\[
M^{-1}TM = \begin{bmatrix} I & X \\ 0 & I \end{bmatrix}
\begin{bmatrix} T_{11} & T_{12} \\ 0 & T_{22} \end{bmatrix}
\begin{bmatrix} I & -X \\ 0 & I \end{bmatrix}
\]

First multiply $$T$$ by $$M$$:
\[
TM = \begin{bmatrix} T_{11} & T_{12} \\ 0 & T_{22} \end{bmatrix}
\begin{bmatrix} I & -X \\ 0 & I \end{bmatrix}
= \begin{bmatrix} T_{11} & T_{11}(-X) + T_{12} \\ 0 & T_{22} \end{bmatrix}
= \begin{bmatrix} T_{11} & -T_{11}X + T_{12} \\ 0 & T_{22} \end{bmatrix}.
\]

Now multiply by $$M^{-1}$$:
\[
M^{-1}(TM) = \begin{bmatrix} I & X \\ 0 & I \end{bmatrix}
\begin{bmatrix} T_{11} & -T_{11}X + T_{12} \\ 0 & T_{22} \end{bmatrix}
\]
\[
= \begin{bmatrix} T_{11} & -T_{11}X + T_{12} + XT_{22} \\ 0 & T_{22} \end{bmatrix}.
\]

Using (4.45): $$-T_{11}X + T_{12} + XT_{22} = -(T_{11}X - XT_{22}) + T_{12} = -T_{12} + T_{12} = 0$$.

Therefore:
\[
M^{-1}TM = \begin{bmatrix} T_{11} & 0 \\ 0 & T_{22} \end{bmatrix}.
\]

This shows the block diagonalization.
2. **Step 2: Deduce formula for $$F_{12}$$**

Since $$M^{-1}TM$$ is block diagonal, we have:
\[
f(M^{-1}TM) = M^{-1}f(T)M.
\]

Compute $$f(M^{-1}TM)$$:
\[
f(M^{-1}TM) = f\left(\begin{bmatrix} T_{11} & 0 \\ 0 & T_{22} \end{bmatrix}\right) =
\begin{bmatrix} f(T_{11}) & 0 \\ 0 & f(T_{22}) \end{bmatrix}.
\]

Compute $$M^{-1}f(T)M$$:
\[
M^{-1}f(T)M = \begin{bmatrix} I & X \\ 0 & I \end{bmatrix}
\begin{bmatrix} f(T_{11}) & F_{12} \\ 0 & f(T_{22}) \end{bmatrix}
\begin{bmatrix} I & -X \\ 0 & I \end{bmatrix}
\]

First multiply $$f(T)$$ by $$M$$:
\[
f(T)M = \begin{bmatrix} f(T_{11}) & F_{12} \\ 0 & f(T_{22}) \end{bmatrix}
\begin{bmatrix} I & -X \\ 0 & I \end{bmatrix}
= \begin{bmatrix} f(T_{11}) & f(T_{11})(-X) + F_{12} \\ 0 & f(T_{22}) \end{bmatrix}
\]
\[
= \begin{bmatrix} f(T_{11}) & -f(T_{11})X + F_{12} \\ 0 & f(T_{22}) \end{bmatrix}.
\]

Now multiply by $$M^{-1}$$:
\[
M^{-1}(f(T)M) = \begin{bmatrix} I & X \\ 0 & I \end{bmatrix}
\begin{bmatrix} f(T_{11}) & -f(T_{11})X + F_{12} \\ 0 & f(T_{22}) \end{bmatrix}
\]
\[
= \begin{bmatrix} f(T_{11}) & -f(T_{11})X + F_{12} + Xf(T_{22}) \\ 0 & f(T_{22}) \end{bmatrix}.
\]
3. **Step 3: Equate the two expressions**

From $$f(M^{-1}TM) = M^{-1}f(T)M$$, we have:
\[
\begin{bmatrix} f(T_{11}) & 0 \\ 0 & f(T_{22}) \end{bmatrix} =
\begin{bmatrix} f(T_{11}) & -f(T_{11})X + F_{12} + Xf(T_{22}) \\ 0 & f(T_{22}) \end{bmatrix}.
\]

Comparing the (1,2) blocks gives:
\[
0 = -f(T_{11})X + F_{12} + Xf(T_{22}).
\]

Solving for $$F_{12}$$:
\[
F_{12} = f(T_{11})X - Xf(T_{22}).
\]

This proves formula (4.46).
4. **Step 4: Verify equivalence with Sylvester equation (4.44)**

To show that (4.46) satisfies (4.44), substitute $$F_{12} = f(T_{11})X - Xf(T_{22})$$ into the left side of (4.44):
\[
T_{11}F_{12} - F_{12}T_{22} = T_{11}[f(T_{11})X - Xf(T_{22})] - [f(T_{11})X - Xf(T_{22})]T_{22}
\]
\[
= T_{11}f(T_{11})X - T_{11}Xf(T_{22}) - f(T_{11})XT_{22} + Xf(T_{22})T_{22}.
\]

Since $$f(T_{11})$$ commutes with $$T_{11}$$ and $$f(T_{22})$$ commutes with $$T_{22}$$, we can rearrange:
\[
T_{11}F_{12} - F_{12}T_{22} = f(T_{11})T_{11}X - T_{11}Xf(T_{22}) - f(T_{11})XT_{22} + XT_{22}f(T_{22})
\]
\[
= f(T_{11})(T_{11}X - XT_{22}) - (T_{11}X - XT_{22})f(T_{22}).
\]

Using (4.45): $$T_{11}X - XT_{22} = T_{12}$$, so:
\[
T_{11}F_{12} - F_{12}T_{22} = f(T_{11})T_{12} - T_{12}f(T_{22}).
\]

This matches the right side of (4.44), confirming the equivalence.

**Conclusion:** We have shown that if $$T_{11}X - XT_{22} = T_{12}$$, then:

1. $$T$$ can be block diagonalized as shown
2. $$F_{12} = f(T_{11})X - Xf(T_{22})$$
3. This expression satisfies the Sylvester equation (4.44)

This provides an alternative method for computing $$F_{12}$$ via block diagonalization.

Loading solution...

Problem #3