Exercise 6.6.3

The first and the third step comes from the Spectral theorem and the fact that these matrices are self-adjoint or at least normal. So we only do the first two steps. Also, we denote the matrix $E_{\mathit{ij}}$ to be a matrix, with suitable size, whose $\mathit{ij}$-entry is $1$ and all other entries are zero. Finally, it’s remarkble that that the matrices $P$ and $D$ are different from the each questions. They are defined in Exercise 6.5.2.

1.

Let $A_3=P^{\text{∗}}{E}_{11}P$ and $A_{-1}=P^{\text{∗}}{E}_{22}P$. Then we have $T_3=L_{A_3}$, $T_{-1}=L_{A_{-1}}$ and

$$L_A=3T_3-1T_{-1}.$$

2.

Let $A_i=P^{\text{∗}}{E}_{11}P$ and $A_i=P^{\text{∗}}{E}_{22}P$. Then we have $T_{-i}=L_{A_{-i}}$, $T_i=L_{A_i}$ and

$$L_A=-i{T}_{-i}+i{T}_i.$$

3.

Let $A_8=P^{\text{∗}}{E}_{11}P$ and $A_{-1}=P^{\text{∗}}{E}_{22}P$. Then we have $T_8=L_{A_8}$, $T_{-1}=L_{A_{-1}}$ and

$$L_A=8T_8-1T_{-1}.$$

4.

Let $A_4=P^{\text{∗}}{E}_{11}P$ and $A_{-2}=P^{\text{∗}}(E_{22}+E_{33})P$. Then we have $T_4=L_{A_4}$, $T_{-2}=L_{A_{-2}}$ and

$$L_A=4T_4-2T_{-2}.$$

5.

Let $A_4=P^{\text{∗}}{E}_{11}P$ and $A_1=P^{\text{∗}}(E_{22}+E_{33})P$. Then we have $T_4=L_{A_4}$, $T_1=L_{A_1}$ and

$$L_A=4T_4+1T_1.$$