Unitary transformation

Question

I have a matrix in following form

$$A=\begin{bmatrix} 1&0&0&0&0&0&0&0\\ 0&-1&0&0&0&0&0&0\\ 0&0&1&0&0&0&0&0\\ 0&0&0&-1&0&0&0&0\\ 0&0&0&0&1&0&0&0\\ 0&0&0&0&0&-1&0&0\\ 0&0&0&0&0&0&1&0\\ 0&0&0&0&0&0&0&-1\\ \end{bmatrix}$$

I need some help to reduce the matrix in following form using unitary transformation

$$C=\begin{bmatrix} 1&0&0&0&0&0&0&0\\ 0&1&0&0&0&0&0&0\\ 0&0&1&0&0&0&0&0\\ 0&0&0&1&0&0&0&0\\ 0&0&0&0&-1&0&0&0\\ 0&0&0&0&0&-1&0&0\\ 0&0&0&0&0&0&-1&0\\ 0&0&0&0&0&0&0&-1\\ \end{bmatrix}$$

I have used the unitary matrix transformation$UAU^*$ method but the result always come in following form

$$UAU^*=\begin{bmatrix} 1&0&0&0&0&0&0&0\\ 0&1&0&0&0&0&0&0\\ 0&0&-1&0&0&0&0&0\\ 0&0&0&-1&0&0&0&0\\ 0&0&0&0&1&0&0&0\\ 0&0&0&0&0&1&0&0\\ 0&0&0&0&0&0&-1&0\\ 0&0&0&0&0&0&0&-1\\ \end{bmatrix}$$

Answer 1

Denote by $e_i$ the standard basis for $\mathbb{C}^8$. You are given a matrix $A$ such that

$$ Ae_i = \begin{cases} e_i & i \text{ is odd}, \\ -e_i & i \text{ is even}. \end{cases} $$

and you look for a unitary matrix $U \in M_{8}(\mathbb{C})$ such that

$$ UAU^{*}(e_i) = \begin{cases} e_i & 1 \leq i \leq 4, \\ -e_i & 5 \leq i \leq 8. \end{cases} $$

We can achieve this by letting $U^{*}$ (and hence also $U$) be a permutation matrix that rearranges the $e_i$'s. If we define $U^{*}$ by

$$ U^{*}e_1 = e_1, U^{*}e_2 = e_3, U^{*}e_3 = e_5, U^{*}e_4 = e_7, U^{*}e_5 = e_2, U^{*}e_6 = e_4, U^{*}e_7 = e_6, U^{*}e_8 = e_8 $$

then you can check that $U^{*}$ is unitary and $U = U^{-1}$ acts on the standard basis vectors by

$$ Ue_1 = e_1, Ue_2 = e_5, Ue_3 = e_2, Ue_4 = e_6, Ue_5 = e_3, Ue_6 = e_7, Ue_7 = e_4, Ue_8 = e_8. $$

Calculating explicitly, we have

$$ UAU^{*}(e_i) = \begin{cases} UAU^{*}(e_1) = UAe_1 = Ue_1 = e_1 & i = 1, \\ UAU^{*}(e_2) = UAe_3 = Ue_3 = e_2 & i = 2, \\ UAU^{*}(e_3) = UAe_5 = Ue_5 = e_3 & i = 3, \\ \ldots \\ UAU^{*}(e_8) = UAe_8 = U(-e_8) = -e_8 & i = 8. \end{cases} $$

as required and

$$ U = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix}. $$

Answer 2

Let $e_1 = (1,0,\dots,0)^T, e_2 = (0,1,0,\dots,0)^T$, and so on. One $U$ that works here is the $U$ whose columns are $$ U = [e_1\quad e_3 \quad e_5 \quad e_7 \quad e_2 \quad e_4 \quad e_6 \quad e_8] $$ Notice that this $U$ is a permutation matrix.