Finding the matrix of transformation for $3\times 3$ matrices.

Question

Consider the matrices $$ X= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}, \qquad Y= \begin{bmatrix} 1 & 0 & 0 \\ 0 & -\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ 0 & \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{bmatrix}, \qquad Z= \begin{bmatrix} 1 & 0 & 0 \\ 0 & -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ 0 & -\frac{\sqrt{3}}{2} & -\frac{1}{2} \end{bmatrix}.$$ Can someone tell me if there is a matrix of transformation $P$ such that $$ PXP^{-1}=\begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix},\quad PYP^{-1}=\begin{bmatrix}0&0&1\\ 1&0&0\\ 0&1&0\end{bmatrix},\quad PZP^{-1}=\begin{bmatrix}0&1&0\\ 0&0&1\\ 1&0&0\end{bmatrix}? $$ I believe that $P$ exists, but I am not entirely sure.

Answer 1

While this problem can be difficult for general $X,Y,Z$, the particular problem is easy because of the following 3 facts:

1. $X$ is the identity matrix, so $PXP^{-1} = X = I$ will be the same for any choice of $P$
2. $Z = Y^{-1}$, so if $PYP^{-1}$ has the desired form, it immediately follows that $$ PZP^{-1} = PY^{-1}P^{-1} = (PYP^{-1})^{-1}, $$ which is the desired outcome
3. The matrix $Y$ is in a "canonical form". In particular, it is a "real Jordan form" (or more specifically, "real block-diagonal" form) matrix.

Fact 3 means that a satisfactory choice of $P$ for $Y$ can be found systematically. Facts 1 and 2 mean that such a $P$ will automatically result in the desired transformations to $X$ and $Z$.

Let $M$ denote the desired form for $PYP^{-1}$, namely $$ M = \pmatrix{0&0&1\\1&0&0\\0&1&0}. $$ We can block-diagonalize the matrix as follows: begin by calculating the eigenvalues of $Y$. We obtain the characteristic polynomial $$ \det(M - \lambda I) = \lambda^3 - 1 = (\lambda - 1)(\lambda - \omega)(\lambda - \overline{\omega}), $$ where $\omega = e^{2 \pi i/3} = -\frac 12 + i \frac{\sqrt{3}}{2}$ and $\bar \omega$ denotes the complex-conjugate of $\omega$. Conclude that the eigenvalues of $M$ are $1, \omega, \bar \omega$. Next, we calculate the associated eigenvectors of $M$. An eigenvector of $M$ associated with $1$ is given by $v_1 = (1,1,1)$. An eigenvector of $M$ associated with $\omega$ is given by $v_{\omega} = (1, \bar \omega, \omega) = x + iy$, where $x$ and $y$ are real vectors (i.e. $x,y$ are the real an imaginary parts of $v_\omega$ respectively). Let $a,b$ denote the real and imaginary parts of $\omega$, so that $\omega = a + bi$. We have $$ Av_\omega = \omega v_\omega \implies\\ A(x + iy) = (a + bi)(x + iy) \implies\\ Ax + iAy = (ax - by) + i(bx + ay) \implies\\ \begin{cases} Ay = ay + bx\\ Ax = -by + ax. \end{cases} $$ Now, consider the basis $\mathcal B = (v_1,y,x)$. Let $P = [v_1\ \ y\ \ x]$. We note that $$ Av_1 = v_1, \qquad Ay = ay + bx,\qquad Ax = -by + ax. $$ It follows that the matrix of $M$ relative to the basis $\mathcal B$ (whose columns correspond to the above equations) is given by $$ [M]_{\mathcal B} = P^{-1}MP = \pmatrix{1&0&0\\0&a&-b\\0&b&a} = Y. $$ That is, we have $P^{-1}MP = Y$. It follows that $M = PYP^{-1}$, as desired.

Thus, the answer to our questions is yes, with one suitable option being $$ P = [v_1\ \ y\ \ x] = \pmatrix{ 1&0&1\\ 1&-\sqrt{3}/2&-1/2\\ 1&\sqrt{3}/2&-1/2} \implies\\ P^{-1} = \frac 13\pmatrix{1&1&1\\0&-\sqrt{3}&\sqrt{3}\\2&-1&-1}. $$ Note that the columns of $P$ are mutually orthogonal. Thus, if we want an orthogonal matrix for our solution $P$, it suffices to normalize the columns of $P$ to obtain $$ P = \pmatrix{ 1/\sqrt{3} & 0 & 2/\sqrt{6}\\ 1/\sqrt{3} & -1/\sqrt{2} & -1/\sqrt{6}\\ 1/\sqrt{3} & 1/\sqrt{2} & -1/\sqrt{6}}^{-1} = \pmatrix{ 1/\sqrt{3} & 0 & 2/\sqrt{6}\\ 1/\sqrt{3} & -1/\sqrt{2} & -1/\sqrt{6}\\ 1/\sqrt{3} & 1/\sqrt{2} & -1/\sqrt{6}} \implies\\ P^{-1} = P^T = \pmatrix{1/\sqrt{3} & 1/\sqrt{3} & 1/\sqrt{3}\\ 0 & -1/\sqrt{2} & 1/\sqrt{2}\\ 2/\sqrt{6} & -1/\sqrt{6} & -1/\sqrt{6}}. $$