I have this matrix, \begin{equation} T=\begin{bmatrix}a&b\\-b&-a\end{bmatrix} \end{equation}

To normalize it, the matrix $T$ must satisfy this condition: $T^2=1$ and $1$ is the identity matrix. To solve that I set $x^2T^2=1$ and solve for x which is $\frac{1}{\sqrt{a^2-b^2}}$. The normalized matrix is \begin{equation} T=\frac{1}{\sqrt{a^2-b^2}}\begin{bmatrix}a&b\\-b&-a\end{bmatrix} \end{equation}

The next matrix P is a bit different, \begin{equation} P=\begin{bmatrix}c+a&b\\-b&c-a\end{bmatrix} \end{equation} Can this matrix P be normalized for the same condition $P^2=1$?

  • $\begingroup$ do you have determinants? $\endgroup$ Nov 17, 2017 at 0:49
  • $\begingroup$ If you take $P$ as it is now, what is $P^2$ exactly? Is it a diagonal matrix? $\endgroup$
    – Will Jagy
    Nov 17, 2017 at 1:23

1 Answer 1


Only if either $c=0$ or $a=b=0$.


If $P^2\varpropto I$, then $P^2_{ij}\varpropto I_{ij}$ for all $(i, j)$

$$I=\begin{bmatrix}1&0\\0&1\end{bmatrix}\\ I_{12}=I_{21}=0$$

So: $$P_{12}=2bc=P_{21}=-2bc\varpropto I_{12}=I_{21}=0$$

And therefore $bc=0$.

Furthermore, $P^2_{11}$ must equal $P^2_{22}$, as $I_{11}=I_{22}$.

$$(c+a)^2-b^2=(c-a)^2-b^2\\ |c+a|=|c-a|\\ c+a=c-a \text{ or } c+a=a-c\\ a=0 \text{ or } c=0$$

In fact, a quick check on Wolfram|Alpha shows that for a 2x2 matrix to be normalizable, the top left index must exactly equal the negative of the bottom right index (among other conditions) unless the top right and bottom left indices are both zero. The matrix $P$ only satisfies this condition when $c=0$ or $a=b=0$.


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