Characteristic polynomial of unitary operator

Question

If I have $f_T(x)=x^2-1$ is that mean that my $T$ is unitary? I tried to show that is not true but I didn't succed so I might missing smoething

Answer 1

Considering that $\dim V = \deg f_T = 2,$ we may restrict our attention to the equivalent question of $2 \times 2$ matrices over $\mathbb R$ or $\mathbb C$ (as $f_T$ splits over both fields).

Given any $2 \times 2$ matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ with characteristic polynomial $x^2 - 1,$ we have that $$\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I = A^2 = \begin{pmatrix} a^2 + bc & b(a + d) \\ c(a + d) & d^2 + bc \end{pmatrix}$$ by the Cayley-Hamilton Theorem. Consequently, if $a = 0,$ we find that $bc = 1,$ $bd = 0,$ $cd = 0,$ and $d^2 + bc = 1.$ But this implies that $d = 0$ and $c = \frac 1 b.$ Ultimately, therefore, we have that $$\begin{pmatrix} b \bar b & 0 \\ 0 &\frac{1}{b} \bar{\frac 1 b} \end{pmatrix} = \begin{pmatrix} 0 & b \\ \frac{1}{b} & 0 \end{pmatrix} \begin{pmatrix} 0 & \bar{\frac 1 b} \\ \bar b & 0 \end{pmatrix} = AA^* = A^*A = \begin{pmatrix} 0 & \bar{\frac 1 b} \\ \bar b & 0 \end{pmatrix} \begin{pmatrix} 0 & b \\ \frac{1}{b} & 0 \end{pmatrix} = \begin{pmatrix} \frac 1 b \bar{\frac 1 b} & 0 \\ 0 & b \bar b \end{pmatrix}$$ so that $AA^* = A^* A = I$ if and only if $b \bar b = \frac 1 b \bar{\frac 1 b} = 1.$ Considering $A$ as a real matrix, then, the only orthogonal $2 \times 2$ matrices with $a = d = 0$ and characteristic polynomial $x^2 - 1$ are $$\pm \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$ since we have that $\bar b = b$ implies that $b^2 = 1$ so that $b = \frac 1 b = \pm 1.$

We will assume now that $A$ is complex so that $b = x + iy$ for some real numbers $x$ and $y.$ Our condition that $b \bar b = 1$ implies that $(x + iy)(x - iy) = x^2 + y^2 = 1.$ Considering that $x$ and $y$ are real, we must have that $b = \pm 1$ or $b = \pm i.$ (We find the same by analyzing $\frac 1 b \bar{\frac 1 b}.$) Consequently, the only unitary $2 \times 2$ matrices with $a = d = 0$ and characteristic polynomial $x^2 - 1$ are $$\pm \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \text{ and } \pm \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}.$$

Ultimately, all of this is to say that $f_T(x) = x^2 - 1$ is not sufficient to conclude that $T$ is unitary. Particularly, the linear operator $T : \mathbb C^2 \to \mathbb C^2$ corresponding to the $2 \times 2$ matrix $$A = \begin{pmatrix} 0 & 1 + i \\ \frac 1 {1 + i} & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 + i \\ \frac{1 - i}{2} & 0 \end{pmatrix}$$ satisfies $f_T(x) = x^2 - 1$ and is not unitary. Explicitly, we have that $$AA^* = \begin{pmatrix} 0 & 1 + i \\ \frac{1 - i}{2} & 0 \end{pmatrix} \begin{pmatrix} 0 & \overline{\frac{1 - i}{2}} \\ \overline{1 + i} & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 + i \\ \frac{1 - i}{2} & 0 \end{pmatrix} \begin{pmatrix} 0 & \frac{1 + i}{2} \\ 1 - i & 0 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & \frac 1 2 \end{pmatrix}.$$