Let \(V\) be the set of complex numbers, regarded as a real vector space.
Show that \((\alpha \mid \beta) = \operatorname{Re}(\alpha \bar{\beta})\) defines an inner product on \(V\).
Exhibit an (inner product space) isomorphism of \(V\) onto \(\RR^2\) with the standard inner product.
For each \(\gamma\) in \(V\), let \(M_\gamma\) be the linear operator on \(V\) defined by \(M_\gamma(\alpha) = \gamma\alpha\). Show that \((M_\gamma)^* = M_{\overline{\gamma}}\).
For which complex numbers \(\gamma\) is \(M_\gamma\) self-adjoint?
For which \(\gamma\) is \(M_\gamma\) unitary?
For which \(\gamma\) is \(M_\gamma\) positive?
What is \(\det(M_\gamma)\)?
Find the matrix of \(M_\gamma\) in the basis \(\{1, i\}\).
If \(T\) is a linear operator on \(V\), find necessary and sufficient conditions for \(T\) to be an \(M_\gamma\).
Find a unitary operator on \(V\) which is not an \(M_\gamma\).
Solution
Write \(\gamma=a+ib\) with \(a,b\in\mathbb R\).
1. The formula \((\alpha\mid\beta)=\operatorname{Re}(\alpha\overline\beta)\) is an inner product on the real vector space \(V\).
Real bilinearity follows from linearity of the real part in real scalars, and symmetry follows from \[ \operatorname{Re}(\alpha\overline\beta) =\operatorname{Re}(\overline{\alpha\overline\beta}) =\operatorname{Re}(\beta\overline\alpha). \] Moreover \[ (\alpha\mid\alpha)=\operatorname{Re}(|\alpha|^2)=|\alpha|^2\ge0, \] with equality exactly when \(\alpha=0\). Thus all inner-product axioms hold.
2. The map \(\Phi:V\to\mathbb R^2\) given by \(\Phi(x+iy)=(x,y)\) is an inner-product-space isomorphism.
It is clearly a real-linear bijection. If \(\alpha=x+iy\) and \(\beta=u+iv\), then \[ (\alpha\mid\beta) =\operatorname{Re}((x+iy)(u-iv)) =xu+yv, \] which is exactly the standard dot product of \((x,y)\) and \((u,v)\).
3. The adjoint of \(M_\gamma\) is \(M_{\overline\gamma}\).
For \(\alpha,\beta\in V\), \[ \begin{aligned} (M_\gamma\alpha\mid\beta) &=\operatorname{Re}(\gamma\alpha\overline\beta)\\ &=\operatorname{Re}(\alpha\overline{\overline\gamma\beta})\\ &=(\alpha\mid M_{\overline\gamma}\beta). \end{aligned} \] Hence \[ \boxed{M_\gamma^*=M_{\overline\gamma}}. \]
4. \(M_\gamma\) is self-adjoint exactly when \(\gamma\) is real.
By step 3, \[ M_\gamma=M_\gamma^* \iff M_\gamma=M_{\overline\gamma}. \] Evaluating both operators at \(1\) gives \(\gamma=\overline\gamma\), which is equivalent to \(\gamma\in\mathbb R\). The converse is immediate.
5. \(M_\gamma\) is unitary exactly when \(|\gamma|=1\).
Using step 3, \[ M_\gamma^*M_\gamma =M_{\overline\gamma}M_\gamma =M_{|\gamma|^2}. \] This is the identity operator exactly when \(|\gamma|^2=1\). Therefore \[ \boxed{M_\gamma\text{ unitary}\iff |\gamma|=1}. \]
6. \(M_\gamma\) is positive exactly when \(\gamma\) is a positive real number.
A positive operator is self-adjoint, so step 4 first forces \(\gamma\in\mathbb R\). For nonzero \(\alpha\), \[ (M_\gamma\alpha\mid\alpha) =\operatorname{Re}(\gamma|\alpha|^2) =\gamma|\alpha|^2. \] This is strictly positive for every \(\alpha\ne0\) exactly when \(\gamma>0\). Thus \[ \boxed{M_\gamma\text{ positive}\iff \gamma\in\mathbb R_{>0}}. \]
7. The determinant of \(M_\gamma\) is \(|\gamma|^2\).
In the real basis \((1,i)\), \[ M_\gamma(1)=a+bi, \qquad M_\gamma(i)=-b+ai, \] so \[ [M_\gamma]_{(1,i)} =\begin{pmatrix}a&-b\\ b&a\end{pmatrix}. \] Therefore \[ \boxed{\det(M_\gamma)=a^2+b^2=|\gamma|^2}. \] This also answers part (h): the displayed matrix is the requested matrix.
8. Characterize which real-linear operators are multiplication operators.
Let \[ J=M_i, \] so \(J(\alpha)=i\alpha\). Every multiplication operator commutes with \(J\): \[ M_\gamma J=JM_\gamma. \] Conversely, suppose a real-linear operator \(T\) commutes with \(J\). Put \[ \gamma=T(1). \] Every \(\alpha=x+iy\) satisfies \[ \alpha=x\cdot1+y\cdot i, \] and hence, by real linearity and commutation with \(J\), \[ T(\alpha) =xT(1)+yT(i) =x\gamma+yJ\gamma =x\gamma+yi\gamma =\gamma\alpha. \] Thus \(T=M_\gamma\). Therefore \[ \boxed{T=M_\gamma\text{ for some }\gamma\iff TJ=JT.} \] Equivalently, in the basis \((1,i)\) the matrix of \(T\) has the form \[ \begin{pmatrix}a&-b\\ b&a\end{pmatrix}. \]
9. Complex conjugation is a unitary real-linear operator that is not any \(M_\gamma\).
Define \[ C(z)=\overline z. \] It is real-linear, and \[ (C\alpha\mid C\beta) =\operatorname{Re}(\overline\alpha\,\beta) =\operatorname{Re}(\alpha\overline\beta) =(\alpha\mid\beta), \] so \(C\) is unitary. But \[ CJ(1)=C(i)=-i, \qquad JC(1)=i, \] so \(CJ\ne JC\). By step 8, \(C\) cannot equal \(M_\gamma\) for any \(\gamma\in\mathbb C\).