Let $H$ and $R$ be two Hilbert spaces and let $T$ be a bounded operator $T: H \to R$. I have been asked to show that there exists a unique operator $T^*: R \to H$ such that \begin{align*} (Tx|y) = (x|T^*y) \ \forall x \in H, \ \forall y \in R. \end{align*} I have been further asked to show that the map $T \to T^*$ is a linear isometry of ${\bf B}(H,R)$ onto ${\bf B}(R,H)$ that satisfies \begin{align*} \left\vert \left\vert {T^*T} \right\vert \right\vert = \left\vert \left\vert {T} \right\vert \right\vert ^2 = \left\vert \left\vert{TT^*} \right\vert \right\vert. \end{align*}
I have shown that $T^*$ exists. The book I am using gives a hint for the latter part for the isometry that says consider the operator \begin{align*} \left(\begin{array}{cc} 0 & 0\\ T & 0 \end{array}\right) \end{align*} in ${\bf B}(H \oplus R)$ and use the fact that to each bounded operator $\phi$ on a Hilbert space there exists a unique $\phi^*$ such that $(\phi x|y) = (x|\phi^* y)$ and the map $\phi \to \phi^*$ is a conjugate linear antimultiplicative isometry of the bounded operators on the Hilbert space onto itself, such that \begin{align*} \left\vert \left\vert{\phi^*\phi} \right\vert\right\vert = \left\vert \left\vert{\phi}\right\vert \right\vert^2. \end{align*} What I am struggling with is I do not know how to find the adjoint to the matrix above. I would be very grateful if someone could give direction on how to find the adjoint, so I can do the calculation.
Here is a possible direction:
Suppose $T$ were a complex valued scalar; then the matrix is just a $2\times 2$ matrix. What would be its adjoint?
Now formulate an educated guess as to what is the adjoint in your case. Verify your guess by writing explicitly down the action of the adjoint on an element of (the dual of ) $H\oplus R$.