1. ## *-isomorphism

I can see that it might be possible to identify $B(\mathbb{C}^{n})$ with $M_{n}(\mathbb{C})$

However, I am finding some trouble showing that
$\varphi:B(\mathbb{C}^{n})\rightarrow M_{n}(\mathbb{C})$
is a *-isomorphism...?

Can you think of a specific *-isomorphism $\varphi$ which will identify $B(\mathbb{C}^{n})$ with $M_{n}(\mathbb{C})$?

$B(\mathbb{C}^{n})$ is the set of all bounded linear operators form the Hilbert space $\mathbb{C}^{n}$ to itself.

$M_{n}(\mathbb{C})$ is the space of all $n\times n$ matrices with complex entries.

2. Here is an idea that might work

let $(e_n)$ the the standard basis for $\mathbb{C}^n$ and take a vector $b\in\mathbb{C}^n$. We can write $b=\beta_1e_1+\dots+\beta_ne_n$ and let $T$ be an operator in $B(\mathbb{C}^n)$, then
$Tb=T(\beta_1e_1+\dots+\beta_ne_n)=\beta_1Te_1+\dot s+\beta_nTe_n$

and then use this to construct a matrix representation for $T$

How then would we proceed to find such a matrix construction?

3. Figured out how to do this (for those who are interested)

let $(e)_{i=1}^{n}$ be the standard orthonormal basis of $\mathbb{C}^{n}$ and let $T\in B(\mathbb{C}^{n})$

we can write any vector $x\in\mathbb{C}^{n}$ as $x=x_1e_1+\dots+x_ne_n$ where the $x_i$`s are scalars, so

$Tx=x_1Te_1+\dots+x_nTe_n$ but $Te_j\in\mathbb{C}^{n}$ so we can write it as follows

$Te_j=t_{j1}e_1+t_{j2}e_2+\dots+t_{jn}e_{n}$ for every $j$

this then enables us to make to following connection
$
\left( \begin{array}{cccc}
t_{11} & t_{12} & \dots & t_{1n} \\
t_{21} & t_{22} & \dots & t_{2n} \\
\vdots & \vdots & \ddots& \vdots \\
t_{n1} & t_{n2} & \dots & t_{nn}\end{array} \right)
\left( \begin{array}{c}
x_1\\
x_2\\
\vdots\\
x_n
\end{array}\right)
=Tx
$

we can then use this to define the mapping $\varphi:B(\mathbb{C}^{n})\rightarrow M_n(\mathbb{C})$ as follows

$\varphi(T)=
\left( \begin{array}{cccc}
t_{11} & t_{12} & \dots & t_{1n} \\
t_{21} & t_{22} & \dots & t_{2n} \\
\vdots & \vdots & \ddots& \vdots \\
t_{n1} & t_{n2} & \dots & t_{nn}\end{array} \right)$

it is easy to thow that this choice then leads to a *-isomorphism.