# *-isomorphism

• Feb 27th 2010, 11:11 PM
Mauritzvdworm
*-isomorphism
I can see that it might be possible to identify $\displaystyle B(\mathbb{C}^{n})$ with $\displaystyle M_{n}(\mathbb{C})$

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

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

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

$\displaystyle M_{n}(\mathbb{C})$ is the space of all $\displaystyle n\times n$ matrices with complex entries.
• Feb 28th 2010, 10:02 PM
Mauritzvdworm
Here is an idea that might work

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

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

How then would we proceed to find such a matrix construction?
• Mar 3rd 2010, 09:40 PM
Mauritzvdworm
Figured out how to do this (for those who are interested)

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

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

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

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

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

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

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

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