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Thread: *-isomorphism

  1. #1
    Member Mauritzvdworm's Avatar
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    *-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.
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  2. #2
    Member Mauritzvdworm's Avatar
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    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?
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  3. #3
    Member Mauritzvdworm's Avatar
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    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.
    Last edited by Mauritzvdworm; Mar 4th 2010 at 11:54 AM. Reason: typing error
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