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Math Help - Standard Basis of the complex filed?

  1. #1
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    Standard Basis of the complex filed?

    Hi,

    What is the standard basis of the complex field, C^n?
    Like C^3 for example. Would it be the same as R^3?
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  2. #2
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    Quote Originally Posted by thejinx0r View Post
    Hi,

    What is the standard basis of the complex field, C^n?
    Like C^3 for example. Would it be the same as R^3?
    You can use \bold{i},\bold{j},\bold{k} as a basis. Do not get confused, i\not = \bold{i}. Those are different things.
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  3. #3
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    Thanks for that, but I'm still having some troubles doing this problem. Or I am a little unsure of my answer.

    So here's the question.
    I attached the image below, because it had all the proper formatting.
    So I calculated that
    T=\left(<br />
\begin{array}{cc}<br />
 1 & 1 \\<br />
 6 & 0<br />
\end{array}<br />
\right)<br />

    And that the change-of-basis matrix is
    P=\left(<br />
\begin{array}{cc}<br />
 i & 0 \\<br />
 0 & 1+i<br />
\end{array}<br />
\right)<br />

    And it follows that
    P^{-1}=\left(<br />
\begin{array}{cc}<br />
 -i & 0 \\<br />
 0 & \frac{1-i}{2}<br />
\end{array}<br />
\right)<br />



    The next part is where I am unsure.

    So the matrix T with respect to the new basis would just be
    [T]_{U,U}=P^{-1}\cdot T \cdot P
    but this is just equal to T.
    Is that possible?
    Attached Thumbnails Attached Thumbnails Standard Basis of the complex filed?-99-6.jpg  
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  4. #4
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    Quote Originally Posted by thejinx0r View Post
    The next part is where I am unsure.

    So the matrix T with respect to the new basis would just be
    [T]_{U,U}=P^{-1}\cdot T \cdot P
    but this is just equal to T.
    Is that possible?
    I did not actually do any computations for this problem (so it is possible that you make a mistake in your computations) but I wanted to say if they are equal then so what? Why is that a problem? If you follow the correct way of getting your computations then your answer would be correct regardless if the new matrix is identical to the old matrix.
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