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Math Help - Kernel, dimension and image of trace ?

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    Kernel, dimension and image of trace ?

    Let F:M(nxn)R->R where F(A) = tr(A).
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    Re: Kernel, dimension and image of trace ?

    The dimention of M_{n \times n}=n^2

    Since the trace of a matrix is the sum of the diagonal any non diagonal entry is mapped to 0. Since there are n entries on the diagonal that means that there are

    n^2-n non diagonal entries.

    The trace of the matrix is

    m_{11}+m_{22}+...+m_{nn}=0 \iff m_{11}=-m_{22}-...-m_{nn}

    So this equation has n-1 free parameters in it.

    So the dimention of the kernal is

    n^2-n +(n-1)=n^2-1.


    What do you think about the image?
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