# Thread: Kernel, dimension and image of trace ?

1. ## Kernel, dimension and image of trace ?

Let F:M(nxn)R->R where F(A) = tr(A).

2. ## 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?

### kernal of trace matrix

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