Let F:M(nxn)R->R where F(A) = tr(A).
The dimention of $\displaystyle M_{n \times n}=n^2$
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
$\displaystyle n^2-n$ non diagonal entries.
The trace of the matrix is
$\displaystyle m_{11}+m_{22}+...+m_{nn}=0 \iff m_{11}=-m_{22}-...-m_{nn}$
So this equation has n-1 free parameters in it.
So the dimention of the kernal is
$\displaystyle n^2-n +(n-1)=n^2-1$.
What do you think about the image?