Dimension of subspace of trace of matrix

[specialnlovin]

Let V=M_n(k) be a vector space of matrices with entries in k. For a matrix M denote the trace of M by tr(M).
What is the dimension of the subspace of {M\inV: tr(M)=0}
I know that I am supposed to use the rank-nullity theorem. However I'm not sure exactly how to use it. I know that the trace is a linear map itself. Since in this case it equals zero would the dim=dim(ker)?

 

[micromass]

so we got that tr:M_n(k)\rightarrow k is linear. Rank-nullity gives us that

dim(ker(tr))+dim(im(tr))=dim(M_n(k))

You need to find dim(ker(tr)). For this you have to figure out the other dimensions, what are they?

 

[HallsofIvy]

The set of all n by n matrices has dimension n^2. From "tr(A)= 0", you have a_{11}+ a_{22}+ \cdot\cdot\cdot+ a_{nn}= 0 so that a_{nn}= -a_{11}- a_{22}- \cdot\cdot\cdot- a_{n-1 n-1}. That is, you can replace one entry in the matrix by a linear combination of the others. That reduces the dimension of the subspace by 1: the dimension is n^2- 1.