Dimension of subspace of trace of matrix

  • #1
Let V=Mn(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[tex]\in[/tex]V: 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)?
 

Answers and Replies

  • #2
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so we got that [tex]tr:M_n(k)\rightarrow k [/tex] is linear. Rank-nullity gives us that

[tex]dim(ker(tr))+dim(im(tr))=dim(M_n(k)) [/tex]

You need to find dim(ker(tr)). For this you have to figure out the other dimensions, what are they?
 
  • #3
HallsofIvy
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The set of all n by n matrices has dimension [itex]n^2[/itex]. From "tr(A)= 0", you have [itex]a_{11}+ a_{22}+ \cdot\cdot\cdot+ a_{nn}= 0[/itex] so that [itex]a_{nn}= -a_{11}- a_{22}- \cdot\cdot\cdot- a_{n-1 n-1}[/itex]. 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 [itex]n^2- 1[/itex].
 

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