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dimension of subspace of trace of matrix |
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| Nov3-10, 08:32 AM | #1 |
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dimension of subspace of trace of matrix
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)? |
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| Nov3-10, 09:00 AM | #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? |
| Nov3-10, 11:15 AM | #3 |
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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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| dimension, linear algebra, nullity, rank, trace |
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