[SOLVED] Linear Algebra, trace similar matrices

Quote:

Originally Posted by JJMC89

a) Show that if $A$ and $B$ are $n$ -by- $n$ matrices with coefficients in an arbitrary field $F$ , then $Tr(AB) = Tr(BA)$ .

b) use (a) to show that if $A$ and $B$ are similar matrices, then $Tr(A) = Tr(B)$ .

c) Show that the matrices of trace zero form a subspace of $M{_n}(F)$ of dimension $n{^2} - 1$ . [Hint: The mapping $Tr: M{_n}(F) \rightarrow F$ is a linear transformation.]

d) Prove that the subspace of $M{_n}(F)$ defined in (c) is generated by the matrices $AB - BA,\ A,\ B \in M{_n}(F)$ .

a) If $A = (a_{ij})$ and $B = (b_{ij})$ then $\text{Tr}(AB) = \sum_{i,j=1}^na_{ij}b_{ji}$ , and $\text{Tr}(BA)$ is given by the same formula.

b) By a), $\text{Tr}(P^{-1}(AP)) = \text{Tr}((AP)P^{-1})$ .

c) Use the hint. The subspace consisting of matrices in $M_n(F)$ with trace 0 is the kernel of the map in the hint.

d) The identity $A = \bigl(A-\tfrac1n\text{Tr}I_n\bigr) + \tfrac1n\text{Tr}I_n$ shows that every n×n matrix can be written as the sum of a matrix with trace 0 (namely the expression in parentheses) and a multiple of the identity.

Write $E_{ij}$ for the matrix having a 1 in the (i,j)-position and zeros everywhere else. If $i\ne j$ then $E_{ij} = E_{ii}E_{ij} - E_{ij}E_{ii}$ . By a linear combination of such matrices you can construct any matrix with zeros along the diagonal as a linear combination of commutators. Next, $E_{11}-E_{ii} = E_{1i}E_{i1} - E_{i1}E_{1i}$ for all $i>1$ . Check that by forming linear combinations of such matrices you can form any diagonal matrix that is not a scalar multiple of the identity.

That shows that any matrix with trace 0 can be expressed as sum of commutators.