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Let A, B and C be nxn matrices such that A=BC-CB. Show that the trace of A (denoted Tr(A)) is 0, where the trace of an nxn matrix is defined as the sum of the entries along the leading diagonal.

This is the type of question which requires some out of the box thinking. A sketch is presented below: We start by defining A = (a_ij), B = (b_ij) and C = (c_ij). We then use the definition of matrix multiplication to write A in ''sigma notation''. The trace of A, we have been told, is defined as sum(a_ii) from i=1 to i=n (or j could equally be used). Taking the sum from i=1 to i=n to our expression for BC-CB, but with i's replaced with j's, and setting equal to sum(a_ij) from i=1 to i=n we end up with the difference of two ''double sums''. The two sums are equal (obviously, or by spotting a pattern after writing out) and so their difference is 0. Hence, the trace of A is 0. i.e Tr(A)=0

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Let A, B and C be nxn matrices such that A=BC-CB. Show that the trace of A (denoted Tr(A)) is 0, where the trace of an nxn matrix is defined as the sum of the entries along the leading diagonal.

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