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Unformatted text preview: By the deﬁnition of the transpose, and what we have just seen, we have: (row k of AB ) = (column k of (AB )t ) = (column k of B t A) = = That is, (row i of AB ) = (row i of B ) which tells us what the rows of AB are. SECOND SOLUTION FOR (a) We can also start by directly working out what the rows of AB are. By the ﬁrst part of Theorem 13, or what is the same, V.I.F. 4, (AB )k, = (row k of A) · (column By the given information, (row k of A) = ei 0 if k = i . if k = i of B ) = Bi, , we see that (∗∗) of B ) . (∗) and all other rows of AB are zero. (column i of B t ) 0 if j = i if j = i (row i of B ) if j = i . 0 if j = i Combining (∗) and (∗∗), and observing that ei · (column (AB )k, = This says that (row i of AB ) = (row i of B ) and Bi, 0 if k = i . if k = i all other rows of AB are zero. We can now use the transpose to handle the part about the columns, using what we have just seen about the rows, or we could handle the rows directly, as in the ﬁrst solution. (b) No. If B commutes with every diagonal matrix, it commutes with each of the matrices A considered in part (a). For such a diagonal matrix A, as we have seen, AB has zero in every entry that is not in the ith row, but the ith column of BA is the ith column of B . But then since AB = BA, the only non zero entry in the ith column of B is in the ith place. Since this is true for each i, B is diagonal. 39 21/september/2005; 22:09 ...
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 Fall '08
 Gladue
 Calculus

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