By Sergei Treil
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Vm is complete in Rn (spanning, generating) iff echelon form of A has a pivot in every row; 3. The system v1 , v2 , . . , vm is a basis in Rn iff echelon form of A has a pivot in every column and in every row. Proof. The system v1 , v2 , . . , vm ∈ Rm is linearly independent if and only if the equation x1 v1 + x2 v2 + . . + xm vm = 0 has the unique (trivial) solution x1 = x2 = . . = xm = 0, or equivalently, the equation Ax = 0 has unique solution x = 0. By statement 1 above, it happens if and only if there is a pivot in every column of the matrix.
1 0 k ... 0 a 0 0 1 .. . 0 0 1 0 0 38 2. Systems of linear equations multiplies the row number k by a. 1 .. . . . . j k ... 0 A way to describe (or to remember) these elementary matrices: they are obtained from I by applying the corresponding row operation to it Finally, multiplication by the matrix .. . .. . 1 ... 0 .. . . . a ... 1 .. 1 0 adds to the row #k row #j multiplied by a, and leaves all other rows intact.
Abr , and that is exactly how the matrix AB was defined! Let us return to identifying again a linear transformation with its matrix. Since the matrix multiplication agrees with the composition, we can (and will) write T1 T2 instead of T1 ◦ T2 and T1 T2 x instead of T1 (T2 (x)). Note that in the composition T1 T2 the transformation T2 is applied first! The way to remember this is to see that in T1 T2 x the transformation T2 meets x fist. Remark. There is another way of checking the dimensions of matrices in a product, different form the row by column rule: for a composition T1 T2 to be defined it is necessary that T2 x belongs to the domain of T1 .