By Lipschitz S.

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A) Starting with the sequence w1 , . . 30 to show that there is a unique linear transformation f : V −→ W with the stated properties. (b) Starting with the sequence w1 , . . , wm , 0, . . 30 to show that there is a unique linear transformation f : V −→ W with the stated properties. 29. The next result provides a convenient way to decide if two vector spaces are isomorphic: simply show that they have the same dimension. 33. Let V and W be finite dimensional vector spaces over the field F . Then there is an isomorphism V −→ W if and only if dimF V = dimF W .

Suppose that f : V −→ W is a linear transformation. Then there are unique scalars tij ∈ F (i = 1, . . , m, j = 1, . . 1) f (vj ) = t1j w1 + · · · + tmj wm . 2) t11 · · ·  .  . T [f ]S = [tij ] =  .. tm1 · · ·  t1n ..  .  . 34. (a) Notice that this matrix depends on the bases S and T as well as f . (b) The coefficients are easily found: the j-th column is made up from the coefficients that occur when expressing f (vj ) in terms of the wi . In the next result we will also assume that there are bases S : v1 , .

Recall that once the reduced echelon matrix A has been found, the general solution can be expressed in terms of n − r basic solutions each obtained by setting one of the parameters equal to 1 and the rest equal to 0. These basic solutions form a spanning sequence for Ker fA which is linearly independent and hence is a basis for Ker fA . 4) row rank of A = r = n − dim Ker fA . On the other hand, if c is the column rank of A, we find that the columns of (A )T are a spanning sequence for the subspace of F m spanned by the columns of A, and in fact, they are linearly independent.

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