By Gallier J.

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11. TRANSPOSE OF A LINEAR MAP AND OF A MATRIX Proof . (a) Consider the linear maps p j E −→ Im f −→ F, p f j where E −→ Im f is the surjective map induced by E −→ F , and Im f −→ F is the injective inclusion map of Im f into F . By definition, f = j ◦ p. To simplify the notation, p j s let I = Im f . Since E −→ I is surjective, let I −→ E be a section of p, and since I −→ F r is injective, let F −→ I be a retraction of j. Then, we have p ◦ s = idI and r ◦ j = idI , and since (p ◦ s) = s ◦ p , (r ◦ j) = j ◦ r , and idI = idI ∗ , we have s ◦ p = idI ∗ , and j ◦ r = idI ∗ .

A 1 n .. .. ... ... = ... . xn an 1 . . a n n xn showing that the old coordinates (xi ) of x (over (u1 , . . , un )) are expressed in terms of the new coordinates (xi ) of x (over (v1 , . . , vn )). Since the matrix P expresses the new basis (v1 , . . , vn ) in terms of the old basis (u1 , . . , un ), we observe that the coordinates (xi ) of a vector x vary in the opposite direction of the change of basis. For this reason, vectors are sometimes said to be contravariant. However, this expression does not make sense!

V ∗ ∈ (f )−1 (U 0 ), proving that f (U )0 = (f )−1 (U 0 ). Since we already observed that E 0 = 0, letting U = E in the above identity, we obtain that Ker f = (Im f )0 . The identity Ker f = (Im f )0 holds because (f ) = f . The following theorem shows the relationship between the rank of f and the rank of f . 31 Given a linear map f : E → F , the following properties hold. (a) The dual (Im f )∗ of Im f is isomorphic to Im f = f (F ∗ ). (b) rk(f ) ≤ rk(f ). If rk(f ) is finite, we have rk(f ) = rk(f ).