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Lecture 6 We can give an alternative characterization of one-to-one and onto: Proposition 0.1. Let T : V → W be linear. 1. T is injective if and only if it maps linearly independent sets of V to linearly independent sets of W . 2. T is surjective if and only if it maps spanning sets of V to spanning sets of W . 3. T is bijective if and only if it maps bases of V to bases of W . Proof. The third part follows from the first two. For the first, assume that T is injective and let S ⊂ V be linearly independent. We will show that T (S) = {T (v) : v ∈ S} is linearly independent. So let a1 T (v1 ) + · · · + an T (vn ) = ~0 . This implies that T (a1 v1 + · · · + an vn ) = ~0, implying that a1 v1 + · · · + an vn = ~0 by injectivity. But this is a linear combination of vectors in S, a linearly independent set, giving ai = 0 for all i. Thus T (S) is linearly independent. Conversely suppose that T maps linearly independent sets to linearly independent sets and let v ∈ N (T ). If v 6= ~0 then {v} is linearly independent, so {T (v)} is linearly independent. But if T (v) = ~0 this is impossible, since {~0} is linearly dependent. Thus v 6= ~0 and N (T ) = {~0}, implying T is injective. For item two, suppose that T is surjective and let S be a spanning set for V . Then if w ∈ W we can find v ∈ V such that T (v) = w and a linear combination of vectors of S equal to v: v = a1 v1 + · · · + an vn for vi ∈ S. Therefore w = T (v) = a1 T (v1 ) + · · · + an T (vn ) , meaning that we have w ∈ Span(T (S)), so T (S) spans W . Conversely if T maps spanning sets to spanning sets, then T (V ) = R(T ) must span W . But since R(T ) is a subspace of W , this means R(T ) = W and T is onto. Isomorphisms Definition 0.2. A linear transformation T : V → W that is bijective (that is, injective and surjective) is called an isomorphism. Generally speaking, we can view a bijection between sets X and Y as a relabeling of the elements of X (to get those of Y ). In the case of an isomorphism, this labeling also respects the vector space structure, being linear. Proposition 0.3. Let T : V → W be an isomorphism. Then T −1 : W → V is an isomorphism. Here, as always, the inverse function is defined by T −1 (w) = v if and only if T (v) = w . Proof. It is an exercise to see that any bijection has a well-defined inverse function and that this inverse function is a bijection. (This was done, for example, in the 215 notes in the first chapter.) So we must only show that T −1 is linear. To this end, let w1 , w2 ∈ W and c ∈ F. Then T (T −1 (cw1 + w2 )) = cw1 + w2 , whereas T (cT −1 (w1 ) + T −1 (w2 )) = cT (T −1 (w1 )) + T (T −1 (w2 )) = cw1 + w2 . Since T is injective, this implies that T −1 (cw1 + w2 ) = cT −1 (w1 ) + T −1 (w2 ). Using the notion of isomorphism, we can see that any n dimensional vector space V over F “is just” Fn . Theorem 0.4. Let V be an n-dimensional vector space over F. Then V is isomorphic to Fn . Proof. Let B = {v1 , . . . , vn } be a basis for V . We will think of B as being ordered. Define the coordinate map TB : V → Fn as before as follows. Each v ∈ V has a unique representation v = a1 v1 + · · · + an vn . So set TB (v) = (a1 , . . . , an ). This was shown before to be a linear transformation. So we must just show it is an isomorphism. Since the dimension of V is equal to that of Fn , we need only show that TB is onto. Then by the rank-nullity theorem, we will find dimN (TB ) = dim(V ) − dim(R(TB )) = dim(V ) − dim(Fn ) = 0 , implying that N (TB ) = {~0}, and that TB is one-to-one. So to show onto, let (a1 , . . . , an ) ∈ Fn . The element v = a1 v1 + · · · + an vn maps to it: TB (v) = TB (a1 v1 + · · · + an vn ) = (a1 , . . . , an ) , so TB is an isomorphism. Matrices and coordinates We will now see that, just as V with dimension n “looks just like” Fn , all linear maps from V to W look just like matrices with entries from F. Suppose that T : V → W is linear and these are finite dimensional vector spaces with dimension n and m respectively. Fix B = {v1 , . . . , vn } and C = {w1 , . . . , wm } to be bases of 2 V and W respectively. We know that T is completely determined by its values on B, and each of these values lies in W , so we can write T (v1 ) = a1,1 w1 + · · · + am,1 wm T (v2 ) = a1,2 w1 + · · · + am,2 wm and so on, up to T (vn ) = a1,n w1 + · · · + am,n wm . Now we take some arbitrary v ∈ V and express it in terms of coordinates using B. This time we write it as a column vector and use the notation [v]B : a1 [v]B = · · · , where v = a1 v1 + · · · + an vn . an Let us compute T (v) and write it in terms of C: T (v) = a1 T (v1 ) + · · · + an T (vn ) = a1 (a1,1 w1 + · · · + am,1 wm ) + · · · + an (a1,n w1 + · · · + am,n wm ) = (a1 a1,1 + · · · + an a1,n )w1 + · · · + (a1 am,1 + · · · + an am,n )wm . Therefore we can write T (v) in coordinates using C as a1,1 · · · a1 a1,1 + · · · + an a1,n ··· ··· = [T (v)]C = am,1 · · · a1 am,1 + · · · + an am,n a1,n am,n a1 · ··· . an Therefore we have found on half of: Theorem 0.5 (Matrix representation). Let T : V → W be linear and B = {v1 , . . . , vn } and C = {w1 , . . . , wm } be (ordered) bases of V and W respectively. There exists a unique matrix, written [T ]B C such that for all v ∈ V , [T (v)]C = [T ]B C [v]B . Proof. We have already shown existence. To show uniqueness, suppose that A is any m × n matrix with entries from F such that for all v ∈ V , A[v]B = [T (v)]C . Choose v = vi for some i = 1, . . . , n (one of the basis vectors in B). Then the coordinate representation of v is [v]B = ei , the vector with all 0’s but a 1 in the i-th spot. Now the product of matrices A[v]B actually gives the i-th column of A. We can see this by using the matrix multiplication formula: if M is an m × n matrix and N is an n × p matrix then the matrix M N is m × p and its (i, j)-th coordinate is given by (M N )i,j = n X k=1 3 Mi,k Nk,j . Therefore as A is m×n and [v]B is n×1, the matrix A[v]B is m×1 and its (j, 1)-th coordinate is n n X X (A[v]B )j,1 = Aj,k ([v]B )k,1 = Aj,k (ei )k,1 = Aj,i . k=1 k=1 This means the entries of A[v]B are A1,i , A2,i , . . . , Am,i , the i-th column of A. However, this B also equals [T (ei )]C , which is the i-th column of [T ]B C by construction. Thus A and [T ]C have the same columns and are thus equal. In fact much more is true. What we have done so far is defined a mapping Φ : L(V, W ) → Mm,n (F) in the following manner. Given fixed bases B and C of sizes n and m respectively, we set Φ(T ) = [T ]B C . This function is actually an isomorphism, meaning that the space of linear transformations is just a relabeling of the space of matrices (after choosing “coordinates” B and C): Theorem 0.6. Given bases B and C of V and W of sizes n and m, the spaces L(V, W ) and Mm,n (F) are isomorphic via the mapping Φ. Proof. We must show that Φ is a bijection and linear. First off, if Φ(T ) = Φ(U ) then for all v ∈ V , we have [T (v)]C = Φ(T )[v]B = Φ(U )[v]B = [U (v)]C . But the map sending vectors in W to their coordinates relative to C is also a bijection, so T (v) = U (v). Since this is true for all v, we get T = U , meaning Φ is injective. To show surjective, let A be any m × n matrix with (i, j)-th entry Ai,j . Then we can define a linear transformation T : V → W by its action on the basis B: set T (vi ) = A1,i w1 + · · · + Am,i wm . By the slogan, there is a unique linear transformation satisfying this and you can then check that [T ]B C = A, meaning Φ is surjective and therefore a bijection. To see that Φ is linear, let T, U ∈ L(V, W ) and c ∈ F. Then the i-th column of [cT + U ]B C is simply the coefficients of (cT + U )(vi ) expressed relative to the basis C. This coordinate map is linear, so [(cT + U )(vi )]C = [cT (vi ) + U (vi )]C = c[T (vi )]C + [U (vi )]C , which is c times the i-th column of Φ(T ) plus the i-th column of Φ(U ). Thus B B [cT + U ]B C = c[T ]C + [U ]C . 4