.6. Restriction of Linear Maps In this section, we restrict linear maps to subspaces. We observe that the notion of linearity still makes sense for maps whose domain and codomain are subspaces of R n, R m, not necessarily R n and R m themselves. We discuss what notions are still preserved and what notions have to be adjusted to account for this increase in generality. Definition.6.. Suppose S R n, V R m are subspaces. A map U : S V is linear if (a) U( x + y) = U( x) + T ( y) (b) U(c x) = c U( x), for all x, y S and scalars c. Why do we assume the domain and codomain are subspaces? To answer this, consider what { [ happens if the domain is a subset of R n which is not a subspace. For example, what would it mean U : 0, R 0]} to be linear? In particular, it would mean ([ U = U ]) ([ 0]), { } which makes no sense, since / 0,. In order for the notion of linearity to make sense, adding or scaling inputs must result in other valid inputs. This means if the domain is a subset of R n, it must be a subspace of R n. Assuming the codomain is a subspace then becomes a natural assumption because the range of any linear map is a subspace, Section., Problem (The same logic applies when the domain is any subspace). In fact, we can make the domain and codomain any set of objects where the operations of addition and scalar multiplication make sense, but this level of generality is a topic for a more advanced linear algebra class. One way to obtain a linear map between subspaces is simply to restrict the domain from R n to some subspace of R n. Restricting the domain is a notion that makes sense for any map, so now define for maps in general. We look at an example restricting a map between finite sets and an example restricting a linear map T : R R. We leave it to the reader to show that restricting a linear map to subspace results in a linear map, Problem. Definition.6.. Given a map f : X Y, and a subset S X, let f restricted to S, denoted f S, be the map f S : S Y given by f S ( x) = f( x) for all x S. Example.6.. Suppose f : {,,, 4, 5} {,,, 4, 5} is visualized below: 4 5 Visualize the restriction f {,,5}. In addition, determine if f, f {,,5} are injective and/or surjective. Solution: We simply remove, 4 and the arrows that go with them to get f {,,5} below: 6 8 9
6 8 5 Note that while f is not injective and surjective, f {,,5} is injective, but not surjective. Suppose S R n, V R m are subspaces. The notion of kernel and range still makes sense for linear maps U : S V. We have ker(u) = { x S U( x) = 0}, range(u) = {U( x) x S}. Restricting linear maps can change the kernel and range. We defined injectivity and surjectivity for maps in general, so these notions apply here as well. For linear U : S V, we have U is injective ker(u) = { 0}, U is surjective range(u) = V, by Theorem..4 and Definition..6. Although Theorem..4 was proven in the context of linear maps T : R n R m, its proof extends to linear maps U : S V as well. Example.6.4. Consider the linear map T : R R with [T ] = (a) Find ker(t ), range(t ), (b) Find ker(t S ), range(t S ), (c) Find a linear map U : R R so that U T, but U S = T S. Solution: (a) We find ker(t ) = ker () = { 0}, range(t ) = span 4 (b) Converting S to relation form gives S = x x =. Thus, we can rewrite the formula for T S : S R as x 9 ([, and let S = span. 4 ]) (, {[ x x ] x = x }. Thus, every x = ) = R. 4 x x () x x x T S ( x) = T S = =. x 4 x x x Therefore, for each x = S, x () x x T S ( x) = 0 T S = 0 = 0 x x x = 0 x = This means ker(t S ) = { 0}. For range(t S ), we have { x range(t S ) = {T S ( x) x S} = x x R} = { } x = span S is of the form [ x x ] = 0. ().
[ x ] (c) We have already seen that for all x = S, x x + x T ( x) = = x + 4x where U : R R has [U] = relation x = x on x = U( x), x 0. Hence, U T, but T 0 S = U S. In addition, one can use to T () x x + x = x x + 4x to get infinitely more linear maps U : R R that agree with T on S. One point to get from Example (c) is the notion of [U] for linear maps U : S V between subspaces is not defined, since U could be written as matrix multiplication for many different matrices. In Example (c), 0 T S : S R could be seen as matrix multiplication by or. 4 0 Suppose S R n is a subspace. Restricting a linear map T : R n R m to S gives us a linear map U : S R m. But are all linear maps U : S R n restrictions of linear maps T : R n R m? Yes, as explained in Proposition 0. We then explain how to find the kernel and range or restricted linear maps in general. Proposition.6.5. Suppose S R n is a subspace, and U : S R m is linear. Then there exists a linear map T : R n R m so that U = T S. Proof. Problem 0. Theorem.6.6. Suppose T : R n R m is a linear map, and S = span( U,..., U k ) is a subspace of R n. Then T S : S R m has ker(t S ) = S ker(t ), ( range(t S ) = span T ( U ),..., T ( U ) k ) Proof. First, x ker(t S ) means T ( x) = 0 and x S. So, ker(t S ) = { x S T ( x) = 0} = { x S x ker(t )} = S ker(t ). As for range(t S ), we see directly that range(t S ) = {T ( x) x S} = {T (c u + + c k u k ) c,..., c k R} = {c T ( u ) + + c k T ( u k ) c,..., c k R} = span (T ( u ),..., T ( u k )). 0 Example.6.. Consider the linear map T : R 4 R with [T ] =, and let S = 0 4 0 span 0 0,. 0 (a) Find ker(t ), range(t ) (b) Find ker(t S ), range(t S ). Solution:
4 (a) We have ( ker(t ) = span, 4 0, range(t ) = span, 0 0 (b) Converting to relation form and using Theorem.6.6, 0,, ) = R. 4 ker(t S ) = S ker(t ) x x = x x x =, x = x x x x = 0, x x 4 = 0 x = x x x =, x =, x x = 0, x x 4 = 0 = span. 0 Using Theorem.6.6 with S = span 0 0, gives 0 0 ( ) ([ range(t S ) = span T 0 0, T = span, = span, 0]) 0 [ () since span, using Section., Problem 4. 0] 0 Although we are studying linear maps U : S V between subspaces in this section, we have hardly mentioned the codomain V. What role does the codomain V play here? The role of the codomain is simply to contain all of the outputs, namely the range. The codomain can be changed without affecting the([ inputs, ]) the [ outputs, ] or Theorem.6.6, as long as the codomain still contain the range. For example, if x x + x U x = x + 4x, then ([ does NOT make sense since U = ]) ([ U : span ]) [ ] / span ([ span ]) ([ ]). However, ([ ([ ([ U : span span, U : span R ]) ]) ]) ( ()) () both make sense, since range(u) = span U = span by Theorem.6.6. The main purpose of changing the codomain is to make maps surjective. One reason to study restricted linear maps is that injectivity and surjectivity are properties of linear maps that we want, and we can always restrict the domain and codomain of a linear map T : R n R m to make it bijective. We first exhibit an example of this phenomenon and then explain how to restrict the domain and codomain of a linear map T : R n R m to make it bijective in general.
Example.6.8. Consider T : R R linear with [T ] = 0 0, and let S = span 0, 0. 0 Show that U : S range(t ) given by U( x) = T ( x) is bijective. Solution: First, we observe ker(t ) = span, Using Theorem.6.6, ker(u) = ker(t ) S = and Thus, U is bijective. 0 range(t ) = span, 0, = span, 0. 0 0 x x x x = x = x range(u) = span U 0, U 0 = span x x x x = 0 = { 0},, 0 = range(t ). 0 5 Theorem.6.9. Suppose T : R n R m is linear, and S R n is a subspace complementary to ker(t ). Then, is bijective. Proof. First, for injectivity, we have U : S range(t ) ker(u) = S ker(t ) = { 0}, since S, ker(t ) are complements in R n. Hence, U is injective. As for surjectivity, for any y range(t ), we can write y = T ( x) for some x R n. This does not mean y = U( x), since U( x) only makes sense when x S, and x need not be in S. But, we know that S + ker(t ) = R n, so Hence, since v S, w ker(t ). Therefore, U is surjective. x = v + w for some v S, w ker(t ). y = T ( x) = T ( v + w) = T ( v) + T ( w) = T ( v) = U( v), Why might we care about a map being bijective? As Theorem tells us, bijective maps preserve dimension. In Section??, we will use Theorem to deduce a special case of the Rank-Nullity Theorem. Theorem.6.0. If U : S V is bijective, then dim(s) = dim(v ). Proof. Suppose ( a,..., a k ) is a basis for S, which means dim(s) = k. Thus, ( a,..., a k ) spans S, so by Theorem.6.6, range(u) = span (U( a ),..., U( a k )). But also, range(u) = V since U is surjective. Putting these facts together, we conclude that (U( a ),..., U( a k )) spans V. On the other hand, ( a,..., a k ) is linearly independent, and U is injective, so by Section., Problem (U( a ),..., U( a k )) is linearly independent as well. Hence, (U( a ),..., U( a k )) is a basis for V, so dim(v ) = k. Therefore, dim(v ) = dim(w ).
6 Finally, we discuss how the Rank-Nullity Theorem, Theorem.6. generalizes to restricted linear maps. The dimension of the domain is still the sum of the dimensions of the kernel and range. Also, this generalized Rank-Nullity Thoerem still characterizes what ranges and kernels are possible for restricted linear maps. Theorem.6. (Rank-Nullity). Suppose V R n, W R m are subspaces. If T : V W is linear, then dim(range(t )) + dim(ker(t )) = dim(v ). Proposition.6.. Suppose V R n, W R m are subspaces. Given any subspaces S V and S W satisfying dim(s ) + dim(s ) = dim(v ), there exists a linear map T : V W so that Proof. Problem 5. Exercises:. Consider the linear map T : R 4 R with [T ] = of R 4, find ker(t S ) and range(t S ). (a) S = R 4 x (b) S = x x x + x + x + = 0 x (c) S = x x 4x 0x + x + 5 = 0 0 (d) S = span 0 0, 0 0 (e) S = span 0 0, 0 (f) S = span 0, 0 5 8 (g) S = span (h) S = span (i) S = { 0} Problems:. () Suppose S R n is a subspace, and ker(t ) = S, range(t ) = S. 5 0. For each of the following subspaces S
. Suppose T, U : R n R m are linear maps, S R n, V R m are subspaces. Show that restriction to S has the following properties: (a) () (T + U) S = T S +U S, (b) () For any scalar c (ct ) S = c(t S ).. () Suppose T : R n R m is a linear map, and S, S R n are subspaces with S S. Show that (T S ) S = T S 4. () Suppose T, U : R n R n are linear maps, and S R n is a subspace. Prove or give a counterexample: (U T ) S = U S T +S. 5. () Suppose T : R n R m is a linear map, and S R n, V R m are subspaces. Show that (T S ) (V ) = T (V ) S. 6. () Suppose T : R n R m is a linear map, and S R n is a subspace. Show that T S : S R m is the zero map if and only if S ker(t ).. () Suppose U, T : R n R m are linear maps, and S, S are complementary subspaces in R n so that U S = T S and U S = T S. Show that U = T. 8. Suppose T : R n R m is a linear map, and S, S R n are subspaces. For each of the following statements, prove it or give a counterexample: (a) () ker(t S ) ker(t S ) = ker(t S S ), (b) () range(t S ) range(t S ) = range(t S S ), (c) () ker(t S ) + ker(t S ) = ker(t S+S ), (d) () range(t S ) + range(t S ) = range(t S+S ). 9. () Suppose T : R n R k, U : R k R m are linear, and S R k is a subspace so that range(t ) S. Show that U T = U S T. 0. () Prove Proposition.6.5.. () Suppose T : R n R m is linear, and S R n is a subspace. Prove or give a counterexample: If U : S range(t ) given by U( x) = T ( x) is bijective, then S, ker(t ) are complementary subspaces of R n.. () Suppose S, S R n are complementary subspaces, and T : S R m and T : S R m are linear maps. Show that there exists a unique linear map T : R n R m so that T S = T, T S = T.. () Suppose S,..., S k R n are subspaces so that every x R n can be written as x = v + + v k, for some unique v S,..., v k S k. Suppose T : S R m,..., T k : S k R m are linear maps. Show that there exists a unique linear map T : R n R m so that T Sj = T j, for all j =,..., k. 4. () Let S, S be subspaces in R n with S + S = R n. Let T : S R m and T : S R m be linear maps satisfying T S S = T S S. Show that there exists a unique linear map T : R n R m so that T S = T, T S = T.
8 5. Prove Proposition.6.. 6. Let T : R n R k and U : R k R m be linear maps, and let S = ker(u T ). Consider the restriction T S : S R k. (a) () Show that range(t S ) = ker(u) range(t ). (b) () Show that nullity(u T ) = nullity(t ) + dim(ker(u) range(t )).. Let T : R n R k and U : R k R m be linear maps, and let V = range(t ). Consider the restriction U V : V R m. (a) () Show that range(u V ) = range(u T ). (b) () Show that rank(u T ) = rank(t ) dim(ker(u) range(t )). 8. () Let S be a subspace of R n, and let k = dim(s). Show that there exists a bijective linear maps T : S R k and U : R k S. 9. () Suppose S R n, V R m are subspaces. Show that exists a surjective linear map U : S V if and only if dim(s) dim(v ). 0. () Suppose S R n, V R m are subspaces. Show that exists a injective linear map U : S V if and only if dim(s) dim(v ).. () Suppose S R n, V R m are subspaces. Show that exists a bijective linear map U : S V if and only if dim(s) = dim(v ).