MATRIX THEORY (WEEK 2)

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1 MATRIX THEORY (WEEK 2) JAMES FULLWOOD Example 0.1. Let L : R 3 R 4 be the linear map given by L(a, b, c) = (a, b, 0, 0). Then which is the z-axis in R 3, and ker(l) = {(x, y, z) R 3 x = y = 0}, im(l) = {(x, y, z, w) R 4 z = w = 0}. The nullity of L is then 1, and the rank of L is 2. Whenever you define a class of maps, the composition of two maps in the class should be of the same class. The case of linear maps follows from the following Proposition 0.2. Let L : V W and M : W Z be linear maps. Then the composition M L : V Z is linear as well. Proof. Let v, w V and c K. Then (M L)(cv + w) = M(L(cv + w)) = M(cL(v) + L(w)) = cm(l(v)) + M(L(w)) = c(m L)(v) + (M L)(w) Dual spaces. Let V and W be vector spaces over a field K, and let Hom(V, W ) = {linear maps L : V W }. Then Hom(V, W ) is a vector space over K. Indeed, given L 1, L 2 Hom(V, W ) and c K, then cl 1 + L 2 is the linear map given by (cl 1 + L 2 )(v) = cl 1 (v) + L 2 (v). If V and W are finite dimensional then is it straightforward to show Hom(V, W ) is finite dimensional as well, and moreover, dim (Hom(V, W )) = dim(v ) dim(w ). (0.1) For any vector space V over K, the vector space Hom(V, K) is referred to as the dual space of V, and is denoted V. From formula (0.1) we see dim(v ) = dim(v ) 1 = dim(v ), so that V is isomorphic to V. Given a basis e = {v 1,..., v 2 } of V, the associated dual basis of V, given by e = {v1,..., vn}, is such that { vi 1 if v = v i (v) = 0 otherwise. An explicit isomorphism between V and V is then induced by mapping v i to v i. 1

2 Example 0.3 (On the meaning of differentials). Let p = (x 0, y 0 ) R 2, and consider the vector space T p R 2 introduced in Example 1.4. Now given an arrow v p T p R 2 whose tip has coordinates (x 1, y 1 ) R 2, we will often write v p = x 1 x 0, y 1 y 0 p, which we refer to as the component notation of v p. Now let (a, b) = (x 1 x 0, y 1 y 0 ), so that v p = a, b p, and let F(R 2 ) be the set of smooth real-valued functions on R 2 (i.e., infinitely differentiable functions with continuous partial derivatives of all orders). Then with v p T p R 2 we can associate a map v p : F(R 2 ) R given by v p (f) = a f (p) + b f x y (p), which we refer to as the vectorial derivative of f with respect to the vector v p. Dynamically, we can think of v p (f) as the rate of change of f along a curve passing through p with velocity v p. By thinking of v p as a real-valued map on differentiable functions, we can then associate with every f F(R 2 ) an element df in the dual space (T p R 2 ), which refer to as the differential of f. More precisely, df acts on a vector v p T p R 2 via the formula df(v p ) = v p (f). (0.2) In particular, from equation (0.2) it immediately follows that {dx, dy} is the dual basis of {, }, where x denotes the function f(x, y) = x and y denotes the function g(x, y) = y. We then have df(v p ) = a f (p) + b f x y (p) = v p (x) f x (p) + v p(y) f y (p) so that = f x (p)dx(v p) + f y (p)dy(v p) ( ) f f = (p)dx + x y (p)dy (v p ), df = f f (p)dx + (p)dy. (0.3) x y Formula (0.3) should be familiar from calculus texts, which now should have a clear meaning The concept of a canonical isomorphism. The isomorphism between a vector space V and its dual induced by sending a basis e = {v 1,..., v n } to its associated dual e = {v1,...vn}, depends manifestly on e. If we consider a basis e = {v 1,..., v n} different from e, then the isomorphism induced by mapping v i to v i will be a different linear map than the one induced by mapping v i to vi. Now let L : V W be a general isomorphism of vector spaces. The map L is said to be a canonical isomorphism if and only if it may be defined without any reference to a basis of V, and in such a case, V and W are said to be canonically isomorphic. As the choice of a basis of a general vector space is arbitrary, a canonical isomorphism between vector spaces V and W is an isomrphism which doesn t depend on arbitrary choices, and thus is more natural 2

3 from an abstract point of view. In particular, the isomorphism between a vector space V and its dual V by sending a basis e of V to its dual e is not a canonical isomorphism, and more generally, one can show that a canonical isomorphism between V and V does not exist. Now consider the double dual V = (V ) of a vector space V. An element λ of V is then a linear map λ : V K. It turns out that V is canonically isomorphic to V, which we now show by constructing a canonical isomorphism L : V V. For this, given an element v V we have to specify a linear map L(v) : V K, without making any reference to a basis of V. This turns out to be easy (because it s canonical). For f V, simply let L(v)(f) = f(v) K. (0.4) Since no reference to basis appears in (0.4), and L is injective, L is in fact a canonical isomorphism, so that a vector space V is always canonically isomorphic to its double dual V. Example 0.4. Let L Hom(V, W ). Then L induces a map L Hom(W, V ), referred to as the dual of L. More precisely, L takes an element f W, which is a linear map of the form f : W K, to the map L (f) = f L : V K, which is linear (since compositions of linear maps are linear). Moreover, the map L L is in fact a canonical isomorphism between Hom(V, W ) and Hom(W, V ) Direct sums. Given two vector spaces V and W over a field K, we can form another vector space V W over K, referred to as the direct sum of V and W. As a set, the direct sum V W is just the cartesian product V W, and the vector space structure is given by c(v 1, w 1 ) + (v 2, w 2 ) = (cv 1 + v 2, cw 1 + w 2 ), for all v 1, v 2 V, w 1, w 2 W and c K. It then follows that 0 V W = (0 V, 0 W ). There are four linear maps canonically associated with the direct sum V W, namely, the two injective maps ι V : V V W and ι W : W V W given by v (v, 0 W ) and w (0 V, w) respectively, and the two surjective maps π V : V W V and π : V W W given by (v, w) v and (v, w) w respectively. The maps ι V and ι W are referred to as the inclusions of V and W into V W, and the maps π V and π W are referred to as the projections from V W onto V and W. The subspaces im(ι V ) and im(ι W ) of V W are then canonically isomorphic to V and W respectively, so that we may naturally identify im(ι V ) with V and im(ι W ) with W. Moreover, im(ι V ) im(ι W ) is canonically isomorphic to V W. As such, when working with a direct sum V W we will often denote im(ι V ) by V and im(ι W ) by W, and in such a case, an element (v, 0 W ) im(ι V ) will then be denoted simply by v and an element (0 V, w) im(ι W ) will be denoted simply by w. Note that after identifying V with ι V (V ) and W with ι W (W ), we have 0 V = 0 W = (0 V, 0 W ) = 0 V W. Proposition 0.5. Let e V and e W be bases of vector spaces V and W over a field K. Then e V e W is a basis of V W. Proof. Suppose I have a linear combination of elements of e V and e W which give 0 V W, namely, (a 1 v a n v n ) + (b 1 w b m w m ) = 0 V W, 3

4 with v i e V, w j e W, and a i, b j K. Now let v = a 1 v a n v n and w = b 1 w b m w m. Then 0 V W = v + w = (v, w). But this implies v = 0 V and w = 0 W, so that e V e W is a linearly independent set. Moreover, any element (v, w) V W may be expressed as the sum (v, w) = (v, 0 W ) + (0 V, w), and since (v, 0 W ) Span(e V ) and (0 V, w) Span(e W ), it follows that Span(e V e W ) = V W. Corollary 0.6. Let V and W be vector spaces over a field K. Then every element of ϑ V W be be expressed uniquely as a sum ϑ = v + w, with v V and w W. Moreover, if V and W are finite dimensional, then dim(v W ) = dim(v ) dim(w ). Proof. Both claims are immediate consequences of Proposition 0.5. In the definition of direct sum, we started with two vector spaces V and W, and cooked up a new vector space V W in such a way that V and W may be canonically identified with subspaces of V W. Moreover, the subspaces V and W of V W are such that V W = {0 V W } and dim(v ) + dim(w ) = dim(v W ). From a dual perspective, given a vector space V, the next proposition says precisely when V may be given the structure of a direct sum V = V 1 V 2, where V 1 and V 2 are subspaces of V. Proposition 0.7. Let V 1 and V 2 be subspaces of a finite dimensional vector space V over a field K. Then V is canonically isomorphic to V 1 V 2 if and only if V 1 V 2 = {0 V } and dim(v 1 ) + dim(v 2 ) = dim(v ). In such a case, we write V = V 1 V 2, and refer to V as a direct sum decomposition of the subspaces V 1 and V 2. Proof. Suppose V is canonically isomorphic to V 1 V 2. Then dim(v ) = dim(v 1 V 2 ) = dim(v 1 ) + dim(v 2 ), where the second equality follows from Corollary 0.6. Moreover, since V 1 consists of pairs (v 1, 0 V2 ) with v 1 V 1 and V 2 consists of pairs (0 V1, v 2 ) with v 2 V 2, it follows that V 1 V 2 = {(0 V, 0 V2 )} = {0 V }. For the reverse implication, suppose V 1 and V 2 are subspaces of V such that V 1 V 2 = {0 V } and dim(v ) = dim(v 1 ) + dim(v 2 ). We then have to show that V is canonically isomorphic to V 1 V 2. Now since any element of V 1 V 2 is of the form (v 1, v 2 ) with v 1 V 1 and v 2 V 2, we let L : V 1 V 2 V be the map given by L((v 1, v 2 )) = v 1 + v 2 V. Then it is clear that the conditions V 1 V 2 = {0 V } and dim(v ) = dim(v 1 )+dim(v 2 ) imply L is a linear isomorphism, and since L doesn t depend on the choice of a basis, it is a canonical isomorphism. Given vector spaces V 1, V 2,...V n, over K, one can iterate the process of taking direct sums to form the vector space V = V 1 V 2 V n. In such a case, each V i is canonically isomorphic to a subspace of V via an associated inclusion map ι Vi : V i V, and we say that V 1 V 2 V n is a decomposition of V into a direct sum of the subspaces im(ι Vi ) = V i V, 4

5 or rather, that V = V 1 V 2 V n is a direct sum decomposition of V. In such a case, it follows by from Proposition 0.7 and induction that V i V j = 0 V for i j and that dim(v ) = dim(v 1 ) + + dim(v n ). Moreover, it follows that every v V may be expressed uniquely as a sum v = v v n, with v i V i Eigenspaces. Let V be a finite dimensional vector space over a field K, and let L : V V be a linear operator (which we recall is just a linear map from a vector space to itself). An element λ K is said to be an eigenvalue of L if and only if there exists a non-zero vector v V such that L(v) = λv. In such a case, v is said to be an eigenvector associated with the eigenvalue λ of L. If λ K is an eigenvalue of a linear operator L : V V, let E λ V be the set given by For any c K and v, w E λ we have E λ = {v V L(v) = λv}. L(cv + w) = cl(v) + L(w) = c(λv) + (λw) = λ(cv + w), so that E λ is in fact a subspace of V, which we refer to as the eigenspace of V associated with the eigenvalue λ of L. Note that since an eigenvector is required to be non-zero, it follows that dim(e λ ) 1 for any eigenvalue λ of a linear operator L : V V. Moreover, it follows from the definition of eigenspace that if λ 1 and λ 2 are distinct eigenvalues of a linear operator L : V V, then E λ1 E λ2 = 0 V, so that L can have no more than dim(v ) distinct eigenvalues in the case that V is finite dimensional. Example 0.8. Let L : K 2 K 2 be the linear operator given by L(x, y) = ( y, x). As such, if λ is a eigenvalue of L then there exists a non-zero (x 0, y 0 ) K 2 such that (x 0, y 0 ) = ( λy 0, λx 0 ), so that x 0 = λy 0, and y 0 = λx 0. Substituting the second equation in the first yields x 0 = λ 2 x 0 = λ = 0. We then see L admits an eigenvalue if and only if the polynomial z = 0 admits zeros in K. In particular, for K = R, L admits no eigenvalues, and for K = C, L admits the eigenvalues ±i, where i is the imaginary unit in C. For K = C, we have E i = {(x 0, ix 0 ) C 2 } and E i = {(x 0, ix 0 ) C 2 }. Moreover, we have {(1, i)} is a basis of E i, {(1, i)} is a basis of E i, and by Proposition 0.7 we have C 2 = E i E i. School of Mathematical Sciences, Shanghai Jiao Tong University, 800 Dongchuan Road, Shanghai, China address: fullwood@sjtu.edu.cn 5