Review++ of linear algebra continues

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1 Review++ of linear algebra continues Recall a matrix A P M m,n pf q is an array n A p j i q n m n m This encodes a map ' P HompF n,f m q via the images of basis vectors of F n, written in terms of basis vectors of F m. Namely, to write ' as a matrix, we first fix ordered bases A ta 1,...,a n u of F n and B tb 1,...,b n u of F m, and then compute the coe cients j i such that 'pa j q ÿ jb i i. Then A M B A p'q. iprms Transpose and symmetric/skew-symmetric matrices. Let A p j i q be a matrix. The transpose of A is A t p i jq (swap the columns and rows). Like inverses, transpose distributes by reversing the order: pabq t B t A t. Think: transpose reverses the side by which the linear transformation acts, i.e. if 'puq v, then pmp'quq t v t u t Mp'q t. We call a matrix symmetric if A t A; or skew-symmetric if A t A.

2 Dual spaces Again, we write Hom F pv,wq HompV,Wq tlinear maps ' : V Ñ W u, (homomorphisms) End F pv q EndpV q HompV,V q, (endomorphisms) AutpV q t' P EndpV q ' is invertible u GLpV q. (automorphisms, general linear group) For a vector space V over F,thedual space as V HompV,Fq. Homework: You show Hom F pv,wq is a vector space and that EndpV q is a ring. So V is a vector space over F as well! Dual spaces For a vector space V over F,thedual space as V HompV,Fq. By the homework exercise, V is a vector space over F as well. In which case, good first questions are: What is V s dimension? What is a natural basis for V? What is pv q? Fix a basis B of V.DefinethedualtoB by B tv v P Bu, where v puq u,v for each u, v P B, and extend linearly. Sanity check: V is a set of linear map from V to F, of which v is one, and so v puq should be an element of F. Proposition If V is finite-dimensional, then B is a basis of V ; inparticular, V V as vector spaces. Otherwise, if V is infinite-dimensional, then B does not span V.

3 Dual spaces For a vector space V over F,thedual space as V HompV,Fq. Fix a basis B of V.DefinethedualtoB by B tv v P Bu, where v puq u,v for each u, v P B. Fix an order on B tv 1,...,v n u. Under the natural identification of HompV,Fq with M 1,n pf q (thm from last time) we identify ' P V given by ' i v i with the row vector p 1, 2,..., n qpm 1,n. Further, the evaluation of ' on u n j 1 jv i is 'puq jv j i j v i pv j q i v i j 1 j 1 i i, which is just the dot product of p 1, 2,..., n q p 1, 2,..., nq. Dual spaces Fix an order on B tv 1,...,v n u. Under the natural identification of HompV,Fq with M 1,n pf q (thm from last time) we identify ' P V given by ' i v i with the row vector p 1, 2,..., n qpm 1,n. Further, the evaluation of ' on u n j 1 jv i is 'puq jv j i j v i pv j q i v i j 1 j 1 i i, which is just the dot product of p 1, 2,..., n q p 1, 2,..., nq. Basically, the isomorphism V V can be thought of as taking the transpose (once you ve picked a basis, of course): M t1u B pvq t M B t1u pv q.

4 Forms Let V 1,...,V m,andw be vector spaces over F. Amap ' : V 1 ˆ ˆV m Ñ V is multilinear if it is linear in every coordinate; namely,foreach i 1,...,n,wehave 'pv 1,...,v i 1, v i,v i`1,...,v m q 'pv 1,...,v i 1,v i,v i`1,...,v m q and 'pv 1,...,v i 1,u i ` v i,v i`1,...,v m q 'pv 1,...,v i 1,u i,v i`1,...,v m q ` 'pv 1,...,v i 1,v i,v i`1,...,v m q for any v j P V j, u i P V i. (Caution: Do not confuse this with linear!! The v i are not basis elements like before they re just any old element of V i.) Forms Let V 1,...,V m,andw be vector spaces over F.Amap ' : V 1 ˆ ˆV m Ñ W is multilinear if it is linear in every coordinate. In particular, if V i V for all i, ' is called (m-)multilinear on V. And if, additionally, W F,then' is called a multilinear form on V. In short, a (multilinear) form on V is a map ' : V ˆ ˆV Ñ F that is linear in each coordinate. For m 2 or 3, wesaybilinear or trilinear, respectively, in place of multilinear. We say a form ' is degenerate (in coordinate i) if there are some v 1,...,v j 1,v j`1,...,v m P V such that 'pv 1,...,v j 1,u,v j`1,...,v m q 0 for all u P V. If for all i, nosuchv j s exist, then we say ' is non-degenerate.

5 Forms A (multilinear) form on V is a map ' : V ˆ ˆV Ñ F that is linear in each coordinate. A form ' on V is called symmetric if 'pv 1,...,v m q 'pv p1q,...,v pmq q for any P S m. For example, dot product is a symmetric bilinear form on R m. Similarly, a form ' on V is called alternating if 'pv p1q,...,v pmq sgnp q'pv 1,...,v m q for any P S m, where sgnp q is the sign of the permutation. (Recall sgnp q p q 1, depending on the number of transpositions needed to build.see 3.5 of D&F). Aform' on V is called alternating if 'pv p1q,...,v pmq sgnp q'pv 1,...,v m q for any P S m, where sgnp q is the sign of the permutation. Properties of alternating forms. If ' is alternating, then 'pv 1,...,v m q 0 whenever v i v j for any i j. Proof: Swapping v i and v j should introduce a factor of sgnppijqq 1, but also leaves the things unchanged. For any i j, replacingv i Ñ v i ` v j fixes ': (See D&F Prop ) 'pv 1,...,v i ` v j,...,v j,...,v m q 'pv 1,...,v i,...,v j,...,v m q ` 'pv looooooooooooooooomooooooooooooooooon 1,...,v j,...,v j,...,v m q. 0

6 Bilinear forms A bilinear form on a vector space V is often thought of (and written) as a pairing Specifically, the bilinearity means x, y : V ˆ V Ñ F. xu ` u 1,vy xu, vy`xu 1,vy, xu, v ` v 1 y xu, vy`xu, v 1 y, and x u, vy xu, vy xu, vy. Favorite example: the dot product on R n, given by a b xpa 1,...,a n q, pb 1,...,b n qy a 1 b 1 ` `a n b n. Again, the dot product is symmetric, sincexa, by xb, ay for all a, b P V. An alternating bilinear form is also called skew-symmetric, sinceit satisfies xa, by xb, ay for all a, b P V. Bilinear forms A bilinear form on a vector space V is a pairing x, y : V ˆ V Ñ F. Fact: Non-degenerate bilinear forms x, y are particularly special because the maps V Ñ V defined by v fiñ pu fiñ xv, uyq and V Ñ V defined by v fiñ pu fiñ xu, vyq are both isomorphisms. Example: Let x, y be the usual dot product on R 2, and consider the map f : V Ñ V defined by v fiñ pu fiñ xv, uyq. We claim f is linear, so we need only compute the images of v 1 p1, 0q and v 2 p0, 1q.

7 Bilinear forms via matrices Fix a basis B of V,andidentifyV with column vectors in M n,1 pf q F n. For a matrix J P M n pf q, define x, y j : V ˆ V Ñ F by pu, vq fiñ u t Jv. Sanity check: u t P M 1,n pf q, J P M n pf q, andv P M n,1 pf q, so u t Jv P M 1 pf q F. Now, since x, y J is the composition of linear maps, it is indeed bilinear. X Trace For a matrix A P M n pf q, thetrace of A p i j q is trpaq i, i the sum of the diagonal entries. For example, ˆˆ 2 1 ˆˆ1 2 tr 2 ` 0 2, and tr Theorem (Properties of trace) 1 ` 1 2. (a) Viewed as a map tr : M n pf qñf, trace is a linear transformation. Namely, for all A, B P M n pf q and P F,we have trp Aq trpaq and trpa ` Bq trpaq` trpbq. However, tr is not a ring homomorphism, since in general, trpabq trpaqtrpbq. (b) Trace is invariant under transpose, i.e. trpa t q trpaq.

8 Theorem (Properties of trace) (a) Trace is a linear transformation, i.e. for all A, B P M n pf q and P F,wehavetrp Aq trpaq and trpa ` Bq trpaq`trpbq. However,tr is not a ring homomorphism, since in general, trpabq trpaqtrpbq. (b) Trace is invariant under transpose, i.e. trpa t q trpaq. (c) Trace is invariant under cyclic permutations, i.e. trpabq trpbaq for all A, B P M n pf q. More generally, for all A i P M n pf q, trpa 1 A 2 A`q trpa 2 A`A 1 q trpa`a 1 A 2 A` 1 q. (d) Trace is invariant under change of basis. Namely, if A and B are both bases for a vector space V,and' P EndpV q, then trpm A A p'qq trpm B B p'qq. By (d), the trace of an endomorphism ' P EndpV q is well-defined by trp'q trpm B B p'qq where B is any basis of V.