Vector Spaces. EXAMPLE: Let R n be the set of all n 1 matrices. x 1 x 2. x n
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1 Vector Spaces DEFINITION: A vector space is a nonempty set V of ojects, called vectors, on which are defined two operations, called addition and multiplication y scalars (real numers), suject to the following axioms (or rules): (A) The sum of u and v, denoted y u+v, is in V (B) The scalar multiple of u y c, denoted y cu, is in V (i) u+v = v+u (ii) (u+v)+w = u+(v+w) (iii) There is a unique element in V, called the zero vector, such that u+ = u (iv) For each u in V, there is a unique vector u in V such that u+( u) = (v) u = u (vi) a(u) = (a)u (vii) a(u+v) = au+av (viii) (a+)u = au+u These axioms must hold for all vectors u, v, and w in V and all scalars a and EXAMPLE: Let R n e the set of all n matrices Define u+v and cu as the vector addition and scalar multiplication defined in Section, that is y +y y +y u+v = + y n = +y n and Then R n is a vector space In fact, (A) u+v is in R n cu = c = c c c (B) cu is in R n
2 (i) Since x i +y i = y i +x i for all real numers x i,y i, it follows that u+v = v+u Indeed, y +y y + y y +y y + y u+v = + y n = = +y n = y n + y n + = v+u In the same way one can prove (ii): Since (x i + y i ) + z i = x i + (y i + z i ) for all real numers x i,y i,z i, it follows that (u+v)+w = u+(v+w) (iii) The zero vector satisfies u+ = u, since (iv) For each u = u+ = + = = in Rn, there is the vector u = u+( u) = + = = +( ) +( ) +( ) = u in Rn such that u+( u) =, since = = (v) u = u, since u = = = = u (vi) a(u) = (a)u, since a(u) = a = a = = a( ) a( ) a( ) (a) (a) (a) = (a) = (a)u
3 (vii) a(u+v) = au+av, since a(u+v) = a (viii) (a+)u = au+u, since + (a+)u = (a+) y y y n = a +y +y +y n a( +y ) a( +y ) = a( +y n ) a +ay a +ay = a +ay n a a ay = + ay a ay n = a (a+) (a+) +a = (a+) a + a + = a + a a = + a = a + y y y n = au+av = au+u
4 EXAMPLE: Let V e the set of all n m matrices Define u+v and cu in the following way: m u+v = m + m and au = a Then V is a vector space In fact, (A) u+v is in V (B) cu is in V The axioms (i) and (ii) are true, since and for all real numers x ij,y ij,z ij (iii) The zero vector satisfies u+ = u (iv) For each u = u+( u) = m m m One can show that (v) (viii) are also true m m m y y m y y m y n y nm m m m = = x ij +y ij = y ij +x ij +y m +y m +y m +y m +y n m +y nm a am a am a am (x ij +y ij )+z ij = x ij +(y ij +z ij ) = in V, there is the vector u = EXAMPLE: The set P n of all polynomials of degree at most n: p(t) = a n t n ++a t +a t+a m m m where the coefficients a n,,a and the variale t are real numers is a vector space EXAMPLE: The set of all real-valued functions defined on R is a vector space in V such that
5 EXAMPLE: Here are some examples of sets that are not vector spaces: The set of all vectors u = is not a vector space, since axiom (B) fails The set of all vectors u = x z where x i where x,z are all real numers is not a vector space (there is no ) One can check, however, that the set of all vectors is a vector space (see Appendix I) v =< x,,z > The set V of all vectors u =< x,,x > where x is any real numer is not a vector space, since axiom (A) fails Indeed, if u =<,, > and u =<,,8 >, then u +u =<,,9 > which is not in V The set of all polynomials of degree n with n > is not a vector space (there is no ) DEFINITION: A suspace of a vector space V is a suset H of V that has properties: The zero vector of V is in H H is closed under vector addition That is, for each u and v in H, the sum u+v is in H H is closed under multiplication y scalars That is, for each u in H and each scalar c, the vector cu is in H THEOREM : A suspace H of a vector space V is a vector space EXAMPLE: Let H e the set of all functions x(t) which satisfy the differential equation d x x = () dt with the sum of two functions and the product of a function y a numer eing defined in the usual manner That is to say, (f +f )(t) = f (t)+f (t) and (cf)(t) = cf(t) It is easy to verify that H is a vector space We first note that H is a suset of the vector space of all real-valued functions Let us show that this is a suspace Indeed, is from H, since x(t) = is a solution of () If (t) and (t) are in H, then (t)+ (t) is in H, since the differential equation () is linear Similarly, if x(t) is in H, then cx(t) is in H So, H is a suspace of the vector space of all real-valued functions Therefore H is a vector space y Theorem aove 5
6 EXAMPLE: Let H e the set of all functions x(t) which satisfy the differential equation d x dt 6x = with the sum of two functions and the product of a function y a numer eing defined in the usual manner H is not a vector space since the sum of any two solutions, while eing defined, is not necessarily in H Similarly, the product of a solution y a constant is not necessarily in H For example, the function x(t) = t is in H since it satisfies the differential equation (see Appendix II), ut the function x(t) = t is not in H since it does not satisfy the differential equation THEOREM : If v,,v p are in a vector space V, then the set of all linear cominations of v,,v p is a suspace of V (and therefore a vector space) c v ++c p v p EXAMPLE: Let v =, v = The set of all linear cominations of v and v is a vector space y Theorem aove c v +c v = c EXAMPLE: Let H e the set of all vectors of the form a c c a +c where a,, and c are aritrary scalars Show that H is a vector space Solution: We have a c c a = a + ( ) + c a + + ( ) c ( ) a + + c a + + c = a v + v +c v Since v,v,v are in the vector space R and H is the set of all linear cominations of v,v,v, it follows that H is a vector space y Theorem aove 6
7 Appendix I Here we show that the set H of all vectors < x,,z >, where x,z are real numers, is a vector space Let u = v = x w = z z z Solution : (A) u+v is in H, since u+v = z + z = + z +z (B) cu is in H, since cu = c z = c cz (i) We have u+v = z + z = + z +z = + z +z = z + z = v+u In the same way one can prove (ii): Since ( + )+x = +( +x ) and (z +z )+z = z +(z +z ), it follows that (u+v)+w = u+(v+w) (iii) The zero vector satisfies u+ = u, since u+ = z + = = + + z + = z = u (iv) For each u = z in H, there is the vector u = u+( u) = z + z = z +( ) + z +( z ) in H such that u+( u) =, since = = (v) u = u, since u = z = z = z = u 7
8 (vi) a(u) = (a)u, since a(u) = a z = a z = = a( ) a(z ) (a) (a)z = (a) z = (a)u (vii) a(u+v) = au+av, since a(u+v) = a z + z = a = = = = a + z +z a( + ) a(z +z ) a +a az +az a az z + +a a az z = au+av (viii) (a+)u = au+u, since (a+)u = (a+) z = = = (a+) (a+)z a + az +z a az = a z + z + z = au+u 8
9 Solution : We first show that the set H of all vectors < x,,z >, where x,z are real numers, is a suspace of R Indeed, The zero vector <,, > of R is oviously in H If u and v are in H, then u+v is in H, since u+v = + z z = + z +z Similarly, if u is in H, then cu is in H, since cu = c z = c cz So, H is a suspace of the vector space R y the definition of a suspace Therefore H is a vector space y Theorem aove Solution : We have x z = x + z x + z x + z = x v +z v Since v,v are in the vector space R and H is the set of all linear cominations of v,v, it follows that H is a vector space y Theorem 9
10 Appendix II One can easily check that the function x(t) = t satisfies d x dt 6x = plugging it into the equation Here we show how to come up with this function without guessing To this end, we first rewrite it as d x dt = 6x Multiplying oth sides y dx, we get dt d By the Chain Rule, the left-hand side is d dt Integrating oth sides, we otain ( d dt x dt dx dt = 6x dx dt ( ( ) ) dx and the right-hand side is d dt dt ( ) ) dx = d ( ) x dt dt ( ) dx = x dt ( ) dx = x dt dx dt = x/ dx = dt x/ dx x = dt / x /+ /+ = t x / / = t x / = t (x / ) = ( t) x = t ( x ) So, we have x = t = t
11 Appendix III EXAMPLE: Let H e the set of all vectors of the form a a a where a and are aritrary scalars Show that H is a vector space Solution: We have a a a = a v + v Since v,v are in the vector space R and H is the set of all linear cominations of v,v, it follows that H is a vector space y Theorem EXAMPLE: Let H e the set of all vectors of the form a+ a 5 where a and are aritrary scalars Show that H is not a vector space Solution: H is not a vector space, since H (the second entry is always nonzero) EXAMPLE: Let V e the set of vectors in R where a and are real numers Then a a A V is a vector space, since is in V B V is not a vector space, since is not in V C V is a vector space, since u+v is from V for any u and v from V D V is a vector space, since cu is from V for any u from V and any c from R E V is not a vector space, since there exist u and v from V such that u+v is not from V F None of the aove Solution: Let u = (here a =, = )
12 and then However, u+v = v = (here a =, = ) 6 = = since 6 = a implies a =, 9 = implies =, therefore 5 a So, the correct answer is E Info: This prolem was given in Spring 7 (Midterm II) The average in the class for this prolem was 6% EXAMPLE: Let V e the set of vectors in R where a and are real numers Show that V is not a vector space Solution: Let and then However, u+v = u = v = + 5 a a a a (here a =, = ) (here a =, = ) = since otherwise a =, = and a = 5, which is impossile a a =
13 EXAMPLE: Let V e the set of vectors a in R where a and are real numers Show that V is not a vector space Solution: Let and then However, u+v = u = v = + (here a =, = ) (here a =, = ) = since otherwise = and a =, which is impossile +( ) + +( ) a = REMARK : Note that if we set a =, = and a =, = for u and v (like we did in the two previous examples), we will not get a contradiction Indeed, in this case we have But which is equal to a u+v = u = + if a = 5/ and = and v = = = 5
14 REMARK : If V is the set of vectors a+ in R where a and are real numers, then V is a vector space Indeed, we have a+ = a + v v Since v,v are in the vector space R and V is the set of all linear cominations of v,v, it follows that V is a vector space y Theorem EXAMPLE: Let W e the set of vectors and let W e the set of vectors where x and y are all real numers Then u = v = x x+ y x y x+y A W is a vector space, ut W is not a vector space B W is not a vector space, ut W is a vector space C W is a vector space and W is a vector space D W is a not vector space and W is not a vector space correct Info: This prolem was given in Fall 7 (Midterm Exam II) The average in the class for this prolem was 7% EXAMPLE: Let W e the set of all polynomials of the form p(t) = at +t+, where a, are real numers Then A W is not a vector space, since it is not closed under multiplication y a scalar B W is not a vector space, since is not in W C W is not a vector space, since it is not closed under addition D All of the aove correct E None of the aove Info: This prolem was given in Fall 7 (Final Exam II) The average in the class for this prolem was %
15 REMARK: If W is the set of all polynomials of the form p(t) = at +t, where a, are real numers, then W is a vector space Indeed, putting v = t and v = t we get p(t) = av +v Since v,v are in the vector space P and W is the set of all linear cominations of v,v, it follows that W is a vector space y Theorem EXAMPLE: Let W e the set of singular matrices under the usual operations Then A W is not a vector space, since it is not closed under addition correct B W is not a vector space, since is not in W C W is not a vector space, since it is not closed under multiplication y a scalar D W is not a vector space, since it is not a suspace of the vector space of matrices E None of the aove Info: This prolem was given in Fall 7 (Final Exam I) The average in the class for this prolem was % REMARK: A singular matrix is a square matrix that does not have a matrix inverse A matrix is singular if and only if its determinant is (see Sections 5, 6) 5
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