in any vector space ax=bx implies that a=b

such that there exists a vector x with Ax = b.Thus we have the following Theorem. (e)A vector in Fn may be regarded as a matrix in M n 1(F). Label the following statements as true or false. First we recall some backgrounds. (c) In any vector space, ax = bx implies that a = b. This de nitely can't happen. We briefly discuss these below. Then (a + ( b))x = 0. Solution: Any Bsatisfying the given equation has to be a 2 3 matrix which implies that the null space of B, N(B), can not be trivial and hence we either have an in nitely many solutions (when [0 1]T lies in the column space of B) or no solution. This implies that P⊥ is the row space of A. 2. x (a - b) = 0 either x = 0 or ( a-b) = 0. take x = 0 and any a≠b for x. a vector space over F. To check that a subset U ⊂ V is a subspace, it suffices to check only a couple of the conditions of a vector space. Given an m-dimensional real vector space V and a basis B of V, any element of ℝ m is the coordinate vector of some vector of V with respect to B; Prove that there are infinitely many bases of ℝ 2 one of whose members is (1, 0). (b) A vector space may have more than one zero vector. Let U ⊂ V be a subset of a vector space V over F. Then U is a subspace of V if and only if 1. additive identity: 0 ∈ U; 2. closure under addition: u,v ∈ U implies u+v ∈ U; In that proof there must be some "other" basis vector- if the vector space is one dimensional, and x is not 0, then Ax=Bx implies A= B! We have U ˆV since U is a subspace of V, and we have just shown that V ˆU. (c) In any vector space ax = bx implies that a = b. A left module of A A is an additive abelian group ( M, +) ( M, +), together with an ring operation A × M → M A × M → M such that ( a + b) x = a x + b x a ( x + y) = a x . A vector in Fn may be regarded as a matrix in Mn×1 (F ). Is V a vector space over R with these operations? In any vector space, ax = ay implies that x = y. I have a test tomorrow and the teacher said we should be able to prove these; Question: Linear Algebra Theory I need solid straight forward proofs of Theorem 1.2 a, b, and c. As you can . A vector in Fn may be regarded as a matrix in Mn×1 (F ). (d) In any vector space, ax = ay implies that x = y. (d) In any vector space, au = av implies u = v. 1.3 Subspaces It is possible for one vector space to be contained within a larger vector space. (c) In any vector space, au = bu implies a = b. Exercise 2. T (b) A vector space may have more than one zero vector. An m × n matrix has m columns and n rows. Since Ax = 0 if and only if −Ax = 0, we see N(A) = N(−A). implies x 1 = x 2 for any x 1;x 2 2X. Symmetric and Hermitian matrices. Proof: If Ais invertible, substituting A 1b into the equation gives A(A 1b) = (AA 1)b = I nb = b so it is a solution. (F) not so when x=0 In any vector space, ax=ay implies that x=y. Any scale or multiple of 3, 1 is the null space. Explanation: (I guess if it were written "properly" it would be ax=bx implies a=b). Therefore V ˆU. Suppose A A is a ring with multiplicative identity 1 A 1 A. 0 = 0 for every α ∈ F. 7. Let P3 be the vector space of all polynomials with degree at most 3. In any vector space, ax = bx implies that a = b. Let V= {( a1, a2) : a1, a2 1,a 2), ( b 1,b 2 (a 1, a2) + (b 1, b2) = (a 1+2b 1 , a 2+3b 2) and c(a 1,a 2) = (ca 1,ca 2). If Ax= Bx for ALL x, then see what happens when you apply A and B to each of (1, 0, 0,. In P(F ), only polynomials of the same degree may be added. d. In any vector space ax = ay implies that x = y. )This subset actually forms a subspace of R . Since A is m by n, the set of all vectors x which satisfy this equation forms a subset of R n. (This subset is nonempty, since it clearly contains the zero vector: x = 0 always satisfies A x = 0. 7 (d) In any vector space, ax = ay implies that x = y. Notation: When the same vector norm is used in both spaces, we write jjAjj c=maxjjAxjj c s.t. Then we have, . In any vector space V, show that (a+b)(x+y) = ax + ay + bx + by for any x, y EV and any a, b e F. This problem has been solved! (b) A vector space may have more than one zero vector. (b)A vector space may have more than one zero vector. (2)(a)If V is a vector space and W is a subset of V that is a vector space, then W is a subspace of V. (b)The empty set is a . Vector Space. We we'll frequently regard vectors as row Let A be a given mxn matrix. A subset of a vector space V that is itself a vector space is a subspace of V. Theorem 2.1.2. (b)A vector space may have more than one zero vector. The function f: R !R given by f(x) = x2 is not injective as, e.g., . Problem 711. (a + b)x = ax + bx if ab = 0 5. The system has exactly one solution, A 1b, i Ais invertible. (a) Every vector space contains a zero vector. This section will look closely at this important . Label the following statements as true or false. If A is right divisible by B, then clearly Bx = 0 implies Ax = 0 for any column vector x ∈ Fn. 1. Obviously, this vector by itself would also be a solution to Ax is equal to b, because you can just set x2 to be equal to 0. An m × n matrix has m columns and n rows. Is the converse true? We define the vector norm of a matrix A by kAk = max kxk=1 kAxk. Label the following statements as being true or false. (a) Every vector space contains a zero vector. If E is any real or complex vector space of finite dimension, then any two norms on E are equivalent. Label the following statements as true or false. (a) Every vector space contains a zero vector. Thus, nding a Bthat Since U is a vector space, any linear combination of elements of U is also in U. In mathematics, physics, and engineering, a vector space (also called a linear space) is a set whose elements, often called vectors, may be added together and multiplied ("scaled") by numbers called scalars.Scalars are often real numbers, but can be complex numbers or, more generally, elements of any field.The operations of vector addition and scalar multiplication must satisfy certain . Let V 1 and V 2 be vector spaces of n-vectors. Note: If A is not square, then this equation is invalid. V called addition and scalar multiplication. Any vector space over a finite field is necessarily finite-dimensional. (a) Every vector space contains a zero vector. (Remember, the empty set is not a vector space.) Solution True. a vector space. Notations and preliminaries. with the results denoted x+y and ax respectively, that satisfy the following axioms: VS 1 (Commutativity of addition) For all x;y 2 V; we have x+y = y +x. vectors spaces: The 3-d space of vectors, the vector space of all polynomials of a fixed degree, and vector spaces of n×n matrices. A1)(a) Let A and B be m×n and k×n matrices, respectively, with entries in a field F. We say that A is right divisibleby B if there exists an m × k matrix C such that A = CB. Nevertheless, if we redefine the vectors as a=3i+5j+0k and b=2i+4j+0k, we could define the cross product as . A related statement -- also listed as false -- is that "in any vector space, ax=ay implies that x=y." Again, given the axioms we have. Answer (1 of 10): The problem itself does not say that A is a square matrix (in fact it may be NOT a square one, one says that a matrix is invertible or not only when the matrix is square), so let's treat it as an m by n matrix, and the vector x should be a n-D column vector, A^T A is a n-squar. Problem 2: (15=6+3+6) (1) Derive the Fredholm Alternative: If the system Ax = b has no solution, then argue there is a vector y satisfying ATy = 0 with yTb = 1. a;b: Rm n!R de ned as jjAjj a;b=max x jjAxjj a s.t. (c)In any vector space, ax = bx implies that a = b. Since a vector space contains a vector v if and only if it contains the vector −v, we see C(A) = C(−A). In particular, any p-norm. (ii) A vector space may have more than one zero vector. (2)(a)If V is a vector space and W is a subset of V that is a vector space, then W is a subspace of V. (b)The empty set is a . 11.Let Ax = b be a . Let A be an m×n matrix. 7 (d) In any vector space, ax = ay implies that x = y. 1 Vector space de-nition De-nition 1 A vector space V over a -eld F is a set together with two opera-tions: + : V V ! (c)In any vector space, ax = bx implies that a = b. C'[R'] is the n-dimensional complex [real] vector space. (a) Every vector space contains a zero vector. (e) A vector in F" may be regarded as a matrix in M.„xi (F). Scalars are usually considered to be real numbers. ), (0, 1, 0.). Here, letting A = 1 1 0 1 and c = 2, we see that cA = cA = 2 2 0 2 2= V Note: I actually wanted this to be a vector space! that any continuous real-valued function on a nonempty compact set has a minimum and a maximum, and that they are achieved. (a) Every vector space contains a zero vector. Then V 1 ∩ V 2 is a vector space. Justify why its true of false vector space ax = bx implies that a=b? Then V 1 ∩ V 2 is a vector space. Now ax,bx,ax+bx and (a+b)x are all in U by the closure hypothesis. Vector Spaces and Linear Transformations Beifang Chen Fall 2006 1 Vector spaces A vector space is a nonempty set V, whose objects are called vectors, equipped with two operations, called addition and scalar multiplication: For any two vectors u, v in V and a scalar c, there are unique vectors u+v and cu in V such that the following properties are satisfled. We used vector norms to measure the length of a vector, and we will develop matrix norms to measure the size of a matrix. Applying Ato w, we get Aw= Ax Ax= 0, so w2Nul(A). Label the following statements as true or false. (d)In any vector space, ax = ay implies that x = y. (b)The only solution to Bx = 0 1 is x = 2 4 1 2 3 3 5. For instance, if Ax = b has no solutions, then Ax = 2b jjxjj c 1: Examples: 4 implies the dependence of row vectors, and vice versa. (f) a 0 = 0 for every scalar a. Note. Check this. Therefore, U = V. Label the following as true or false: 1)In any vector space, ax=bx implies a=b 2)In any vector space, ax=ay implies that x=y. (F) not so when a=0 18. True. This is NOT a vector space. The solution sets of homogeneous linear systems provide an important source of vector spaces. Mono-tonicity of complex matrices was studied independently in [4]. Let V= {( a1, a2) : a1, a2∈ℝ }. Let S 1 be the set of rst{order sta-tionary points for the problems Pand let S 2 be the set of rst{order stationary points for the problem minfh(z) : z2IRng. Terms in this set (14) 1. (c) Ax = b has no solutions for some b 2Rm and one solution for every other b 2Rm. In any vector space V, show that (a + b)(x + y) = ax + ay + bx + by for any x;y 2V and any a;b 2F. In any vector space, ax = bx implies that a = b. Problem 3. Additional Structure on Vector Spaces. False. F (c) In any vector space, ax = ba implies that arb. Let A be an m by n matrix, and consider the homogeneous system. (a) Every vector space contains a zero vector. It is straightforward to show that this definition yields a norm on the vector space of all m × n matrices. jjxjj b 1; where jj:jj a is a vector norm on Rm and jj:jj b is a vector norm on Rn. T (b) A vector space may have more than one zero vector. Answer (1 of 2): (a) The cross product of the two vectors is not defined as these vectors are defined only in two dimensional space, and three dimensions is required for the cross product. (F) not so when a=0 18. In any vector space, ax = ay implies that x = y. Label the following statements as true or false. (I guess if it were written "properly" it would be a x = b x implies a = b ). The equality is due to vector space properties of V.Thus(i)holdsforU.Each of the other axioms is proved similarly. Proof: By de nition, F is closed under addition, so write a+ b := c 2F. The Vector Space of 3-Dimensional Vectors The vectors in <3 also form a vector space, where in this case the vector operations of addition and scalar multiplication are done . Given the axioms we were given, it would seem that the statement should be true, no? Academia.edu is a platform for academics to share research papers. The actual proof of this result is simple. (e) A vector in Rn may be regarded as a matrix in M n 1(R). Define T:Rn 6 Rm by, for any x in Rn, T(x) = Ax.Then T is a linear transformation. An m × n matrix has m columns and n rows. Justig why ture of [2] In any ; Question: True of False , Justify why true of false DIF va vector space and wis a subset of v that is a vector space then w is a subspace of ev? with vector spaces. Let U ⊂ V be a subset of a vector space V over F. Then U is a subspace of V if and only if 1. additive identity: 0 ∈ U; 2. closure under addition: u,v ∈ U implies u+v ∈ U; In any vector space ax = bx implies that a = b. A vector in Fn may be regarded as a matrix in Mn×1 (F ). 7 (d) In any vector space, ax = ay implies that x = y. Justify why its true of false vector space ax = bx implies that a=b? TRUE FALSE Solution: From ax = bx, we can add (bx) = bx to both sides, and since bx + ( bx) = 0, we have ax + ( bx) = 0. particular subset of a vector space is in fact a subspace. V: F V ! (iv) In any vector space, ax = ay implies that x = y (for vectors x and y and a scalar a). Label the following statements as true or false. Problem 4. Definition. Note. (a) Every vector space contains a zero vector. A vector space may have more than one zero vector. If f and g are polynomials of degree n, then f + g is a polynomial of degree n. It is only true if we take x = 0. i.e ; ax = bx \. Then kAk= max kxk (n)=1 kAxk (m) is a matrix norm called theinduced matrix norm. 1)The set of all polynomials of degree at most 3 such that p(1) = 0. If f and g are polynomials of degree n, then f + g is a polynomial of degree n. One could also take the determinant of this matrix and nd that it is which implies dependence. Lemma 6. Show that the rst-order . It's only true if you add the hypothesis that x 6=0. Linear Algebra Theory I need solid straight forward proofs of Theorem 1.2 a, b, and c. As you can see there is a proof in the book but I do not understand them. Further Examples. In P (F ), only polynomials of the same degree may be added. Still stuck? Last edited by a moderator: Oct 4, 2005 Forums Mathematics Linear and Abstract Algebra For number 1, the answer is no because if you let e be the zero vecto, we have 1e=2e. V s 1: ( a 1, a 2) + ( b 1, b 2) = ( a 1 + b 1, a 2 + b 2) = ( b 1 + a 1, b 2 + a 2) = ( b 1, b 2 . Remark Here thevector norm could be any vector norm. If f and g are polynomials of degree n, then f + g is a polynomial of degree n. (c) Set h(z) = f(x 0 + Pz) where x 0 is any point satisfying Ax 0 = b. The size of a matrix is used in determining whether the solution, x, of a linear system Ax = b can be trusted, and determining the convergence rate of a vector sequence, among other things. Lemma 6. Thus, we can express any v 2V as a linear combination of u 1;:::;u n. But each u i is an element of U. (c) The matrices A and −A have the same four subspaces. 7 (e) A vector in F may be regarded as a matrix in Mexi(F). Solution: In any vector space ax = bx implies that a = b false. (2) Ax e S implies Bx e T. In the real case, for the choice of B = I, S = Rm4 and T = R' , (2) reduces to (1) and our results reduce accordingly to the above mentioned real results. A union of two vector spaces of n-vectors need not be a vector space, as you will show by example in Exercise 2.2. In any vector space, ax=bx implies that a=b. F (g) In P (F), only polynomials of the same degree may be added. Thus any v 2V is also an element of U. Use Problem 2 to show that any line passing through the origin of R3 is a vector space. [FIS P12 1.] Then apply Theorem 1.2 (c) (d) In any vector space, ax= ayimplies that x= y. - False - Corollary 1 (of Theorem 1.1) (c) In any vector space, ax= bx implies that a= b. Given the axioms we were given, it would seem that the statement should be true, no? It's going to be that. In any vector space, ax = bx implies that a = b. Label the following statements as true or false. Prove that the inverse element defined in (VS4) is unique. Here we have: detA ≠ 0, so the system is consistent. The equality is due to vector space properties of V.Thus(i)holdsforU.Each of the other axioms is proved similarly. Due to a planned power outage on Friday, 1/14, between 8am-1pm PST, some services may be impacted. 7 (e) A vector in F may be regarded as a matrix in Mexi (F). Before I give the formal de nition of a vector space, I rst need to de ne the concept of a eld of numbers2; these will be the numbers allowed as coe cients (R in problems 1 and 2 above; C in problem 3). (f) Anm * matrix has m columns and n rows. (d) In any vector space, ax = ay implies that x = y. To show (i), note that if x ∈U then x ∈V and so (ab)x = ax+bx. A T(AA ) 1Ax. For (a 1,a 2), ( b 1,b 2)∈V and c ∈ℝ , define (a 1, a2) + (b 1, b2) = (a 1+2b 1 , a 2+3b 2) and c(a 1,a 2) = (ca 1,ca 2). In any vector space, ax=bx implies that a=b. (c) In any vector space, ax = bx implies that a = b. If a 6= b, then a + ( b) 6= 0 since otherwise by adding b to both sides and using properties of elds we . But there are few cases of scalar multiplication by rational numbers, complex numbers, etc. (c) In any vector space,ax˘bximplies thata˘b. Problem 4. Solution. (f) An m n matrix has m columns and n rows. Problem 2. (b) A vector space may have more than one zero vector. 1. u+v = v +u, False. but it can be any norm. (b) A vector space may have more than one zero vector. V is said to be a vector space over R ( or an abstract R- vector space) if and only if there exists a mapping : R x V → V ( the image of any (a,x) in R x V will be denoted by ax such that for all x,y ∈ V and a,b ∈ R , the following properties hold 1. a2 (x + y) = ax + ay 2. a(bx) = (ab)x if a2 = a 3. 1. 1. For example, we could have kAk2 = max kxk2;(n)=1 kAxk 2;(m): To keep notation simple we often drop indices. Show that the solution set of Ax = 0 is a vector space, but that the solution set of Ax = b for b 6= 0 is not. See the answer See the answer See the answer done loading State why or why not. a vector space over F. To check that a subset U ⊂ V is a subspace, it suffices to check only a couple of the conditions of a vector space. (d) In any vector space ax = ay implies that x = y. Show that S 1 = x 0 + P(S 2). 1 The above statement is listed as false in my text, and I wanted to be sure I understood why that is. In any vector space, ax = ay implies that x = y. particular subset of a vector space is in fact a subspace. (b) Ax = b has one solution for every b 2Rm. Check this. In any vector space, ax = ay implies that x = y. We add c=atimes the rst equation to the second, yielding the two new equations ax 1 + bx 2 = 0 (d cb a)x 2 = 0 As a6= 0, we note that this has a unique solution (namely x 1 = x 2 = 0) if and only if d cb a 6= 0, if and only if ad bc6= 0. Explanation: The system of equations Ax = B is consistent if detA ≠ 0. - False - Take x= 0, a= 0, and b= 1. (n) be vector norms on R m and Rn, respectively, and let A be an m n matrix. In any vector space, ax = bx implies that a = b. (a) Every vector space contains a zero vector. (d)In any vector space, ax = ay implies that x = y. Note. 1x = x 4. A union of two vector spaces of n-vectors need not be a vector space, as you will show by example in Exercise 2.2. Note: a and b are the members of a field, and x is a member of the vector space. Obviously a basis of P⊥ is given by the vector v = 1 1 1 1 . De nition: A eld is a set F of numbers with the property that if a;b2F, then a+b;a b;aband a=bare also in F (assuming, of course, that b6= 0 in the Use Problem 2 to show that any line passing through the origin of R3 is a vector space. (e)A vector in Fn may be regarded as a matrix in M n 1(F). I just goofed and de ned scalar multiplication wrong. A subset of a vector space V that is itself a vector space is a subspace of V. Theorem 2.1.2. (e) A vector in Fn may be regarded as a matrix in Mn×1 (F ). Let V 1 and V 2 be vector spaces of n-vectors. Problem 2. We say that the vector norm kAk is "induced" by the norm k k. It is a measure of the "size" of the operator. Problem 3. (e) 0 v = 0 for every v ∈ V, where 0 ∈ R is the zero scalar. abelian group augmented matrix basis basis for a vector space characteristic polynomial commutative ring determinant determinant of a matrix diagonalization diagonal matrix eigenvalue eigenvector elementary row operations exam finite group group group homomorphism group theory homomorphism ideal inverse matrix invertible matrix kernel linear . Justig why ture of [2] In any Theorem 0.8 Let Ax = b be a system of nlinear equations in nunknowns. Let A be a given mxn matrix. b)) = ax2 + bx: This is a linear transformation as T((a;b) + . Label the following statements as true or false. (d) In any vector space, ax = ay implies that x = y. The Axioms of a Vector Space. It fails many of the properties, in particular, property (B) which states that c~x must be in V for any ~x 2V. 2. a vector space. Show that the solution set of Ax = 0 is a vector space, but that the solution set of Ax = b for b 6= 0 is not. The actual proof of this result is simple. vector space W. But the only full-dimensional subspace of a nite-dimensional vector space is itself, so this happens if and only if the image is all of W, namely . (iii) In any vector space, ax = bx implies that a = b (for scalars a and b and a vector x). 1. This will be true for any nonsingular n n matrix. Solution : V is not a vector space, since . A vector in F^n F n may be regarded as a matrix in (b) A vector space may have more than one zero vector. Using these facts, we can prove the following important theorem: Theorem 4.3. EXERCISES 1. (d) In any vector space,ax˘ayimplies thatx˘y. To show (i), note that if x ∈U then x ∈V and so (ab)x = ax+bx. Determine whether or not the following subsets are subspaces of P3. Furthermore, the kernel of T is the null space of A and the range of T is the column Let us suppose ( a 1, a 2), ( b 1, b 2) ∈ V then. (a) If u + v = u + w, then v = w. (b) If v + u = w + u, then v = w. (c) The zero vector 0 is unique. F (c) In any vector space, ax = ba implies that arb. (i) Every vector space contains a zero vector. (d) For each v ∈ V, the additive inverse − v is unique. - True - Axiom (VS 3) (b) A vector space may have more than one zero vector. If s is any other solution, then As = b, and consequently s = A 1b, so the solution is unique. (Hint: b is not in the column space C(A), thus b is not orthogonal to N(AT).) (F) not so when x=0 In any vector space, ax=ay implies that x=y. (b) A vector space may have more than one zero vector. 7 (e) A vector in F may be regarded as a matrix in Mexi(F). (e) An element of F^ may be regarded as an element of Mx1 (F). Example. F (c) In any vector space, ax = ba implies that arb. And so all of these other solution sets are just some particular vector, some x particular, plus the null space. (4): If V is a vector space over F, with a;b 2F and x 2V with x 6=0, then ax = bx implies a = b. We define a matrix norm in the same way we defined a vector norm. I just showed it right there. Definition. Is V a vector space over R with these operations? (c) In any vector space, ax = bx implies that a = b. Case 2: If a6= 0, then we consider the two equations ax 1 + bx 2 = 0 and cx 1 +dx 2 = 0. It's only true if you add the hypothesis that a 6= 0. e. A vector in Fn may be regarded as a matrix in M n 1(F). Why Does a Vector Space Have a Basis (Module Theory) Module and vector space. An n × n n\times n n × n matrix with entries in R \mathbb{R} R with the property that the sum of entries along each row, column and diagonal is constant and equals a ∈ R a\in\mathbb{R} a ∈ R is called a matrix magic square of order n n n with line-sum a a a.. An example of a matrix magic square of order 3 3 3 is the matrix (8 1 6 3 . Note. Next, we will consider norms on matrices. In P(F ), only polynomials of the same degree may be added. A vector space or a linear space is a group of objects called vectors, added collectively and multiplied ("scaled") by numbers, called scalars. (Remember, the empty set is not a vector space.) The most basic example is m = n = 1 and A = [1]. The most general equation is [math]Ax=b[/math] where A is a square matrix, x is the unknown vector, and b is a vector that matches the inner dimension of the A square matrix. Now ax,bx,ax+bx and (a+b)x are all in U by the closure hypothesis.

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