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# ELEMENTARY LINEAR ALGEBRA HOWARD ANTON PDF

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Howard Anton obtained his B.A. from Lehigh University, his M.A. from the This edition of Elementary Linear Algebra gives an introductory treatment of linear. Elementary Linear Algebra (9th Edition) - Howard Anton e Chris ecogenenergy.info Abir Roy. P R E F A C E This textbook is an expanded version of Elementary Linear. Elementary linear algebra: applications version / Howard Anton, Chris Rorres. . learn key linear algebra concepts by using MATLAB and is available in PDF. Author: JOSE MCTHAY Language: English, Spanish, Dutch Country: Canada Genre: Science & Research Pages: 341 Published (Last): 17.10.2015 ISBN: 679-7-15342-555-2 ePub File Size: 23.42 MB PDF File Size: 15.55 MB Distribution: Free* [*Regsitration Required] Downloads: 21346 Uploaded by: LESHA Chapter 1 Systems of Linear Equations and Matrices Section Exercise Set 1. (a), (c), and (f) are linear equations in x1, x2, and x3. (b) is not linear. Elementary Linear Algebra: Applications Version, 11th Edition. Pages·· MB·4, Howard Anton, Chris Rorres Elementary Linear Algebra.  Howard Anton. Elementary Linear Algebra. MSRI Publications, www ecogenenergy.info  Richard G. Baraniuk.

There is. Thus, its solution space should have dimension n — 1. Since AT is also invertible, it is row equivalent to In. It is clear that the column vectors of In are linearly independent.

Hence, by virtue of Theorem 5. Therefore the rows of A form a set of n linearly independent vectors in Rn, and consequently form a basis for Rn. Any invertible matrix will satisfy this condition. The nullspace of D is the entire xy-plane. Use Theorems 5. However A must be the zero matrix, so the system gives no information at all about its solution. That is, the row and column spaces of A have dimension 2, so neither space can be a line. Rank A can never be 1.

Thus, by Theorem 5. Hence, Theorem 5. Verify that these polynomials form a basis for P 1. Exercise Set 6. So is Axiom 2, as shown: To prove Part a of Theorem 6. To prove Part d , observe that, by Theorem 5. By inspection, a normal vector to the plane is 1, —2, —3. From the reduced form, we see that the nullspace consists of all vectors of the form 16, 19, 1 t, so that the vector 16, 19, 1 is a basis for this space.

Conversely, if a vector w of V is orthogonal to each basis vector of W, then, by Problem 20, it is orthogonal to every vector in W. In fact V is a subspace of W c True. The two spaces are orthogonal complements and the only vector orthogonal to itself is the zero vector. For instance, if A is invertible, then both its row space and its column space are all of Rn. See Exercise 3, Parts b and c.

The set is therefore orthogonal. It will be an orthonormal basis provided that the three vectors are linearly independent, which is guaranteed by Theorem 6. Note that u1 and u2 are orthonormal. Thus we apply Theorem 6. By Theorem 6. But v1 is a multiple of u1 while v2 is a linear combination of u1 and u2. This is similar to Exercise 29 except that the lower limit of integration is changed from —1 to 0.

Then if u is any vector in V, we know from Theorem 6. Moreover, this decomposition of u is unique. Theorem 6. If the vectors vi form an orthogonal set, not necessarily orthonormal, then we must normalize them to obtain Part b of the theorem.

However, although they are orthogonal with respect to the Euclidean inner product, they are not orthonormal. However, they are neither orthogonal nor of unit length with respect to the Euclidean inner product.

Suppose that v1, v2, …, vn is an orthonormal set of vectors. Thus, the orthonormal set of vectors cannot be linearly dependent.

The zero vector space has no basis 0. This vector cannot be linearly independent. If A is a necessarily square matrix with a nonzero determinant, then A has linearly independent column vectors.

Thus, by Theorem 6. Hence the error vector is orthogonal to the column space of A. Therefore Ax — b is orthogonal to the column space of A. Since the row vectors and the column vectors of the given matrix are orthogonal, the matrix will be orthogonal provided these vectors have norm 1.

Note that A is orthogonal if and only if AT is orthogonal. Since the rows of AT are the columns of A, we need only apply the equivalence of Parts a and b to AT to obtain the equivalence of Parts a and c. If A is the standard matrix associated with a rigid transformation, then Theorem 6. But if A is orthogonal, then Theorem 6. Exercise Set 7. By Theorem 7. Thus by Theorem 7. Since A has no real eigenvalues, there are no lines which are invariant under A. Let aij denote the ijth entry of A.

Since the eigenvalues are assumed to be real numbers, the result follows. The converse of this result is also true. This is a straightforward computation and we leave it to you. The matrices 3I — A and 2I — A both have rank 2 and hence nullity 1.

Thus A has only 2 linearly independent eigenvectors, so it is not diagonalizable. Any matrix Q which is obtained from P by multiplying each entry by a nonzero number k will also work. Suppose that A is invertible and diagonalizable. In addition, Theorem 7. In other words, Dk displays the eigenvalues of Ak along its diagonal. The sequence diverges for all other values of a. Thus each eigenvalue is repeated once and hence each eigenspace is 1-dimensional. By the result of Exercise 17, Section 7.

Moreover, if A has nonnegative eigenvalues, then the diagonal entries of D are nonnegative since they are all eigenvalues. But there is no elementary product which contains exactly n — 1 of these factors Why? Supplementary Exercises 7 7. We know that A has at most n eigenvalues, so that this expression can take on only finitely many values.

Thus the only possible eigenvalues of A are zero and tr A. It is easy to check that each of these is, in fact, an eigenvalue of A. Since every odd power of A is again A, we have that every odd power of an eigenvalue of A is again an eigenvalue of A. This says that T leaves every point in the x-y plane unchanged. All linear transformations have this property, and, for instance there is more than one linear transformation from R2 to R2.

But there is only one linear transformation which maps every vector to the zero vector. Exercise Set 8. Theorem 5. Hence 5, 0 is not in R T. By Theorem 8. Since the only subspaces of R3 are the origin, a line through the origin, a plane through the origin, or R3 itself, the result follows. It is clear that all of these possibilities can actually occur.

These are parametric equations for a line through the origin. That range, which we can interpret as a subspace of R3, is a plane through the origin. Thus, by the Dimension Theorem Theorem 8. Therefore it is a plane through the origin. Hence, it is a line through the origin. Alternative Solution: Suppose that w1 and w2 are in R T. This follows immediately from the linearity of the transformation T. It is easy to show that T is linear. The transformation is linear and only 0, 0, 0 maps to the zero polynomial.

Clearly distinct triples in R3 map to distinct polynomials in P2. T is a linear operator by Theorem 3. That is, T maps a to the zero vector, so if T is one-to-one, a must be the zero vector. But then T would be the zero transformation, which is certainly not one-to-one. But since B is indeed the standard basis for Rn, the matrices are the same. Thus A and C are matrices for the same transformation with respect to different bases. But from Theorem 8. Alternate Solution: Thus the rank of the transformation represented by the matrix CP is the same as that of C.

Since P—1 is also invertible, its null space contains only the zero vector, and hence the rank of the transformation represented by the matrix P—1 CP is also the same as that of C. Thus the ranks of A and C are equal. Again we use the result of Theorem 8. Second Alternative: Since the assertion that similar matrices have the same rank deals only with matrices and not with transformations, we outline a proof which involves only matrices.

But multiplication of the matrix C by an elementary matrix is equivalent to performing an elementary row or column operation on C. From Section 5. Thus A and C must have the same rank. But Cei and Dei are just the ith columns of C and D, respectively.

By Table 1, A is invertible if and only if B is invertible, which guarantees that A is singular if and only if B is singular. Thus, if B is singular, then so is A. Otherwise, B would be the product of 3 invertible matrices. Now, we show that the trace is a similarity invariant. Thus, the image of T is a one-dimensional subspace of R3 and cannot be all of R3.

Thus, T is not onto. Thus, T is onto. In particular, it fails being one-to-one. The maximal rank of AT is m, so the dimension of the image of T is at most m. Since the dimension of the image of T is smaller than the dimension of the codomain Rn, T is not onto. We know from Example 7 in Section 8. Let a1, a2, … , an be any point in Rn. So, T is one-to-one and is thus an isomorphism.

Differentiation is a linear transformation see Example 11, Section 8. In this case, D maps functions in V into other functions in V. Thus, T e3 and any two of the remaining three columns of A is a basis for R T. We can use the method of Section 5. Thus the standard basis for R3 is also a basis for R T.

Thus the vector —1, 1, 0, 1 forms a basis for the kernel.

## Elementary Linear Algebra Books

Supplementary Exercises 8 9. Thus the nullity is 1.

Since the dimension of M22 is 4, the rank of T must therefore be 3. We compute the above result more directly. Exercise Set 9. Case II: Case III: Subcase 1: Subcase 2: Subcase 3: However it sends 1, 0, 0 to 0, 1, 0 and 0, 1, 0 to —1, 0, 0.

However it sends 1, 0, 0 to 0, 0, —1 and 0, 0, 1 to 1, 0, 0.

We use the notation and the calculations of Exercise Thus it also must pass through the origin. The two column vectors of M are linearly independent if and only if neither is a nonzero multiple of the other. To shortcut some of the work, we derive a different expression for the mean square error m.

Therefore, m. These vectors are orthogonal. Since the left side of this equation cannot be negative, then there are no points x, y which satisfy the equation.

Thus the graph consists ofthe single point 0, 0. Thus it represents the point 1, 2. We must also normalize u3. Thus, by Theorem 7. Now let T be the matrix whose column vectors are the 3 linearly independent eigenvectors of A. It follows from the proof of Theorem 7. Theorem 7. To do this, orthonormalize the basis of each eigenspace before using its elements as column vectors of S. Furthermore, by Theorem 6. It follows from Theorem 6. Thus, P represents a rotation.

## Elementary Linear Algebra: Applications Version, 11th Edition

Therefore we need mnp multiplications and m n — 1 p additions to compute C. This requires n n — 1 multiplications and the same number of additions again ignoring the operations whose results we already know. The total number of multiplications so far is n2 and the total number of additions is n n — 1. To reduce the second column to that of In, we repeat the procedure, starting with Row 2 and ignoring Column 1.

Thus n — 1 multiplications assure us that there is a one on the main diagonal, and n — 1 2 multiplications and additions will make all n — 1 of the remaining column entries zero.

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This requires n n — 1 new multiplications and n — 1 2 new additions. See the matrices at the very end of Section 9. By Exercise 27 of Section 2. Now the matrices Ei are all lower triangular and invertible by their construction. Hence L, as the product of lower triangular matrices, must also be lower triangular. We know that A can be reduced to row-echelon form and that this may require row interchanges. The proof can be broken into four cases: Thus, it is a circle of radius 2 and center at the origin.

Geometrically, then, z can be any point on the real axis. We now show this result analytically. Exercise Set We use Formula We show all six 6 3 sixth roots in the diagram. The fourth roots of 16 are 2, 2i, —2, and —2i. Case 1. Case 2. Now suppose that n is a positive integer and hence that —n is a negative integer.

Thus Axiom 6 fails. However, as seen above, the closure property of scalar multiplication may fail, so the vectors need not be in V. Thus, this set is not a vector space because Axiom 6 fails. The second equation will then be valid for all such x1 and x3. That is, x2 is arbitrary. It is closed under scalar multiplication by a real scalar, but not by a complex scalar. Hence, this is an inner product on C2.

We check Axioms 1 and 4, leaving 2 and 3 to you. Axioms 1—3 are easily checked. Hence, Axiom 4 fails. Axioms 1 and 4 are easily checked. Thus both Axioms 2 and 3 fail. Finally, using the result of Problem 37 of Section Thus all four axioms hold. Hence the set is not 6 6 6 orthonormal.

We check Axioms 2 and 4. This shows that the 2 eigenvalues of a symmetric matrix with nonreal entries need not be real. Theorem That is, det A is the sum of the conjugates of the terms in det A. Call the diagonal matrix D. Show the eigenvalues of a unitary matrix have modulus one. Let A be unitary. Using Eq. These together with the original equation form a homogeneous linear system with a non- trivial solution for c1, c2, … , c6.

Thus the determinant of the coefficient matrix is zero, which is exactly Eq. Thus the determinant of this system is zero, which is Eq.

Upon substitution of the coordinates of the three points x1, y1 , x2, y2 and x3, y3 , we obtain the equations: The values of the objective function are shown in the following table: Let x1 be the number of pounds of ingredient A used and x2 the number of pounds of ingredient B. The number of oxen is 50 per herd, and there are 7 herds, so there are oxen.

Substituting for c1, an—1, bn—1, cn—1 from Eqs. Note that the matrix has the same number of rows and columns as the graph has vertices, and that ones in the matrix correspond to arrows in the graph. These are: Inspection shows that this is indeed a clique. Inspection shows that it is indeed a clique. But note that P8 can be added to the first set and we still satisfy the conditions. Theorem 2 says there will be one linearly independent price vector for the matrix E if some positive power of E is positive.

Since E is not positive, try E2: II Step 1: Step 2: Step 4: Assume the result true for n — 1. Or the package textbook and solutions manual together: Get it only at our. Klugman solutions manual accompany introduction linear regression.

## Elementary Linear Algebra Books

Elementary Linear Algebra 11th edition gives. Ideas and Applications, 4th Edition cover image. Download link: Howard Anton: Books - Amazon. I am selling the Elementary linear algebra 9th edition math textbook. The textbook comes with the solution manual as well and I am throwing in the previous years practice exams booklet. Calculus and its Applications 11th Ed. Elementary linear algebra howard anton.

Calculus and analytic geometry solutions manual. Calculus by anton 8th. Elementary Linear Algebra 10th edition gives an elementary treatment of linear Algebra Applications Version: Student Solutions Manual, Howard Anton. Elementary Number Jr. For All Practical Purposes 7th Ed. Solutions Manual, Heidi A. Howard Conceptual Physics, Paul G. Hewitt, 11th Edition.This is a vector space. Since the product of integers is always an integer, each elementary product is an integer. Skip to main content. Hence 5, 0 is not in R T.

Let A be unitary. However, for Axiom 4 to hold, we need the zero vector 0, 0 to be in V. Therefore, E3 must be the matrix obtained from I3 by replacing its third row by —2 times Row 1 plus Row 3, i.

Thus A and C are matrices for the same transformation with respect to different bases. But multiplication of the matrix C by an elementary matrix is equivalent to performing an elementary row or column operation on C.

This proof uses it twice.

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