# Read e-book Student Solution Manual for Elementary Linear Algebra Contents:

In he became a research professor of mathematics at Drexel University in Philadelphia, where he taught and did mathematical research for 15 years. In he left Drexel as a Professor Emeritus of Mathematics to become a full-time writer of mathematical textbooks. Undetected location. NO YES. Selected type: Paperback. The aim is to present the fundamentals of linear algebra in the clearest possible way; pedagogy is the main consideration.

Calculus is not a prerequisite, but there are clearly labeled exercises and examples which can be omitted without loss of continuity for students who have studied calculus. Howard Anton obtained his B. He worked in the manned space program at Cape Canaveral in the early 's. In he became a research professor of mathematics at Drexel University in Philadelphia, where he taught and did mathematical research for 15 years.

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However, the following axioms fail for this set V: Axiom 1: Clearly if A is invertible, then so is —A. Thus, V is not a vector space. Since we are using the standard operations of addition and scalar multiplication, Axioms 2, 3, 5, 7, 8, 9, 10 will hold automatically. However, for Axiom 4 to hold, we need the zero vector 0, 0 to be in V. Thus, the set of all points in R2 lying on a line is a vector space exactly in the case when the line passes through the origin.

Exercise Set 5. However, for Axiom 4 to hold, we need the zero vector 0, 0, 0 to be in V. Thus, the set of all points in R3 lying on a plane is a vector space exactly in the case when the plane passes through the origin.

## Elementary Linear Algebra, Student Solutions Manual, 8th Edition

Planes which do not pass through the origin do not contain the zero vector. Since this space has only one element, it would have to be the zero vector. In fact, this is just the zero vector space. Suppose that u has two negatives, —u 1 and —u 2. We have proved that it has at most one.

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Thus it is not a subspace. Therefore, it is a subspace of R3. The same is true of a constant multiple of such a polynomial. Hence, this set is a subspace of P3. Hence, the subset is closed under vector addition. Thus, the subset is not closed under scalar multiplication and is therefore not a subspace. Thus the set is a subspace. Thus 2, 2, 2 is a linear combination of u and v. Thus, the system of equations is inconsistent and therefore 0, 4, 5 is not a linear combination of u and v.

Since the determinant of the system is nonzero, the system of equations must have a solution for any values of x, y, and z, whatsoever. Therefore, v1, v2, and v3 do indeed span R3. Note that we can also show that the system of equations has a solution by solving for a, b, and c explicitly. Since this is not the case for all values of x, y, and z, the given vectors do not span R3. Hence the given polynomials do not span P2. The set of solution vectors of such a system does not contain the zero vector.

Hence it cannot be a subspace of Rn. Alternatively, we could show that it is not closed under scalar multiplication. Let u and v be vectors in W. Let W1 and W2 be subspaces of V. This follows from the closure of both W1 and W2 under vector addition and scalar multiplication. They cannot all lie in the same plane. Hence, the four vectors are linearly independent.

This implies that k3 and hence k2 must also equal zero. Thus the three vectors are linearly independent. Thus they do not lie in the same plane. Suppose that S has a linearly dependent subset T. Denote its vectors by w1,…, wm. Since not all of the constants are zero, it follows that S is not a linearly independent set of vectors, contrary to the hypothesis.

That is, if S is a linearly independent set, then so is every non-empty subset T. This is similar to Problem Use Theorem 5. The set has the correct number of vectors. Thus the desired coordinate vector is 3, —2, 1. For instance, 1, —1, —1 and 0, 5, 2 are a basis because they satisfy the plane equation and neither is a multiple of the other.

For instance, 2, —1, 4 will work, as will any nonzero multiple of this vector. Hence it is a basis for P2. There is. Thus, its solution space should have dimension n — 1. Since AT is also invertible, it is row equivalent to In.

## Student Solution Manual for Elementary Linear Algebra - PDF Free Download

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. 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.

## Solution Manual Of Elementary Linear Algebra By Howard Anton

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.

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See Exercise 3, Parts b and c. The set is therefore orthogonal.