Introduction to linear algebra by bernard kolman 8th edition solution

If the sum of the entries of each column is 1, it does not follow that the sum of the entries in each column of AT will also be 1. Enter the matrix T and initial state vector x 0 into Matlab. The command sum operating on a matrix computes the sum of the entries in each column and displays these totals as a row vector.

If the output from the sum command is a row of ones, then the matrix is a Markov matrix. Productive Final average: Supplementary Exercises, p. The rectangle has vertices 0, 0 , 4, 0 , 4, 2 , and 0, 2. Supplementary Exercises 57 4. Let u, v be vectors in R2. Chapter 3 Determinants Section 3. The number of inversions are: a 9, 6. Suppose jp and jq are separated by k intervening numbers. Then k interchanges of adjacent numbers will move jp next to jq.

One interchange switches jp and jq. Parallel to proof for the upper triangular case. By n applications of Theorem 3. If A is nonsingular, by Corollary 3. Follows immediately from Theorem 3.

When all the entries on its main diagonal are nonzero. Ten have determinant 0 and six have determinant 1. There are many sequences of row operations that can be used. Here we record the value of the determinant so you may check your result. Section 3. Let A be upper triangular. Since A is symmetric, that submatrix is the transpose of Mij.

Thus adj A is symmetric. By Theorem 3. Since the entries of A are integers, the cofactors of entries of A are integers and adj A is a matrix of integer entries. We present a sample of the cofactors. Before using the expression for the inverse in Corollary 3. From Theorem 3. Thus AB is singular. Chapter 4 Vectors in Rn Section 4. Impossible Locate the point A on the x-axis which is x units from the origin. Construct a perpendicular to the x-axis through A. Locate B on the y-axis y units from the origin.

Construct a perpendicular through B. The intersection of those two perpendiculars is the desired point in the plane.

The value of the inventory of the four types of items. Section 4. See solution to Exercise T. By a of Theorem 4. The result follows by Exercise T. By the remark following Example 10 in Section 1. As in the solution to Exercise T. The negative of each vector in B 3 is itself. The negative of each vector in B 4 is iteself. Just inspect the sets, they include every vector in B 3.

Just inspect the sets, they include every vector in B 4. Then determine the lengths of the vectors. We have, by Exercise T. Hence, L is a linear transformation. Part a of Theorem 4. Using Theorem 5. Following Example 6 we proceed as follows in Matlab. Reversing these steps proves the converse. Thus 5. The noncolinearity of the three points insures that the three cofactors A11 , A12 , A13 are not all zero. The determinant has two equal rows, and so has the value zero. Thus the point Pi lies on the plane whose equation is 5.

Chapter 6 Real Vector Spaces Section 6. P is a vector space. Pr is contained in P. Thus the additive inverse of p t , zero polynomial, etc. Hence P is a vector space. Not a vector space; e , f , and h do not hold.

Vector space. Not a vector space; h does not hold. By Theorem 6. Let 01 and 02 be zero vectors. Let u1 and u2 be negatives of u. The sum of any pair of vectors from B n is, by virtue of entry-by-entry binary addition, a vector in B n. Thus B n is closed. Both 0 and v are in B n , so B n is closed under scalar multiplication.

Section 6. Since Pn is a subset of P and it is a vector space with respect to the operations in P , it is a subspace of P. Hence, W is a subspace of R3. So W is a subspace. Similarly for b. Thus d holds. Finally, e , f , g , h follow for W because those properties hold for any vectors in V and any scalars.

Let W be a subspace of V , let u and v be vectors in W , and let a and b be scalars. Thus W is a subspace by Theorem 6. Otherwise, let x0 be a solution. Hence, the set of all solutions fails to be closed under either vector addition or scalar multiplication. We assume S is nonempty.

Thus span S is a subspace of V. W must be closed under vector addition and under multiplication of a vector by an arbitrary scalar. No, it is not a subspace. Similarly, or from 6. Let w be any vector in W. But, since 0 is not in W , this implies that W is not closed under scalar multiplication, so W cannot be a subspace of V. Note that since the vectors are rows we need to convert them to columns to form this matrix. Next we obtain the reduced row echelon form of the associated linear system.

We use the transpose operator to conveniently enter the vectors. There are many other linear combinations that work. Follow the method in ML. Associate a column vector with each polynomial as in ML. Con- tradiction. This implies that c1 , c2 , and c3 are not all zero.

Let a1 ,. Thus, in the summation 6. Hence 6. Since A is nonsingular, Theorem 1. Since S2 contains the zero vector for R2 it is linearly dependent. Thus in this case S1 is linearly independent. Now equation 6. Hence, v is a linear combination of w1 , w2 ,. Thus, the set of all such vectors v is W. Thus S is linearly dependent. Form the augmented matrix A 0 and row reduce it.

Hence the set of four matrices is linearly independent. Possible answer: ,. The result follows from Theorem 6. First note that any set of vectors in W that is linearly independent in W is linearly independent in V. Suppose now that W is a nonzero subspace of V. If S1 is linearly independent, then it is a basis for V which contains S. Otherwise some vector in S1 is a linear combination of the preceding vectors Theorem 6. Delete it to form a new set S2 with one fewer element than S1 which also spans V.

Either S2 is a basis or else another vj can be deleted. Let W be any subspace of R3. Hence W contains the zero vector 0. Thus the two sets span V. Since the second set has n elements, it is also a basis for V. Hence T is a basis for V. Since every vector in V can be written as a linear combination of the vectors in S, it follows that S spans V.

Hence, S is a basis for V. Since A is singular, Theorem 1. Hence v1 , v2 ,. Follow the procedure in Exercise ML. Proceed as in ML. We proceed as we did in ML. We need only determine if S is a linearly independent subset of V. In Exercises ML. No basis. Since the dimension of the null space of A is 3, the null space of A is R3. Hence, the dimension is either 1 or 2. Hence the null spaces of A and B are the same.

We can compute such a basis directly using the command homsoln as shown next. We can compute such a basis directly using command homsoln as shown next. Similarly, a basis for the row space of AT consists of the transposes of a corresponding basis for the column space of A. Linearly independent. The three column vectors of A span the column space of A and are then a basis for the column space. Hence, they are linearly independent.

Yes, linearly independent. Only the trivial solution. Has a solution. Has no solution. Then Theorem 6. Let B be the matrix whose jth column is xj. If the columns of A are linearly dependent, then by Corollary 6. It follows by Corollary 6. Conversely, suppose the columns of A are linearly independent. Conversely, if the n columns of A span Rn then by Theorem 6.

Thus the rows of A are linearly independent. Then the columns of A span Rm. Thus m columns of A are a basis for Rm and hence all the columns of A span Rm. Suppose that the columns of A are linearly independent. Since A has n columns which span its column space, it follows that they are linearly independent. If b is in the column space of A, then B can be written as a linear combination of the columns of A in one and only one way. Since the rank of a matrix is the same as its row rank and column rank, the number of linearly independent rows of a matrix is the same as the number of linearly independent columns.

We must show that the rows v1 , v2 ,. Since AAT is nonsingular, Theorem 1. Suppose that w1 S , w2 S ,. Using Exercise T. Since v1 , v2 ,. Let v be a vector in V. Since S is a set consisting of three vectors in a 3-dimensional vector space, we can show that S is a basis by verifying that the vectors in S are linearly independent.

It follows that if the reduced row echelon form of the three columns is I3 , they are linearly independent. We can do all three parts simultaneously as follows. Associate with each vector v a column. Form a matrix B from these columns.

Associate a column with each matrix and proceed as in ML. An orthonormal set in Rn is an orthogonal set of nonzero vectors. Since an orthonormal set of vectors is an orthogonal set, the result follows by Theorem 6.

Let w be in span S. We show W is closed under addition of vectors and under scalar multiplication. It follows that kv is in W. Thus W is a subspace of Rn. Hence, A is nonsingular. Since some of the vectors vj can be zero, A can be singular. None of the vectors in A is the zero vector.

Since A contains more than n vectors, Q is a linearly dependent set. Thus one of the vectors is not orthogonal to the preceding ones. See Theorem 6. What remains will be a set of n orthogonal vectors since A originally contained a basis for V.

Use the following Matlab commands. Apply routine gschmidt to the vectors of S. The zero vector is orthogonal to every vector in W. By Theorem 4. Let v be a vector in Rn. Let W be a subspace of Rn. Moreover, w and u are unique. We now show that S is linearly independent. Thus, S is also linearly independent and is then a basis for Rn. Hence x is orthgonal to every row vector of A, so x is orthogonal to every vector in the row space of A.

If all the nontrivial solutions of the homogeneous system are multiples of each other, then the dimension of the solution space is 1. Thus the dimension of the solution space is zero. Thus A is nonsingular. To show that T is a basis we need only show that it spans Rn and then use Theorem 6. Let v belong to Rn. Next we show that the members of T are orthogonal. Hence T is an orthogonal set.

Supplementary Exercises T. Since the columns are orthonormal they are linearly independent. There can be at most m linearly independent vectors in Rm. But row i of B T is just bTi and column j of B is bj. Section 7. Theorem 7. Enter the data into Matlab. Data for quadratic least squares: Sample of cos on [0, 1.

But the weight of b, b , using binary addition is the sum of its bits. Not all columns are distinct. D Thus, the column space of C is in the null space of G.

It follows that if y is a column vector then RT y is a column vector with its entries rearranged in the same order as the columns of Q when Qp is formed. Thus the null space of Qp consists of the vectors in the null space of Q with their entries rearranged in the same manner as the columns of A when Qp was formed.

Hence all the code words have weight greater than or equal to 3. Their respective weights are 3, 3, and 4. There are 15 code words, hence there are pairs of vectors. Using the following Matlab commands we can determine the minimum Hamming distance as the smallest nonzero entry in d. We get the minimum Hamming distance to be 3. Both codes have Hamming distance 3 so each can detect two or fewer errors. Use bingen to generate all the binary representations of integers 0 through 8 using 4 bits and then multiply by the code matrix C using binprod.

Use bingen to generate all the binary representations of integers 0 through 15 using 11 bits and then multiply by the code matrix C using binprod. Chapter 8 Eigenvectors, Eigenvalues, and Diagonalization Section 8. Section 8. An eigenvector must be a nonzero vector, so the zero vector must be included in S.

It follows that the eigenvalues of A are the diagonal elements of A. Associated eigenvectors need not be the same. Thus, Tr A is the sum of the eigenvalues of A. The result follows from part c. Thus the zero vector is the only vector in both S1 and S2. Therefore L w is in W since W is closed under scalar multiplication. Enter each matrix A into Matlab and use command poly A. The eigenvalues of matrix A will be computed using Matlab command roots poly A.

The result follows by Theorem 8. Not diagonalizable. Associated eigenvectors 0 1 1 are the columns of P. P is not unique. Not possible. Associated eigenvectors are 0 1 1 the columns of P. Other answers are possible. By Theorem 8. Not defective. See also Example 6 in Section 8. Hence, Ak and B k are similar. The proof proceeds as in the proof of Theorem 8. The result follows at once from Theorems 8.

Hence A is not diagonalizable. In Exercise T. That is, the dot product of the ith and jth columns of A. Thus the columns of A form an orthonormal set in Rn.

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