Contemporary linear algebra solutions manual pdf


 

(Solution Manual) Contemporary Linear Algebra by Howard Anton, Robert C. Busby - Free ebook download as PDF File .pdf), Text File .txt) or read book online. (Solution Manual) Contemporary Linear Algebra by Howard Anton, Robert C. Busby - Ebook download as PDF File .pdf), Text File .txt) or read book online. Get instant access to our step-by-step Contemporary Linear Algebra solutions manual. Our solution manuals are written by Chegg experts so you can be.

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Contemporary Linear Algebra Solutions Manual Pdf

Matrices And Matrix Algebra . Can you find your fundamental truth using Slader as a completely free Contemporary Linear Algebra solutions manual? Unlock your Contemporary Linear Algebra PDF (Profound Dynamic Fulfillment) today. 6 days ago Contemporary Linear Algebra Howard Anton Solution Manual - [Free] Solution Manual [PDF] [EPUB] Solution Manual For Elementary Linear. Linear Algebra by Howard Anton, Robert C. Busby - Ebook download as PDF File .pdf), Text. File .txt) or read book online. - za, 30 mrt GMT.

The three lines do not intersect in a. This system has no solution. This leads to: This is a linear equation with the given solution set. We can find parametric equations fN the line of intersection by for example solving the given equations for x and y in terms of z, then making z into a parameter: We can find parametric equations for the line of intersection by for example solving the given equations for x and y in terms of z, then making z into a parameter:

Gr in column vector form. Since span S is a subspace already closed under scalar multiplication and addition. Thus WI n w2 is closed under scalar multiplication and addition.

If W is a subspace. Chapter 3 c True. The vector w can be expressed as a linear combination of v The reduced row echelon form of the augmented matrix of this system is 0 1 0 0 -2] 0 1 3 1 From this we conclude that the system has a.

The vector w can be expressed as a linear combination of v 1. The reduced row echelon form of the augmented matrix of this system is 0 2 -1 0 0 From this we conclude that the system has infinitely many solutions This corresponds to the line through the origin wit.

The row reduced echelon form of the augmented mo. Note that t he vectors 5,-7, 1,0 and 4 ,-6,0, 1 are orthogonal to both v 1 and v2. T he general solution will have at least 3 free variables. Thus, assuming A i: For example, the system: Each hyperplane corresponds to a single homogeneous linear equation in four variables, and there must be a least four equations in order to have a unique solution.

Every pla ne in R3 corrr.: This argument c2-n easily be adapted to prove the following:. There are infinitely ma. Thus the factors ;n a do not commute, whe,eas the ractocs ;n h do commute. J o] -1 -1 0 and from this it is easy to see tha t the system has the fixed poi nts of A are vectors of t he form x The lower tria.. The augmented matrix of this system is.

Following are the matric: All vectors of thcform x AT is skew-symmetric. Using Formula 9. Thus the maximum number of distinct entries in a skew-symmetric matrix is n n. There are a.

For a. In a skew-symmetric matrix A is skewsymmroetric. If A is both symmetric and skew symmetric. I --' The maximum nwnber of distinct entries can be attained by selecting distinct positive entries for the p. In fact. In a symmetric matrix. There are a t. Au n x n matrix has n entriP.

Solution manual for linear algebra david lay

If A Thus. Thus a symmetric matrix is completely determined by the entries that lie on or above the main rliagonal. If A is invertible then Ak is invertible. If A is not square then A is not invertible.

This shows that an invertible matrix cannot be nilpotent.. Thus is it not possible to have A.. But if A is square and AAT is invertible.

If A is both symmetric and triangular. For example.. If A and 1J are symmetric. See Theorem 3. We will show t hat if A is symmetric i. Our p roof is by induction on the exponent k. Our proof is by induction on the exponent n. These two steps complete the proof by induction.. These two steps complete the proof by ind uction Step L Since A is symmetric. It is easy to check that thi! T he matrix A can be reduced to row echelon form by the following operations: The mult.

The solution of Ly l X The system Ly. It is easy to check t ha. The multipliers associated wilh these operations arc thus Jt is easy to check that this is the solution of [-! The matrix A can be reduced to row echelon form by the following sequence of operations: Chapter 3 9. Y3 The sy.

An LU-decomposition of! The The soluUon of Ux y 1 Let e 1. The sdu tion of L Chapter 3 from whieh we obtain y 1 ""' 0. Thus x. The nta. Using the given LU-decomposition. This leads t. Using the given decomposition. If we interchange rows 2 and 3 of A. This is equivalent to first multiplying A on the left by the corresponding permutation matrix P: Thus P is a per mutation mn. Multiplication of A on the left!..

VS of the identity matrix 14 4th. T he signed product is 11 8. The signed product. The Exercise Set 4. Using row 2: Using column 2: Thus the determinant. If we e xpand a long the first row. A be a diag0nal matrix with nonzero diagonal entries.

By expanding along the t hird column. Jet A Thus there are a.

We wish to pwve that for each positive integer n. These two steps complete the proof by induetion. Xl The latter condition is equivalent to Y3. Step 2 induction srep. Our proof is by induction on n. Yt x2. If x A and. Corresponding row operations are as indicated.

If we replace the first row of the matrix by the sum of the first and second rows. The second and third rows were interchanged. We use the properties of determinants stated in Theorem 4. Chapter 4 LO. We use the properties of determinants stated in Theorerr1 4. Corresponding row opera. We usc the properties of deter minants stated in T heorem 4.

If we add the first row of A to the second row, the rMult is a lfla. If we add of the first four rows of A t o the last row, t. Thus, from Theorem 4. The matrices sin-gular if and only if the corresponding determ inants are zero.

This leads to the system of equations. This pattern continues and can be summarized as follows:. Let A be an n x n matrix, and let B be the matrix that results when the rows of A are written in reverse order. Then the matrix B can be reduced to A by a series of row interchanges.

This pattern continues;. F'tom Theorem 4. From Theorem 4. Thus the: If the reduced row echelon form of A has a row of zeros, then A is n ot invertible. If det. A f 0 then A is invertible, and an invertible matrix can always be written as a prod uct of elementary matrices. Each elementary product of this matrix must include a factor that comes from the 3 x 3 block of zeros on the upper right. Thus all of the elementary products are zero.

This permutation of the columns of an n x n matrix A can be attained via a sequence of n- 1 column interchanges which successively move the first column to the right by one position i. Thus the determinant of the resulting matrix is equal to -1 "- 1 det A.

Since C has two identical rows. Suppose A is a square matrix. Chapter 4 6. Thus A is invertible if a. In this case the solution is given by: X 3 co The formula for the inverse 1 In this example. We have det A is t2 2. Thus A is invertible if. These vectors lie in the same plane..

The vectors lie in the same plane if and only if the parallelepiped that they determine is Jegenerate in the sen. Let A. Chapter 4 Recall that the dot product distributes across addition. I Using properties of cross products stated in Theorem 4.

The vcc: A proof can be found in any standard calculus text. If A is upper triangular. It follows that the cofactor matrix C is lower triangular. Thus a vector lies in the plane determined by v and w if and only if it is orthogonal to v x w. If A is lower triangular and invertible. We have det AJ. I is lower triangular. B where is the angle between u and w. If either u or v is the zero vector. As was shown in the proof of Theorem 4. The associative law of multiplication is not valid for the cross product.

If u and v are nonzero then. In addition. On the ot. I eigenvalue. Thus area 6. The characteristic equation of Thus. Thus multiplicity 1. The characteristic equation is 1. A-4 b The characteristic equation is -t 5 5. This consist of a ll vectors of the form.

Chapter 4 5. X is the only c is det. J found by solving the system [ t.. The eigenspace con. Thus the eigenvalues of B are. XThe eigenvalues are. The eigenvalues a. Using t he block diagonal structure. Chapter4 Using the block triangular structure. Thoeigenvalues of A are!. The eigenvalues are. Thus the eigenspaces correspond to t he perpendicular li nes y: Thus the eigenspace The characteristic polynomial of A is p.

Corresponding eigenvectors arc the same as above. The eigenspace correspontJing to. Correspcnding eigenvectors same as above. The eigenvalues are.. The constant t. For t hese values of x..

The chl'l. Accorrling to Theorern 4. Note that the second factor in this polynomial cannot. T he characteristic polynomial of A i:. Sirni la dy. On the other har.. This leads to the equations 6a. This leads to the equations. Ax is a.

Thus t he o nly possible repeated eigenvalue of A is. AIIxll 2 and so l2. Using Formul a The correct statement is t. But it is true tha t a:.

If A is:. The statement becomes true if "independent" is replaced by "dependent". Thus We havP. The characteristic polynomial of A fac tors a. The eigenvalues of A wit h multiplicity arc 3.

Chapter4 Continuing in this fashion for n. J d False. The characteristic polynomial of A is a. The eigenspace of A conesponrling to. Xx where x i- 0 and A is invert. Thus is an eigenvalue of. Then sA x value of sA and x is a. T hus t he o nly sytnructric 2 x 2 matrices wi Lh repeated eigenvalues arc those of the form A """ al. This proves part b of Theorem 4.

The reduced row echelon form of the augmented matrix of the system is 0 1 0. If T is defined hy T x. The given equations can we writt.. If Tis defined by the formula Such a transformation cannot be linear since T O D From familiar trigonometric identities.

Chapter 6 d c X Such a. Since Tis linear. Since T O Exercise Set 6. The action of T on the standard unit. Tbe axis o f rotation. A x ofth.. Choosing the positive orientation and comparing with Table Exercise Set The matrix A is a.

T 1 x and x-T1x are or- 0. The matrix A is a rotation matrix since it is orthogonal and found by solving the of t his system is: Writing w in column vector form.

Simllarly for T2 and T3. Yll tho. I n order for the m a t rix to be orthogonal we rnust have and from this it follows that o. If A is an o rthogonal matrix and Ax: Vectors perpendicular to y x will be eigenvect ors corres p o nding to the eigenvalue. From thP. The shel. The t. The transfonnotion T 2. The kernP. The transformation Tis both and onto. The augmented matrix of thi..

The kernel of the tro. The kernel of the t ransformation TA: R 3 -t R4 is the solution set of the linear system [: Chapter6 From this we conclude that the system is con'listent. The augmented matrix of t. The operator can be written as [:: Since det A. The vector w is in the range of the linear operator T if and only if the linear system 2x.

The operator can he written as: Jn particular. TA is not onto. The range of T. TA is not 2 w1. In Th. The augmented matrix of this system can be row reduced as follows: Wt] 0: W2 3 This is just one possibility. The augmented matrix of the system Ax I Wz Wt. WJ] [1. See Theorem 6. If T is one-to-one.

T he transformation TA: No assu ming vo is not a scalar multiple of v. The transformation is not one-to-one since T v to a. If TA is not one-to-one.

If0 0] [00 -3 0! The fac toriut. Reflection of R. Rotation of R 2 about the origin through an angle of Contraction of R 2 by a factor of Expansion of R2 in the y-direc tion wjth factor 2. The standard matrix for the opP. A is invertible. Since A is invertible. The standard mat r ix for the operator T is A r. The standard matrix for 7 ' is A wl. The st andan. T 1s one-t rone and Since A is not invertible. The standard matrix for the operator Tis the sto. Tis one-to-one and thus r.

Tis one-to-one and! Tis not one-to-one. T is one. This l:: The image of the unit square is the parallelogram having the vectors T e! We have H: T int: This also follows from the computation carried out in Example l. ChapterS From Example 2. F Rk then ra.. The ar. If xis in Rn. Tho augmented mat. The augmented matrix of tl1e T he solution space is 2-dimensional wil.. Thus the general solution is [ 0 0 0 0 o: The solution space is 1-dimcnsional with cnnonic. T hus thevcctors v 1 form a basis for the hyperplane.

Thus the vectors v 1 form a basis for the hyperplane. Thus the vect. The augmented matrix of the system is 01 T he subspaces of R5 ha. A hyperplane in R! A basis for R3 must centum three vect ors two nre not eno11gh. Thus v 1. It follows t hat if S' is a ny nonempty subs et of S then S' must. The solution space bas dimension 1 in each case. Such a set is linearly dependent is a linear combination of its predecessors.

The matrix having v 1. Chapter 7 4 Oo tbe other hand. T he matrix having v 1. In pa r ticular. Let A be the matrix having v 11 v 2. Since del 0 I J R3. Thus Vis contained in IV. Vis not contained in W. Thus Vis not contained in W. Thus Vis contained a il' w. Chapter7 Thus a. Thus 4. Thus t here a rc a t. Let V b e a nonzero subspace uf R". Any set of more than n vectors in Rn is linearly depe ndent.

Since det -5 -! Let A be the matrix having the vectors v 1 and dct A ""' sin 2 o. Thus the number of clements in a spanning set must be at least k. Suppose W is a. E ach such operat r corresponds to and is determined by a permutation of the vectors in t he basis B. Any set of less than n vectors cannot be spanning set for Rn. Since VJt v2. Then Sis a linearly independent set in W T vn are inearly independent. Suppose c 1. Then the vectors x f.

Thus it makes sense to define a transformation T: Vn for exactly one choice of scalars c Thus the vectors Vt.

Suppose x is an eigenvector of A. Then x: Thus T v1. Yn span Rn and are linearly independent. Parametric equations for this plane are: Thus WJ. A vector that is orthogonal to both v 1 and v 2 is w!. Thus u is orthogonal to any linear combination of the vectors vlt v 2. Thus u is orthogonal to any linear combination of the vectors v Here W J.

(Solution Manual) Contemporary Linear Algebra by Howard Anton, Robert C. Busby

Note the vector w is parallel to the one obtained In our first solution. Thus SJ. Let A be t he matr ix having the given vectors as its rows. Let A be the matrix having the given vectors as its rows. We have w.. Let A bt! We have Wl.

We have W.

CHEAT SHEET

This matrix can be row reduced to! We have w. The augmented matrix of this system is 1 5 7l bl] -1 -4 -6! Let A be the matrix having the given as its columns. Let A be the matrix having the given vectors s. Then a column vector b is a linear combination of v1.

Solution 1. Solution 2. Solution 3. Solution l. Then a column vector b is a. Then a column vector b is a linear combination of these vectors if a. Chap1er 7 Solution 1.

Solutwn 3. Let A be the matrix having the given vectors as its columns. The matrix v3! The augmented matrix of this system is 0 -2 4 0 1 0 2 1 2 2 -1 -1 b.. The augmented mat. Let A be the matrix having the gi ven vectors as its columns.! From this we conclude Uwt the vectors h 1 2 4.

The aug mented matrix lv 1 v2 V:. Thus the 0 0 0 0 vectors r Thus row A and null A are orthogonaf. It and OQI J respectively. The reduced row forms of A and B are n form 00 13 a basis for both of the row spaces..

We also conclude from a. It is easy to check that r. The vectors r. Thus the 0 0 0 1 vectors r l T his follows from the fact T heorem 7. T his statement is t. On the othe:. If A ha. Let B be the matrix having the given vectors as its rows. Thus S l. The null space of A corresponds to the kernel of TA.

This is in fact true for any invertible matrix E. The first. From Theorem 7. From T heorem 7. The rows of EA are linear combinations of t. The row space of an invertible n x n matrix is all of Rn since its rows form a ba.

The mill matri. On the other hancl. Thus a holds. If Pis invertible. The row vectors of an invertible matrix are linearly independent and so. The reduced row echelon form for A is 1 ll -tl 0 0 0 0:! T he reduced row echelon form for A. It follows t hat nullit. If 4 1s a 6 x I then lhe largest possible value for rank. If A is a 7 x 9 matrix having rank 5. Thus the vectors. Thus there a.

The matrix A can be row reduced to upper triangular form as follows: T hen the reduced row echelon form of A is 0 0 -. Let A be the 3 x 5 matrix having v Thus the vectors 1. If lhe matrix Thus x.

The suh. For all other z] has rank 1 then the first row must be a scalar multiple of the second row. B has only one nonzero column or row. Thus B. The reduced row echelon form. B has only one nonzero entry hence only one nonzero row. Then B.. A has rank 1. A fails to be jnvertible if and only if there is a nonzero. Assuming u and v are nonzero vect. Thus if A Let A be the s tandard matrix ofT. If the additional row is a linear combination of the existing rows.

Note thnt. A is symmetric. Thus it is sufficient to prove that the latter is equivalent to the condition that Suppose that one of the rows of A is a scalar multiple of the other.. Since rank B. Since the set V U W contains n vectors. It follows that 2. A uvA Note. If vTA. Since the vectors v1. This shows that V U W is a linearly independent set. The number of parh. The number of parameters in a general solution is n. The number of parameters in a.

This This corresponds t.. This corresponds to the fact. This corresponds to the fact that A hu. This corresponds to t he fact that A does not have full column rank but does have full row rank.

Th1s corresponds to the fact see Exercise lOa that A has full column rank but not full row rank b ciet AT. This corresponds to the fact see Exercise 9a that A has fuJl column rank but not full row rank. Thus the system [ 0 0 1 b If an m x n matrix A has full row rank u. Frnm T heorem 7. Thus if A. Under these assumptions. If A has rank k the n. A has full column f rank ami full row rank.

As above. This docs not violate T heorem 7. A has full row rank. The row space and column space always have t he same dimension.

Let 8 be the m x r submatrix of A having these vectors as its columns. Let C be an invertible k x k submatrix of A. Let C be the k x k submatrix of B having these vectors as its rows. Let B be the m x k submatrix of A having these vectors as tts columns. Then B also has rank r and thus has r lmearly independent rows.

If A is an m x n matrix with rank k. Then the columns of C are linearly independent and so the columns of A that contain the columns of C are also linearly independent. Thus D is singular Conversely. The first column of A forms a basis for col A. Suppose now that y belongs to null A n col A. The proof is organized as suggested. Let C be the submatrix of B having these vec-torc.

The ma. Step 2. First we prove that if A is a nonzero matrix with rank k. If A is invertible then so is AT. It follows that bne columns of D are linearly dependent since a nontrivial linear dependence arx1ong the containing columns results in a nontrivial linear dependence among the columns of D. This matrix also has rank k and thus bas k linearly independent rows. Then Cis an invertible k x k submatrix of A.

Proceeding as in Example 2. The redur. Thus the first two columns of 0 0 0 0 0 and the reduced row echelon form of AT A form a basis for col A.

From this we conclude that all three columns of A are pivot columns. Let A be t he matrix having the given vectors as its columns. Chapter7 8. From this we conclude that the 1st and 3rd columns of A are the pivot columns. The redu-: The reduced row echelon form of the partitioned matrix [A I! Thus 1.

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