### Matrix Analysis and Applied Linear Algebra Book and Solutions Manual

The fact that eigenvalues associated with diagonal matrices have index 1 while eigenvalues associated with triangular matrices can have higher indices is no accident. Consequently, A and AD are one-to-one map k Exercise 4. The desired result follows because the general solution is any particular solution plus the general solution of the associated homogeneous equation.

In parts a and b the identity element is the ordinary identity matrix, and the inverse of each member is the ordinary inverse. Statement c now follows from 5. The fact that E is the desired projector follows from 5. Notice that the group inverse agrees with the Drazin inverse of A described in Example 5. However, the Drazin inverse exists for all square matrices, but the concept of a group inverse makes sense only for group matrices—i.

Proceed as described on p. T According to 5. According to the discussion of projectors on p.

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Use the results of Exercise 5. The example in the solution to Exercise 5. To prevent A from being normal, simply 0 0 1 2 choose C to be nonnormal. Y P Solutions for exercises in section 5. To resolve the inequality with what it means for points to be on an ellipsoid, realize that the surface of a degenerate ellipsoid one having some semiaxes with zero length is actually the set of all points in and on a smaller dimension ellipsoid.

For example, visualize an ellipsoid in 3 , and consider what happens as one of its semiaxes shrinks to zero. The skin of the three-dimensional ellipsoid degenerates to a solid planar ellipse. Since the 2-norm is unitarily invariant Exercise 5. The development of the more general bound is the same as for 5. You may wish to computationally verify thatthis is indeed the case. The other part is similar. The other parts are similar. So, by 5.

The other part is argued in the same way. The Pythagorean theorem Exercise 5. Equation 5. The Householder or Givens reduction technique can be employed as described in Example 5. If either u or v is the zero vector, then L is a one-dimensional subspace, and the solution is given in Example 5. Suppose that neither u nor v is the zero vector, and let p be the orthogonal projection of b onto L. Example 5. It follows from 5. Sketch a picture similar to that of Figure 5. Refer to Figure 5.

According to part c of Exercise 5. It is now straightforward to verify that the points created by the algorithm are exactly the same points described in Steps 1, 2,. It can be argued that the analogous situation holds at each step of the process—i. G I R Solutions for exercises in section 5. Now use the linearity of trace and expectation together with the result of Exercise 5.

You can use either 5. It was argued in the proof of 5. Combine this with the formula for the rank of a product 4. Consequently, 6. Use the product rule 6. Use 6. According to 3. Therefore, the results of Exercise 6. The fact that each pivot is positive follows from Exercise 6. It was argued in Example 4. This is equivalent to saying that if S is a linearly dependent set, then the Wronski matrix W x is singular for all values of x. But 6. The converse of this statement is false Exercise 4. Recall from Example 5. According to 6. Expand T both of the ways indicated in 6.

The result follows from Example 6. Consequently, approximately 1. The 2 For this matrix, a , c , and d are eigenvectors associated with eigenvalues 1, 3, and 3, respectively. This follows directly from 6. Zero is not in or on any Gerschgorin circle. You could also say that A is nonsingular because it is diagonally dominant—see Example 7.

But, as discussed on p. So, the root in the isolated circle must be real and there must be at least one real root in the union of the other three circles. Use Exercise 7. D E Therefore, 7. Almost any example with rather random entries will do the job, but avoid diagonal or triangular matrices—they are too special. Proceed by induction. Since no eigenvalue is repeated, 7. A similarity transformation P that diagonalizes A is constructed from a complete set of independent eigenvectors. Consider the matrix Exercise 7.

Of course, you could compute A, A2 , A3 ,. A better technique is to diagonalize A with a similarity transformation, and then use the result of Exercise 7. The result of Exer0. From Example 7. Follow the procedure described in Example 5. We know from 7. Use an indirect argument for the converse. If A has k distinct eigenvalues, then the desired conclusion is attained after k repetitions.

This follows from the eigenvalue formula developed in Example 7. Exercise 5. Conversely, if this equation holds, then Exercise 5. Let p. Continuing in this manner produces the desired conclusion. This matrix is nonsingular because Exert cise 6. D E Induction can now be used. Solutions for exercises in section 7. Note that A is diagonalizable because no eigenvalue is repeated.

The trace is the sum of the eigenvalues, and the determinant is the product of the eigenvalues p. When A is diagonalizable, 7. Let xk be the fraction of switches in the ON state and let yk be the fraction of switches in the OFF state after k clock cycles have elapsed. The spectral theorem on p. We already know from 7. See the solution to Exercise 5. Consider the identity matrix—every nonzero vector is an eigenvector, so not every complete independent set of eigenvectors needs to be orthonormal.

Repeating this argument for each row produces the conclusion that T must be diagonal. If A is normal, then so is T. Exercise 7. Conversely, if T is diagonal, then it is normal, and thus so is A.

## Matrix Analysis and Applied Linear Algebra Book and Solutions Manual

The solution for Exercise 7. Consequently, the spectral decomposition p. Use the results on p. The 2-norm condition number is the ratio of the largest to smallest singular values. The procedure is essentially identical to that in Example 7. If the O h4 terms are neglected, and if the boundary values gij are taken to the right-hand side, then, with the same ordering as indicated in Example 7. This follows from the result on p. As noted in Example 7. Use the procedure on p. You might also determine this just by inspection.

Thus every Jordan block is similar to its transpose.

## Matrix analysis - Wikipedia

It was established in Exercise 7. The eigenvalues of An were determined in Exercise 7. The same argument given in the solution of the last part of Exercise 7. It follows from 7. A is the matrix in Example 7. This follows because, as explained in Example 7. Because every square matrix is similar to its transpose recall Exercise 7. So f A exists if and only if f AT exists. As proven in Example 7. The disadvantage is that a higher-degree polynomial might be required, so a larger system might have to be solved. But using fi in 7. Solutions C 0 7.

O C Solutions for exercises in section 7. Since A is convergent, 7. Use 7.

Depending on your own implementation, your answers may vary slightly. Jacobi with 21 iterations: 1 1. Use the result of Example 7. Note: The same proof works for vectors and matrices by replacing with a vector or matrix norm. It was shown in Example 7. We already know from the development of 7.

Similar matrices have the same minimum polynomial because similar matrices have the same Jordan form, and hence they have the same eigenvalues with the same indicies. The result for column sums follows by considering AT. Use Example 7. The lower bound follows from the Collatz—Wielandt formula.

Let p and qT be the respective the right-hand and left-hand Perron vectors for A associated with the Perron root r, and use 8. If A is nonsingular then there are either one or two distinct nonzero eigenvalues inside the spectral circle. Since all eigenvalues on the spectral circle are simple recall 8. A is irreducible because the graph G A is strongly connected—every node is accessible by some sequence of paths from every other node.

A is imprimitive. Since S is irreducible, the result in Example 8. Construct the Boolean matrices as described in Example 8. According to the discussion on p. Thus the limiting distribution is the uniform distribution, and in the long run, each state is occupied an equal proportion of the time.

Exercise 8. We know from Exercise 8. Using the eight-state chain yields the following mean-time-to-failure vector. This is a Markov chain with nine states c, m in which c is the chamber occupied by the cat, and m is the chamber occupied by the mouse. There are three absorbing states—namely 1, 1 , 2, 2 , 3, 3. Related documents. Books Spring Jorge Nocedal Professor of Industrial Engineering.

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