LINEAR ALG
Solucionario-de-algebra-lineal-Kolman-octava-edicion.pdf

5 not required 6 answer each of the following as true

• 60

This preview shows pages 54–58. Sign up to view the full content.

5. Not required. 6. Answer each of the following as true or false. (a) All vectors of the form ( a, 0 , - a ) form a subspace of R 3 . (b) In R n , k c x k = c k x k . (c) Every set of vectors in R 3 containing two vectors is linearly inde- pendent. (d) The solution space of the homogeneous system A x = 0 is spanned by the columns. (e) If the columns of an n × n matrix form a basis for R n , so do the rows. (f) If A is an 8 × 8 matrix such that the homogeneous system A x = 0 has only the trivial solution then rank( A ) < 8. (g) Not required. (h) Every linearly independent set of vectors in R 3 contains three vec- tors. (i) If A is an n × n symmetric matrix, then rank( A ) = n . (j) Every set of vectors spanning R 3 contains at least three vectors. Solution. (a) True. Indeed, these vectors form exactly

This preview has intentionally blurred sections. Sign up to view the full version.

54 CHAPTER 6. VECTOR SPACES span { (1 , 0 , - 1) } which is a subspace (by Theorem 6.3, p. 285). (b) False. Just take c = - 1 and x 6 = 0 . You contradict the fact that the length (norm) of any vector is 0. (c) False. x = (1 , 1 , 1) and y = (2 , 2 , 2) are linearly dependent in R 3 because 2 x - y = 0 . (d) False. The solution space of the homogeneous system A x = 0 is spanned by the columns which correspond to the columns of the reduced row echelon form which do not contain the leading ones. (e) True. In this case the column rank of A is n . But then also the row rank is n and so the rows form a basis. (f) False. Just look to Corollary 6.5, p. 335, for n = 8. (h) False. For instance, each nonzero vector alone in R 3 forms a linearly independent set of vectors. (i) False. For example, the zero n × n matrix is symmetric, but has not the rank = n (it has zero determinant). (j) True. The dimension of the subspace of R 3 spanned by one or two vectors is 2, but dim R 3 = 3.
Chapter 8 Diagonalization Page 421. T1. Let λ j be a particular eigenvalue of A . Show that the set W of all the eigenvectors of A associated with λ j , as well as the zero vector, is a subspace of R n (called the eigenspace associated with λ j ). Solution. First, 0 W and so W 6 = . Secondly, if x , y W then A x = λ j x , A y = λ j y and consequently A ( x + y ) = A x + A y = λ j x + λ j y = λ j ( x + y ). Hence x + y W . Finally, if x W then A x = λ j x and so A ( c x ) = c ( A x ) = c ( λ j x ) = λ j ( c x ). Hence c x W . T3. Show that if A is an upper (lower) triangular matrix, then the eigenvalues of A are the elements on the main diagonal of A . Solution. The corresponding matrix λI n - A is also upper (lower) tri- angular and by Theorem 3.7 (Section 3.1, p. 188), the characteristic poly- nomial f ( λ ) is given by: λ - a 11 - a 12 ... - a 1 n 0 λ - a 22 ... - a 2 n . . . . . . . . . 0 0 ... λ - a nn = ( λ - a 11 )( λ - a 22 ) ... ( λ - a nn ) 55

This preview has intentionally blurred sections. Sign up to view the full version.

56 CHAPTER 8. DIAGONALIZATION (expanding successively along the first column). The corresponding char- acteristic equation has the solutions a 11 , a 22 , ..., a nn . T4. Show that A and A T have the same eigenvalues. Solution. Indeed, these two matrices have the same characteristic poly- nomials: det( λI n - A T ) = det(( λI n ) T - A T ) = det(( λI n - A ) T ) Th . 3 . 1 = det( λI n - A ).
This is the end of the preview. Sign up to access the rest of the document.
• Fall '16
• jjaa
• Diagonal matrix, Det, Solution.

{[ snackBarMessage ]}

What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern