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

5 not required 6 answer each of the following as true

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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
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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.
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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
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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 ).
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