close
標題:
basis for a subspace
發問:
which of the following sets of vectors can be a basis for a subspace of R^3? when the set can be a basis, describe the subspace geometrically. (1) { (1,0,0), (0,1,0), (0,0,1), (1,2,3) } (2) { (1,0,1) }
最佳解答:
(1) The set of vectors must not be a basis as any 4 vectors in R^3 must be linearly dependent. (2) The set is not a basis in R^3 as there is only one R^3 vector (Actually a set consisting one R^3 vector can only span a line)
其他解答:
They both cannot be a basis. Remember that if a vector space has n-dimension, the basis set should also have n-dimemsion. Namely, n number of vectors in the basis set. For example, R^3 has a dimension of 3, its basis set must have three vectors. R^6 has a dimension of 6, its basis set must have six vectors. You can recongnize the dimension of an R^n vector space by reading the number of n. If n = 6, the dimension of this vector space is 6. If n= 2, the dimension of this vector space is 2, etc. For polynomials, you have to recognize the dimension by reading the value of n and added up 1. For example, P3 has the dimension of 3 plus 1=4. P5 has the dimension of 5 plus 1=6, etc. For matrices, you have to read the dimension by multiplying the number of rows and columns. For example, M32 has a dimension of 3*2=6. M48 has a dimension of 4*8=32 2008-08-12 11:31:08 補充： A set of 3 vectors can be a basis for R^3 provided with one of the following conditions satisified: 1) It is a linearly independent set 2) It spans R^3 So if the number of vectors in a set doesn't equal to the dimension of the vector space, it must bot be a basis. 2008-08-12 11:34:34 補充： But if the number of vectors = the dimension of the vector space, it doesn't mean it is a basis. The set needs to have three vectors and also satisfy one of the above conditions in order to be qualified to be a basis. 2008-08-12 11:34:46 補充： A set consists of one R^3 vector cannot span a line. For example, (1,0,1) cannot span the line (1,2)
basis for a subspace
發問:
which of the following sets of vectors can be a basis for a subspace of R^3? when the set can be a basis, describe the subspace geometrically. (1) { (1,0,0), (0,1,0), (0,0,1), (1,2,3) } (2) { (1,0,1) }
最佳解答:
(1) The set of vectors must not be a basis as any 4 vectors in R^3 must be linearly dependent. (2) The set is not a basis in R^3 as there is only one R^3 vector (Actually a set consisting one R^3 vector can only span a line)
其他解答:
They both cannot be a basis. Remember that if a vector space has n-dimension, the basis set should also have n-dimemsion. Namely, n number of vectors in the basis set. For example, R^3 has a dimension of 3, its basis set must have three vectors. R^6 has a dimension of 6, its basis set must have six vectors. You can recongnize the dimension of an R^n vector space by reading the number of n. If n = 6, the dimension of this vector space is 6. If n= 2, the dimension of this vector space is 2, etc. For polynomials, you have to recognize the dimension by reading the value of n and added up 1. For example, P3 has the dimension of 3 plus 1=4. P5 has the dimension of 5 plus 1=6, etc. For matrices, you have to read the dimension by multiplying the number of rows and columns. For example, M32 has a dimension of 3*2=6. M48 has a dimension of 4*8=32 2008-08-12 11:31:08 補充： A set of 3 vectors can be a basis for R^3 provided with one of the following conditions satisified: 1) It is a linearly independent set 2) It spans R^3 So if the number of vectors in a set doesn't equal to the dimension of the vector space, it must bot be a basis. 2008-08-12 11:34:34 補充： But if the number of vectors = the dimension of the vector space, it doesn't mean it is a basis. The set needs to have three vectors and also satisfy one of the above conditions in order to be qualified to be a basis. 2008-08-12 11:34:46 補充： A set consists of one R^3 vector cannot span a line. For example, (1,0,1) cannot span the line (1,2)
此文章來自奇摩知識+如有不便請留言告知
文章標籤
全站熱搜
留言列表
