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Determine if each statement a. through e. below is true or false. Justify each answer a. A linearly independent set in a subspace H is a basis for H A. The statement is true by the Spanning Set Theorem B. The statement is false because the set must be linearly dependent. C. The statement is false because the subspace spanned by the set must also coincide with H D. The statement is true by the definition of a basis b. If a finite set S of nonzero vectors spans a vector space V, then some subset of S is a basis forV O A. The statement is false because the subspace spanned by the set must also coincide with V O B. The statement is false because the subset must be independent. O C. The statement is true by the Spanning Set Theorem O D. The statement is true by the definition of a basis c. A basis is a linearly independent set that is as large as possible O A. The statement is false because a basis is a linearly dependent set. O B. The statement is true by the Spanning Set Theorem C. The statement is false because a basis is the smallest independent set that spans the subspace O D. The statement is true by the definition of a basis d. The standard method for producing a spanning set for Nul A sometimes fails to produce a basis for Nul A. O A. The statement is true because the set produced may not be independent. O B. The statement is false because the method always produces an independent set. O C. The statement is true because the only set produced may be the trivial solution O D. The statement is false because a spanning set for Nul A also spans A. e. If B is an echelon form of a matrix A, then the pivot columns of B form a basis for Col A. O A. The statement is true by the Invertible Matrix Theorem O B. The statement is true by the definition of a basis ° C. The statement is false because the columns of an echelon form B of A are not necessarily in the column space of A. O D. The statement is false because the pivot columns of A form a basis for Col B

## Answer

(a) A linearly independent set in a subspace H is a basis for H. The statement does not mention the fact that the linearly independent set spans the subspace H. Therefore, it is not necessary that it is basis for H. Example: Rºis a subspace Take the linearly independe To span R’ at least 3 linearly independent vectors are required. But the linearly independent set has only 2 Li vectors. Thus, the statement is FALSE. (b) If a finite set S of non-zero vectors spans a vector space V, then some subset of Sis also basis of V. The set Sis linearly independent, and the set S is a subset of itself then S is basis of V Again, if the set is not Ll then some vectors can be written as linear combination of other vectors then if we remove that vectors then the remaining set is Ll and spans the vector space. Hence, basis of V. Continue in this way then there is a subset of set S which is linearly independent and spans V. Thus, the subset of Sis basis of vector space V.

The statement is TRUE. (c) A basis is a linearly independent set that is as large as possible. Consider the set of vectors S = {v, V, V3…v} which is a basis of vector space V. Then if we add some vector y, from vector space V into the set S then the set S’={v, V., V…V, V} is not a linearly independent set because the vector y can be written as linear combination of other vectors. Thus, the set S is the largest possible set of vectors which is basis of vector space V. The statement is TRUE.

(d) The standard method for producing a spanning set of Null A sometimes fail to produce a basis for Null A. The method of finding basis for Null A where the Null A consists of non-zero vectors gives linearly independent set of vectors. This is always true. Thus, the statement is FALSE. (e) If B is an echelon form of matrix A, then the pivot columns of B form basis for Col A. The pivot columns of B give the information that which columns of A forms a basis for Col A. But with the columns of matrix B, the basis for matrix A cannot be formed. The basis for Col A will correspond to columns of matrix A. Hence, the statement is FALSE.