Linear Algebra Linear Independence and Basis TF T F II A is

Linear Algebra: Linear Independence and Basis (T/F)

T F: II A is an n x n matrix and it is invertible, then its columns are linearly independent. T F: Suppose A and B are n x n matrices. If A is upper triangular and B is lower triangular, then A and B are linearly independent. T F: If the Wronskian of f_1(x), f_2{x),...,f_n(x) is zero, then f_1(x), f_2(x) ,...,f_n(x) are linearly dependent. T F: Two real-valued functions f_1 and f_2 are linearly independent if and only if f_1{x) and f_2(x) are linearly independent for each x. T F: If V = span{v_1,...,v_n} then {v_i,...,v_n} is a basis for V. T F. If {v_i,..., v_n} is a basis for a vector space V, then every vector in V can be expressed as a linear combination of v_1,..., v_n. T F: If {v_1,..., v_n} is a basis for a vector space V, then every subset of {v_1,..., v_n} is linearly independent. T F: Every basis of P_2 must contain a polynomial of degree less than 2.

Solution

b) let the columns of A are a_1,a_2 - - - - a_n(Vectors in IRⁿ). If you consider a linear combination of them(a potential linear dependance relation)
   X₁a_1+X₂a_2+ - - - - +X_na_n=0

This is equivalent to the matrix Vector product Ax=0 where x=(x₁,x₂,- - - -x_n)^T.

Multiply A^(-1) to get

A^(-1)Ax=0===>x=0

that is, the vector equation has only one trivial solution providing that the vectors(a₁,a₂,- - - -a_n) are linearly independent.

f) True

A basis is a minimal spanning set therefore any spanning set contains basis.

g) True

Consider any vector x in V

since S spans V

there is atleat one way to express x as linear combination of elements of S.

now suppose that there are two different ways of doing this.

suppose that there are coefficients a_i and b_i(not all equal ) such that

x=a₁v₁+a₂v₂+- - - - +a_nv_n

and

x=b₁v1+b₂v₂+- - - +b_nv_n

subrracting we have

0=(a₁-b₁)v₁+(a₂-v₂)v₂+- - - - +(a_n-v_n)v_n

since theremust be some i such that a_i is different from b_i

we know that atleast one of there(a_i-b_i) coefficients is non zero.

thus we have written 0 as nontrivial linear combination of members of S, so they are not linearly independent.

this controdicts the given informaiton that is a basis.

our assumption that x could be written in multiple ways as a linear combination of v_i resulted in a controdiction,

so it must not be true.

so there is one and only one way to write x as a linear combination of vectors.

h) let S={v₁,v₂,-----vn} is a basis of Vector space V

===> S spans V

===>S is n dimensional sub space of V which is also n dimensional.

====> S={v₁,v₂,-----vn} is linearly independent.

g