Find a subset of [tex]$S=\left\{\bar{v}_1, \bar{v}_2, \bar{v}_3, \bar{v}_4\right\}$[/tex] that forms a basis for the space spanned by the vectors in [tex]$S$[/tex]; then express each vector not in the basis as a linear combination of the basis vectors.
(a) [tex]$\bar{v}_1=(-3,5,1,-5), \bar{v}_2=(-3,3,-1,-7), \bar{v}_3=(3,-1,3,9), \bar{v}_4=$ $(0,1,1,1)$[/tex].
(b) [tex]$\bar{v}_1=(1,1,0,1), \bar{v}_2=(-1,3,2,6), \bar{v}_3=(-1,5,3,-2), \bar{v}_4=(3,0,5,-1)$[/tex].
Answer (1)
For part (a), the RREF of the matrix formed by the vectors has pivot columns 1 and 2. Thus, a basis is { v ˉ 1 , v ˉ 2 } , and v ˉ 3 = v ˉ 1 − 2 v ˉ 2 , v ˉ 4 = 2 1 v ˉ 1 − 2 1 v ˉ 2 .
For part (b), the RREF of the matrix formed by the vectors has pivot columns 1, 2, 3, and 4. Thus, a basis is { v ˉ 1 , v ˉ 2 , v ˉ 3 , v ˉ 4 } .
(a) Basis: { v ˉ 1 , v ˉ 2 } , v ˉ 3 = v ˉ 1 − 2 v ˉ 2 , v ˉ 4 = 2 1 v ˉ 1 − 2 1 v ˉ 2 .
(b) Basis: { v ˉ 1 , v ˉ 2 , v ˉ 3 , v ˉ 4 } .
(a) Basis: { v ˉ 1 , v ˉ 2 } , v ˉ 3 = v ˉ 1 − 2 v ˉ 2 , v ˉ 4 = 2 1 v ˉ 1 − 2 1 v ˉ 2 ; (b) Basis: { v ˉ 1 , v ˉ 2 , v ˉ 3 , v ˉ 4 }
Explanation
Problem Analysis We are given a set of vectors S and asked to find a basis for the space spanned by these vectors, and to express any vectors not in the basis as a linear combination of the basis vectors. We will analyze the two cases separately.
Setting up the Matrix (a) For part (a), we have the vectors v ˉ 1 = ( − 3 , 5 , 1 , − 5 ) , v ˉ 2 = ( − 3 , 3 , − 1 , − 7 ) , v ˉ 3 = ( 3 , − 1 , 3 , 9 ) , v ˉ 4 = ( 0 , 1 , 1 , 1 ) . We form a matrix A with these vectors as columns: A = [ − 3 − 3 3 0 5 3 − 1 1 1 − 1 3 1 − 5 − 7 9 1 ] .
Finding the Basis (a) We find the reduced row echelon form (RREF) of A . The RREF of A is: RREF ( A ) = [ 1 0 1 1/2 0 1 − 2 − 1/2 0 0 0 0 0 0 0 0 ] . The pivot columns are the first and second columns. Therefore, a basis for the span of the vectors in S is given by the set { v ˉ 1 , v ˉ 2 } .
Expressing Non-Basis Vectors (a) Now we express the non-basis vectors v ˉ 3 and v ˉ 4 as linear combinations of v ˉ 1 and v ˉ 2 . From the RREF, we have: v ˉ 3 = 1 ⋅ v ˉ 1 − 2 ⋅ v ˉ 2 , which means v ˉ 3 = v ˉ 1 − 2 v ˉ 2 .
v ˉ 4 = 2 1 ⋅ v ˉ 1 − 2 1 ⋅ v ˉ 2 , which means v ˉ 4 = 2 1 v ˉ 1 − 2 1 v ˉ 2 .
Setting up the Matrix (b) For part (b), we have the vectors v ˉ 1 = ( 1 , 1 , 0 , 1 ) , v ˉ 2 = ( − 1 , 3 , 2 , 6 ) , v ˉ 3 = ( − 1 , 5 , 3 , − 2 ) , v ˉ 4 = ( 3 , 0 , 5 , − 1 ) . We form a matrix B with these vectors as columns: B = [ 1 − 1 − 1 3 1 3 5 0 0 2 3 5 1 6 − 2 − 1 ] .
Finding the Basis (b) We find the reduced row echelon form (RREF) of B . The RREF of B is: RREF ( B ) = [ 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 ] . The pivot columns are the first, second, third, and fourth columns. Therefore, a basis for the span of the vectors in S is given by the set { v ˉ 1 , v ˉ 2 , v ˉ 3 , v ˉ 4 } .
Expressing Non-Basis Vectors (b) Since all vectors are in the basis, there are no vectors to express as a linear combination of the basis vectors.
Final Answer In summary: (a) A basis is { v ˉ 1 , v ˉ 2 } . v ˉ 3 = v ˉ 1 − 2 v ˉ 2 and v ˉ 4 = 2 1 v ˉ 1 − 2 1 v ˉ 2 .
(b) A basis is { v ˉ 1 , v ˉ 2 , v ˉ 3 , v ˉ 4 } .
Examples
Understanding vector spaces and bases is crucial in various fields like computer graphics, physics, and engineering. For instance, in computer graphics, vectors are used to represent 3D objects, and finding a basis helps in efficiently storing and manipulating these objects. If some vectors are linearly dependent, they can be expressed as a combination of the basis vectors, reducing the storage space required. Similarly, in structural engineering, identifying a basis for forces acting on a structure helps in simplifying the analysis and ensuring stability.
