How To Tell If A Set Of Vectors Spans R3

Linear Independence and Span

Span

We take seen in the concluding word that the span of vectors five₁ , v₂ , ... , five _northward is the set of linear combinations

c₁ v₁ + c_ii five₂ + ... + c_n v _n

and that this is a vector space.

We now take this idea further.  If Five is a vector space and S  =  {v_ane , v₂ , ... , five _n ) is a subset of 5, and so is Span(S) equal to V?

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Definition

Let V be a vector space and let S  =  {v_ane, v₂, ... , 5_north) be a subset of V. We say that S spans Five if every vector v in V tin be written as a linear combination of vectors in S .

        v  =  c₁five_ane + c_iiv₂ + ... + c_nv_north

Example

Bear witness that the set up

   South =  {(0,1,i), (1,0,1), (1,i,0)}

spans R³ and write the vector (2,4,eight) as a linear combination of vectors in Southward.

Solution

A vector in R^three has the form

v  =  (x, y, z)

Hence nosotros demand to evidence that every such five tin be written as

   (x,y,z)  =  c₁(0, i, i) + c_ii(1, 0, i) + c_three(1, ane, 0)

        =  (c₂ + c₃, c_ane + c₃, c₁ + c_two)

This corresponds to the system of equations

        c₂ + c₃  =  ten
c_ane +       c_iii  =  y
c_one + c₂         =  z

which tin can be written in matrix form

We can write this equally

        Ac  = b

Notice that

        det(A)  =  2

Hence A is nonsingular and

c  =  A^-1 b

And then that a nontrivial solution exists.  To write (two,4,eight) every bit a linear combination of vectors in S, we find that

and then that

We have

     (2,4,8)  =  v(0,1,1) + three(1,0,1) + (-ane)(ane,1,0)

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Example

Prove that if

v₁   =  t + ii   and five₂   =  t² + ane

and S  =  {v₁ , v_two }

and so

S does non bridge P₂

Solution

A general element of P₂ is of the form

v   =  at^two + bt + c

We set up

v  =  c_one five₁ + c₂ v₂

or

    atⁱⁱ + bt + c  =  c_one(t + two) + c₂(t² + 1)  =  c_iit² + c_anet + c₁ + c_two

Equating coefficients gives

a  =  c_ii
      b  =  c₁
      c  =  c₁ + c₂

Notice that if

     a  =  one        b  =  i        c  =  1

there is no solution to this.  Hence S does not span Five.

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Example

Let

Detect a spanning prepare for the aught space of A.

Solution

Nosotros desire the set of all vectors x with

     Ax  = 0

We discover that the rref of A is

The parametric equations are

     x₁  =  7s + 6t
x_ii  =  -4s - 5t
x₃  =  south
x₄  =  t

We can become the span in the following way.  We first permit

southward  =  1        and  t  =  0

to get

v₁  =  (7,-four,ane,0)

and let

due south  =  0        and t  =  1

to go

five₂   =  (6,-5,0,1)

If nosotros let South  =  {5_i ,v_ii } then S spans the cipher space of A.

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Linear Independence

Nosotros now know how to notice out if a drove of vectors span a vector space.  It should exist clear that if South  =  {v₁ , five₂ , ... , v _n ) and so Bridge(Due south) is spanned by S.  The question that nosotros next ask is are in that location any redundancies.  That is, is at that place a smaller subset of S that as well span Bridge(S).  If so, then one of the vectors tin exist written as a linear combination of the others.

v_i   =  c_ane five_one + c₂ v₂ + ... + c_{i -1} v_{i -1} + c_i+1 5_i+1 + ... + c_n v _n

If this is the example then we call S a linearly dependent set up.  Otherwise, we say that Due south is linearly independent.  At that place is another way of checking that a set of vectors are linearly dependent.

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Theorem

Let S  =  {v_i, v₂, ... , v_northward) exist a set up of vectors, then Southward is linearly dependent if and only if 0 is a nontrivial linear combination of vectors in S .  That is, there are constants c_ane, ..., c_north with at least one of the constants nonzero with

         c_one v₁ + c₂ v₂ + ... + c_north v _{due north}   = 0

Proof

Suppose that Southward is linearly dependent, then

v_i =  c₁ v₁ + c₂ v_ii + ... + c_{i -one} five_{i -1} + c_i+1 v_i+one + ... + c_n v _n

Subtracting v_i from both sides, nosotros become

c_ane v₁ + c₂ 5_two + ... + c_{i -1} 5_{i -1} + five_i + c_i+1 v_i+1 + ... + c_n 5 _north   = 0

In the above equation c_i  =  1 which is nonzero, so that 0 is a nontrivial linear combination of vectors in Due south.

Now let

c₁ five₁ + c₂ 5₂ + ... + c_{i -i} v_{i -1} + c_i five_i + c_i+1 v_i+1 + ... + c_n five _n   = 0

with c_i nonzero.  Divide both sides of the equation past c _i and permit a_j  =  -c_j / c_i to go

-a₁ v_ane - a₂ five_ii - ... - a_{i -1} v_{i -1} + five_i- a_i+i 5_i+1 - ... - a_northward five _n   = 0

finally move all the terms to the other right side of the equation to get

v_i =  a₁ v_i + a_ii v_two + ... + a_{i -1} 5_{i -one} + a_i+1 v_i+1 + ... + a_n 5 _n

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Example

Prove that the fix of vectors

        S  =  {(1, 1, 3, 4),  (0, 2, iii, 1),  (four, 0, 0, 2)}

are linearly independent.

Solution

We write

        c₁(1, 1, 3, iv) + c₂(0, 2, 3, 1) + c_three(4, 0, 0, 2)  =  0

We become four equations

        c₁ + 4c₃  =  0
c_i + 2c_two  =  0
3c₁ + 3c_ii  =  0
4c₁ + c_ii + 2c₃  =  0

The matrix corresponding to this homogeneous system is

and

Hence

     c_one  =  c₂  =  c₃  =  0

and we can conclude that the vectors are linearly independent.

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Case

Let

     S  =  {cosⁱⁱ t, sin² t, 4)

then S is a linearly dependent fix of vectors since

   4  =  4cos^twot + 4sin²t

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Source: https://ltcconline.net/greenl/courses/203/Vectors/linIndSpan.htm

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