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Home / Linear Algebra / Linear combination, Spanning set| Linear Algebra

Linear combination, Spanning set| Linear Algebra

Linear combination: Let V be a vector space over a field K. A vector v in V is a linear combination of vectors u1, u2,—–, um in V if there exist scalars a1, a2, —–, am in K such that

v= a1u1+ a2u2+—+ amum

Example: (Linear combination in Rn): Suppose we want to express v= (3, 7, -4) in R3 as a linear combination of the vectors

u1=(1, 2, 3) u2= (2, 3, 7)   u3= (3, 5, 6)

We suppose scalars x, y, z such that

v=xu1+ yu2+ zu3; that is

=x+y+z

⇒x+2y+3z=3

2x+3y+5z=3

3x+7y+6z=-4

Reducing the system to echelon form yields

x+2y+3z=3

-y   -z   =1

Y  -3z =-13

Again, reducing the system to echelon form yields

x+2y+3z=3

-y   -z   =1

-4z =-12

From last equation we get, z=3

From second equation we get, y = -4

And from first equation we get, x=2

Thus, v= v=2u1-4u2+3u3

Spanning set: Let V be a vector space over K. Vectors u1, u2, …, um in V are said to span V or to form a spanning set of V if every v in V is a linear combination of the vectors u1, u2, …, um that is, if their exist scalars a1, a2, …, am in K such that

v= a1u1+ a2u2+—+ amum

Subspace: Let V be a vector space over a field K and let W be a subset of V. Then W is a subspace of V if W is itself a vector space Over K with respect to the operations of vector addition and scalar multiplication on V.

Prove that, every basis of a vector space V has the same number of elements.

Problem: Prove that, every basis of a vector space V has the same number of ...