- #1

- 6

- 0

## Homework Statement

Let V be a vector space, and suppose that [itex]\vec{v_1}[/itex], [itex]\vec{v_2}[/itex], ... [itex]\vec{v_n}[/itex] is a basis of V. Let [itex]c\in\mathbb R[/itex] be a scalar, and define [itex]\vec{w}[/itex] = [itex]\vec{v_1} + c\vec{v_2}[/itex]. Prove that [itex]\vec{w}, \vec{v_2}, ... , \vec{v_n}[/itex] is also a basis of V.

## Homework Equations

If two of the following conditions hold, the third holds automatically, and the list is a basis for a vector space V.

1. If dimV = the number of vectors in the list.

2. If the list of vectors spans V.

3. If the list of vectors is linearly independent.

## The Attempt at a Solution

So, I feel like I did this right, but I have no way to check because we don't use a textbook for the class and I couldn't find it online anywhere. I did the following.

Since [itex]\vec{v_1}[/itex], [itex]\vec{v_2}[/itex], ... [itex]\vec{v_n}[/itex] is a basis of V, [itex]\vec{v_1}[/itex], [itex]\vec{v_2}[/itex], ... [itex]\vec{v_n}[/itex] are linearly independent and span V. Also, dimV = n, because n vectors form the basis of V.

[itex]\vec{w}, \vec{v_2}, ... , \vec{v_n}[/itex] is also a list of n vectors, so it will be sufficient to show that this list either spans V or is linearly independent in order for it to be a basis of V.

Since [itex]\vec{v_1}[/itex], [itex]\vec{v_2}[/itex], ... [itex]\vec{v_n}[/itex] span V, that means there exists some scalars a

_{i}(where i is between 1 and n inclusive) such that

a

_{1}v

_{1}+ a

_{2}v

_{2}+ ... + a

_{n}v

_{n}= [itex]\vec{x}[/itex], where [itex]\vec{x}[/itex] is in V.

So, for [itex]\vec{w}, \vec{v_2}, ... , \vec{v_n}[/itex] to span V, there must exist some scalars b

_{i}such that b

_{1}w + b

_{2}v

_{2}+ ... + b

_{n}v

_{n}= [itex]\vec{x}[/itex].

Then I basically expand this, and set it equal to the previous expression with all the a's. I show that in order for them to be equal, choose b

_{3}= a

_{3}, b

_{4}= a

_{4}, ... b

_{n}= a

_{n}.

Then all we have to show is that

b

_{1}v

_{1}+ (b

_{1}c + b

_{2})v

_{2}= a

_{1}v

_{1}+ a

_{2}v

_{2}.

Choose b

_{1}= a

_{1}, and b

_{2}= a

_{2}- a

_{1}c.

Then a

_{1}w+ (a

_{2}- a

_{1}c)v

_{2}+ a

_{3}v

_{3}+ ... + a

_{n}v

_{n}= [itex]\vec{x}[/itex], so it spans V.

Thus, since the dimension if n and it spans V, it also forms a basis of V.

Is this right, though?