# Basis of a Vector Space

1. Mar 16, 2015

### Izzy

1. The problem statement, all variables and given/known data
Let V be a vector space, and suppose that $\vec{v_1}$, $\vec{v_2}$, ... $\vec{v_n}$ is a basis of V. Let $c\in\mathbb R$ be a scalar, and define $\vec{w}$ = $\vec{v_1} + c\vec{v_2}$. Prove that $\vec{w}, \vec{v_2}, ... , \vec{v_n}$ is also a basis of V.

2. Relevant 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.

3. 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 $\vec{v_1}$, $\vec{v_2}$, ... $\vec{v_n}$ is a basis of V, $\vec{v_1}$, $\vec{v_2}$, ... $\vec{v_n}$ are linearly independent and span V. Also, dimV = n, because n vectors form the basis of V.
$\vec{w}, \vec{v_2}, ... , \vec{v_n}$ 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 $\vec{v_1}$, $\vec{v_2}$, ... $\vec{v_n}$ span V, that means there exists some scalars ai (where i is between 1 and n inclusive) such that
a1v1 + a2v2 + ... + anvn = $\vec{x}$, where $\vec{x}$ is in V.
So, for $\vec{w}, \vec{v_2}, ... , \vec{v_n}$ to span V, there must exist some scalars bi such that b1w + b2v2 + ... + bnvn = $\vec{x}$.

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 b3 = a3, b4 = a4, ... bn = an.

Then all we have to show is that
b1v1 + (b1c + b2)v2 = a1v1 + a2v2.
Choose b1 = a1, and b2 = a2 - a1c.

Then a1w+ (a2 - a1c)v2 + a3v3 + ... + anvn = $\vec{x}$, 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?

2. Mar 16, 2015

### fourier jr

Looks ok to me....

3. Mar 17, 2015

### Ray Vickson

There is a theorem that a set of $n$ linearly-independent vectors in an $n$-dimensional space $V$ does, indeed, span $V$. If you know and are allowed to use that theorem, you can shorten your argument a bit: just show that $\vec{w}, \vec{v}_2, \ldots, \vec{v}_n$ are linearly independent.

4. Mar 17, 2015

### HallsofIvy

The OP said in his first post
"
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. "
so, yes, he does know that.

5. Mar 17, 2015

### Ray Vickson

I missed the opening sentence.

6. Mar 17, 2015

### Izzy

I tried it like that first, but I couldn't actually figure out how to show that $\vec{w}, \vec{v}_2, \ldots, \vec{v}_n$ are linearly independent. Can I have a hint for how to do that?

Also, I'm a girl btw. :P

7. Mar 17, 2015

### Ray Vickson

Anyway, you want to know if there are $c_1, c_2, \ldots, c_n$, not all zero, such that $0 = c_1 w + c_2 v_2 + \cdots + c_n v_n = 0$. Take it from there.

8. Mar 17, 2015

### Izzy

I think I got it.
$0 = c_1 w + c_2 v_2 + \cdots + c_n v_n = 0$
Let $c_3, ... c_n = 0$, and now we'll show that $c_1w + c_2v_2$ = 0.
$c_1(v_1+cv_2) + c_2v_2 = 0$
$c_1v_1+c_1cv_2 + c_2v_2 = 0$
$c_1v_1 + (c_1c+c_2)v_2 = 0$
$c_1$ must be 0, because it's the only way to get rid of $v_1$, so then $c_2$ must also be 0.
Thus call the constants must be zero, so it's linearly independent?

9. Mar 17, 2015