# Thread: Basis & Dimension Question

1. ## Basis & Dimension Question

Let u and v distinct vectors of a vector space V. Show that if {u,v} is a basis for V and a and b are nonzero scalars, then both {u + v}, au} and {au,bv} are also bases for V.

2. Originally Posted by leungsta
Let u and v distinct vectors of a vector space V. Show that if {u,v} is a basis for V and a and b are nonzero scalars, then both {u + v, au} and {au,bv} are also bases for V.
Suppose that $w = \alpha u + \beta v$ then $w = \beta \left( {u + v} \right) + \frac{{\alpha - \beta }}{a}\left( {au} \right)$.
In other words, if w is in span{u,v} then w is in the span{(u+v),au}.
Now do the converse.

Do the same two proofs for {au,bv}

3. Hmmm...I'm not understanding your solution....where is that equation from?

4. Originally Posted by leungsta
OK, How much do understand about what it means to be a basis of a space?
If {u,v} is a basis for a v-space then the dimension is 2.
What does that mean for any other spanning set of two vectors?

Originally Posted by leungsta
Where is that equation from?

If you have no clue as to what is going on here, then it is time for you have a one-on-one with your course instructor. These ideas are fundamental to your study.

5. I understand that a subset B of vector space V over R is a basis o V if B is linearly independant and Span B = V... I also understand dimensions. I'm just not very clear on proofs. And as for your question of "If {u,v} is a basis for a v-space then the dimension is 2. What does that mean for any other spanning set of two vectors?"

This means that for any other spanning set of two vectors, the dimension is also 2

6. Originally Posted by leungsta
"If {u,v} is a basis for a v-space then the dimension is 2. What does that mean for any other spanning set of two vectors?"
This means that for any other spanning set of two vectors, the dimension is also 2
Good. It does mean that any spanning set has at least two vectors.
It also means that any two independent vectors will be a basis.

So we have two approaches to doing this problem. I chose the ‘spanning’ option.
But can you show that {(u+v),au} are independent?

7. Originally Posted by Plato
Good. It does mean that any spanning set has at least two vectors.
It also means that any two independent vectors will be a basis.

So we have two approaches to doing this problem. I chose the ‘spanning’ option.
But can you show that {(u+v),au} are independent?
So if w is in span {u,v} then w is in span {u + v, au}. And if {u,v} are linear independant then {u + v, au} must also be independant?

Sorry i'm really not good with proofs...

8. Originally Posted by leungsta
And if {u,v} are linear independant then {u + v, au} must also be independant?
NO! That is completely wrong.
The fact that you are given that {u,v} is a basis means they are independent.
Now I asked you if you understood the definition of basis.
I don’t think that you do.

But here goes if $\lambda \left( {u + v} \right) + \eta \left( {au} \right) = 0$ then because u & v are independent we get $\left( {\lambda + a\eta } \right)u + \lambda v = 0\; \Rightarrow \;\left( {\lambda + a\eta } \right) = 0\;\& \;\lambda = 0$.
But that means $a\eta = 0\;\& \;a \ne 0\; \Rightarrow \;\eta = 0$.
Now if you do understand about independence then you see that u+v & au are independent.

Now, I cannot go beyond this in helping you.
If this still leaves you clueless, please go to your instructor. Look, simply using the old excuse “I am not good at proofs” does not cut it. The simple fact is: People who cannot do proofs have no business in mathematics courses. There is a reason why people who do well in mathematics course are rewarded.