# Two by two matrix {{a,b},{c,d}}

1. Jan 1, 2009

### soandos

sorry if this is basic, but what is the proof that if there is a two by two matrix
{{a,b},{c,d}}, then it is a basis if ad-bc = -+1

2. Jan 1, 2009

Re: basis

What do you mean by "it is a basis"? Basis of what?

3. Jan 1, 2009

### soandos

Re: basis

i mean that using that set of vectors one can define every other possible vector.

4. Jan 1, 2009

### tiny-tim

happy new year!

Hi soandos!
But that is another way of saying that the two vectors are independent.

They are independent provided ad - bc ≠ 0.

You should be able to prove that fairly easily.

5. Jan 1, 2009

### soandos

Re: basis

sorry if this is stupid, as i do not take linear algebra.
does that mean that any two vectors that are not the same can define every other vector?
or does independent mean something else.

6. Jan 1, 2009

Re: basis

A set of vectors {X1, ..., Xn} is linearly independent if a1X1 + ... + anXn = 0 implies a1 = ... = an = 0, i.e. the coefficients are all equal zero. Since you have a set of 2 vectors in R^2, and since every set of 2 vectors which is linearly independent in R^2 forms a basis, you have to prove that your two vectors are independent.

7. Jan 1, 2009

### tiny-tim

In two dimensions, yes.

In three dimensions, it's any three vectors not in the same plane.

And so on …

8. Jan 1, 2009

### soandos

Re: basis

forgive me, but is this the same as saying that any two vectors representing different slopes in R^2 are independent?
or are you saying something else.
also, when you say that there are no coefficients, what do you mean exactly. can you please give an example?
sorry to put you through this.

9. Jan 1, 2009

Re: basis

Yes, basically it is. They must not "lie on the same line". For example, (1, 0) and (2, 0) are not independent.

What I wrote above about the coefficients is simply the definition of linear independence of a set of vectors. You can try to experiment a bit. For example do there exist non trivial (i.e. non zero) numbers a, b such that a(1, 0) + b(2, 0) = 0? What about the pair of vectors (1, 0), (0, 1)?

10. Jan 1, 2009

### soandos

Re: basis

for the first one, a=-2, b=1
for the second, nothing (non-zero that is) will work.
is that right?

11. Jan 1, 2009

Re: basis

Yes, that's right. So, the first one is no basis, and the second one is.

Theoretically, when you want to prove that a set of vectors is a basis for some vector space, you have to prove 2 things:

1) the set must be linearly independent
2) the set must span the space (i.e. every vector of the space can be written as a combination of vectors from the set)

However, if you have a vector space of n dimensions, then any linearly independent set of n vectors is a basis, so you needn't check 2). This was our case - you had 2 vectors in R^2 - you only needed to show they are independent.

12. Jan 1, 2009

### HallsofIvy

Re: basis

But "not the same" doesn't necessarily mean "independent". If one vector is a multiple of the other, they are not independent and do not span R2

Yes, if they "have different slopes", do not point in the same direction, then they are independent.

He did not say "no coefficients". He said no non-zero coefficients. If there were coefficients a, b with either or both non-zero such that au+ bv= 0, we can write u= -(b/a)v or v= -(a/b)u depending on whether a or b is non-zero. That is, one is just a multiple of the other. More generally, a set of vectors, $\left{v_1, v_2, \cdot\cdot\cdot, v_n\right}$ is called "independent" if there is no non-zero coefficients such that $a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n= 0$. That says immediately that you cannot write the 0 vector in two different ways as a combination of those vectors and it follows that you cannot write any vector in two different ways using those vectors.