Is a Two by Two Matrix a Basis if ad - bc = ±1?

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Discussion Overview

The discussion revolves around the conditions under which a two by two matrix can be considered a basis for a vector space, specifically focusing on the determinant condition \( ad - bc = \pm 1 \). Participants explore concepts of linear independence and spanning sets within the context of linear algebra.

Discussion Character

  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants inquire about the proof that a two by two matrix is a basis if \( ad - bc = \pm 1 \).
  • Others clarify that a basis allows for the definition of every possible vector, implying that the vectors must be independent.
  • It is noted that two vectors in \( \mathbb{R}^2 \) are independent if \( ad - bc \neq 0 \).
  • Some participants express confusion about the definition of linear independence, questioning whether any two distinct vectors can define every other vector.
  • One participant explains that a set of vectors is linearly independent if the only solution to the equation \( a_1X_1 + a_2X_2 = 0 \) is when all coefficients are zero.
  • Examples are provided to illustrate the concept of linear independence, including specific vector pairs and their relationships.
  • Clarifications are made regarding the necessity of vectors not being scalar multiples of each other to maintain independence.

Areas of Agreement / Disagreement

Participants generally agree on the definition of linear independence and its implications for forming a basis. However, there are varying interpretations of the conditions under which vectors can be considered independent, leading to some unresolved questions about specific cases.

Contextual Notes

Some assumptions about the definitions of independence and basis are not fully explored, and the discussion does not resolve all mathematical steps related to proving independence or spanning.

soandos
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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
 
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soandos said:
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

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


i mean that using that set of vectors one can define every other possible vector.
 
happy new year!

Hi soandos! :smile:
soandos said:
i mean that using that set of vectors one can define every other possible vector.

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. :wink:
 


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.
 


soandos said:
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.

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.
 
soandos said:
does that mean that any two vectors that are not the same can define every other vector?

In two dimensions, yes.

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

And so on … :smile:
 


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.
 


soandos said:
forgive me, but is this the same as saying that any two vectors representing different slopes in R^2 are independent?

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

soandos said:
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.

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


for the first one, a=-2, b=1
for the second, nothing (non-zero that is) will work.
is that right?
 
  • #11


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

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


soandos said:
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.

tiny-tim said:
In two dimensions, yes.

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

And so on … :smile:
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

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

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.
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, [itex]\left{v_1, v_2, \cdot\cdot\cdot, v_n\right}[/itex] is called "independent" if there is no non-zero coefficients such that [itex]a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n= 0[/itex]. 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.
 

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