Prove Sp{(a,b),(c,d)} = R^2 if and only if ad-bc≠0

[peripatein]

Hi,

Homework Statement

I am trying to prove that Sp{(a,b),(c,d)} = R^2 if and only if ad-bc≠0.
I am wondering whether the proof below would be considered rigorous.
NB. I am not permitted to make use of number of dimentions.

Homework Equations

The Attempt at a Solution

First direction:
Let Sp{(a,b),(c,d)} = R^2
Hence, (x,y) = alpha(a,b) + beta (c,d)
Hence, x=a*alpha + c*beta; y=b*alpha + d*beta
Hence, beta(ad-bc) = xb-ya
In order that Sp{(a,b),(c,d)} = R^2 and any vector in R^2 could be represented as a linear combination of ((a,b),(c,d)) there has to be a solution for beta, i.e. ad-bc cannot be zero.

Second direction: I am going to try to show that provided that ad-bc≠0, Sp{(a,b),(c,d)}=R^2.
Supposing ad=bc, then (c,d) = (c/a)(a,b) for a≠0, or (c,d) = (d/b)(a,b) for b≠0.
Hence, {(a,b),(c,d)} is linearly dependent over R^2.
Hence, Sp{(a,b),(c,d)} = Sp{(a,b)} whilst a≠0, which is not equal to R^2, OR, Sp{(a,b)} whilst b≠0, which is also not equal to R^2.
Hence, ad-bc≠0 => Sp{(a,b),(c,d)} = R^2

Is the above sufficient? Is the proof correct?

 

[tt2348]

I'm assuming Sp{x,y} means span of.
Is invertibility allowed? Since given [x y] invertible, you can show that (0,1) and (1,0) are in the span of {x,y} and thus all combinations of (0,1) and (1,0) are in the span (R2)

 

[peripatein]

Invertibility is allowed.
Nevertheless, could you please comment on the proof suggested above?

 

[Mark44]

Hi,

Homework Statement

I am trying to prove that Sp{(a,b),(c,d)} = R^2 if and only if ad-bc≠0.
I am wondering whether the proof below would be considered rigorous.
NB. I am not permitted to make use of number of dimentions.

Homework Equations

The Attempt at a Solution

First direction:
Let Sp{(a,b),(c,d)} = R^2
Hence, (x,y) = alpha(a,b) + beta (c,d)

For the above, you should add something like
"Every vector (x, y) in R² can be written as a linear combination of (a, b) and (c, d)."
Hence (x, y) = α(a, b) + β(c, d) for some constants α and β.

Hence, x=a*alpha + c*beta; y=b*alpha + d*beta
Hence, beta(ad-bc) = xb-ya

It might be helpful here to explain what you're doing, which is to multiply the equation for x by b, and multiply the equation for y by a, and then subtracting the two equations.

Then you could solve for β, getting
β = (ax - by)/(ad - bc)

In order that Sp{(a,b),(c,d)} = R^2 and any vector in R^2 could be represented as a linear combination of ((a,b),(c,d)) there has to be a solution for beta, i.e. ad-bc cannot be zero.

Second direction: I am going to try to show that provided that ad-bc≠0, Sp{(a,b),(c,d)}=R^2.
Supposing ad=bc, then (c,d) = (c/a)(a,b) for a≠0, or (c,d) = (d/b)(a,b) for b≠0.
Hence, {(a,b),(c,d)} is linearly dependent over R^2.
Hence, Sp{(a,b),(c,d)} = Sp{(a,b)} whilst a≠0, which is not equal to R^2, OR, Sp{(a,b)} whilst b≠0, which is also not equal to R^2.
Hence, ad-bc≠0 => Sp{(a,b),(c,d)} = R^2

Is the above sufficient? Is the proof correct?

 

[peripatein]

Thank you very much! Yet what about the second part of the proof?

 

[Mark44]

The second part looks OK.

 

[Michael Redei]

Second direction: I am going to try to show that provided that ad-bc≠0, Sp{(a,b),(c,d)}=R^2.
Supposing ad=bc, . . .

That's the wrong way to start. You need to assume that ad-bc≠0, not =0.

Actually it might be better to assume that Sp{(a,b),(c,d)} = R^2 and deduce that ad-bc can't have been ≠0 after all, i.e. instead of proving

ad-bc ≠ 0 $\Rightarrow$ Sp{(a,b),(c,d)} = R^2

you prove

Sp{(a,b),(c,d)} ≠ R^2 $\Rightarrow$ ad-bc = 0

 

[peripatein]

But wouldn't then be exactly like the first direction?

 

[Mark44]

That's the wrong way to start. You need to assume that ad-bc≠0, not =0.

Peripatain is doing a proof by contradiction, which is a valid way to do things.

Actually it might be better to assume that Sp{(a,b),(c,d)} = R^2 and deduce that ad-bc can't have been ≠0 after all, i.e. instead of proving

ad-bc ≠ 0 $\Rightarrow$ Sp{(a,b),(c,d)} = R^2

you prove

Sp{(a,b),(c,d)} ≠ R^2 $\Rightarrow$ ad-bc = 0

 

[Michael Redei]

Peripatain is doing a proof by contradiction, which is a valid way to do things.

True, but Peripatain has still proven the same thing twice, once
Sp{(a,b),(c,d)}=R^2 $\Rightarrow$ ad-bc≠0
and then
ad-bc=0 $\Rightarrow$ Sp{(a,b),(c,d)}≠R^2.

Adding "Hence, ad-bc≠0 => Sp{(a,b),(c,d)} = R^2" was wrong, this does not follow from either proof.

This situation is like first proving
"if you read good books $\Rightarrow$ you will grow wise"
and then
"if you're not wise $\Rightarrow$ you obviously didn't read good books".

These are not equivalent though, since you could be wise without reading books at all (for instance by listening to advice from others).

 

[peripatein]

Wouldn't proving it the way you suggested be the same proof as the in first direction?

 

[Mark44]

True, but Peripatain has still proven the same thing twice, once
Sp{(a,b),(c,d)}=R^2 $\Rightarrow$ ad-bc≠0
and then
ad-bc=0 $\Rightarrow$ Sp{(a,b),(c,d)}≠R^2.

Adding "Hence, ad-bc≠0 => Sp{(a,b),(c,d)} = R^2" was wrong, this does not follow from either proof.

This situation is like first proving
"if you read good books $\Rightarrow$ you will grow wise"
and then
"if you're not wise $\Rightarrow$ you obviously didn't read good books".

These are not equivalent though, since you could be wise without reading books at all (for instance by listening to advice from others).

You are correct. I confess that I didn't look at the 2nd part that carefully.

Peripatain, you need to start by assuming that ad - bc ≠ 0, and then showing that the two vectors span R².

 

[Michael Redei]

Assume ad-bc≠0. Then you can solve any equation of the kind you mentioned in the first part, that is anything like β(ad-bc) = xb-ya, no matter how you choose x and y. And that means that every vector (x,y) lies in Sp{(a,b),(c,d)}. Actually the two proofs are very similar, only the wording differs a bit:

(1) R^2=Sp{...} => every point (x,y) fulfils the equations... => β(...) = ... must have a solution => ad-bc≠0.

(2) ad-bc≠0 => β(...)=... always has a solution, no matter how x,y are chosen => any point (x,y) lies in Sp{...} => Sp{...}=R^2.