Find a basis for the following subspaces

[Jamin2112]

Homework Statement

Homework Equations

In linear algebra, a basis is a set of linearly independent vectors that, in a linear combination, can represent every vector in a given vector space or free module, or, more simply put, which define a "coordinate system".[1] In more general terms, a basis is a linearly independent spanning set. http://en.wikipedia.org/wiki/Basis_(linear_algebra)"

The Attempt at a Solution

I'm a little confused by these.

Take (b). First, I suppose, I'll see how many of the functions {cos(x), cos(x+1), cos(x+2)} are linearly independent. I'll set a linear combination of them equal to zero:

a * cos(x) + b * cos(x+1) + c * cos(x+2) = 0.

I could take c = 0 and then have b = -a * cos(x+1)/cos(x). So there definitely is a nontrivial solution. ...

Anyways, I'm confused. Help me out in detail.

 

[LCKurtz]

Homework Statement

I'm a little confused by these.

Take (b). First, I suppose, I'll see how many of the functions {cos(x), cos(x+1), cos(x+2)} are linearly independent. I'll set a linear combination of them equal to zero:

a * cos(x) + b * cos(x+1) + c * cos(x+2) = 0.

I could take c = 0 and then have b = -a * cos(x+1)/cos(x). So there definitely is a nontrivial solution. ...

Anyways, I'm confused. Help me out in detail.

Why do you say there is definitely a non-trivial solution?

I'll give you a hint. What happens if you use the addition formula on cos(x+1) and cos(x+2)?

 

[Jamin2112]

Why do you say there is definitely a non-trivial solution?

I'll give you a hint. What happens if you use the addition formula on cos(x+1) and cos(x+2)?

You mean let A = 1+x, B = 1, so that we have

cos(A + B) = cos(x+2) = cos(x+1)cos(1) - sin(x+1)sin(1) ?

Which one are you talking 'bout?

 

[LCKurtz]

You mean let A = 1+x, B = 1, so that we have

cos(A + B) = cos(x+2) = cos(x+1)cos(1) - sin(x+1)sin(1) ?

Which one are you talking 'bout?

Both of them. Apply the formula directly to them.

cos(x+1) = ?
cos(x+2) = ?

 

[wisvuze]

Homework Statement

Homework Equations

In linear algebra, a basis is a set of linearly independent vectors that, in a linear combination, can represent every vector in a given vector space or free module, or, more simply put, which define a "coordinate system".[1] In more general terms, a basis is a linearly independent spanning set. http://en.wikipedia.org/wiki/Basis_(linear_algebra)"

The Attempt at a Solution

I'm a little confused by these.

Take (b). First, I suppose, I'll see how many of the functions {cos(x), cos(x+1), cos(x+2)} are linearly independent. I'll set a linear combination of them equal to zero:

a * cos(x) + b * cos(x+1) + c * cos(x+2) = 0.

I could take c = 0 and then have b = -a * cos(x+1)/cos(x). So there definitely is a nontrivial solution. ...

Anyways, I'm confused. Help me out in detail.

If you think that there is a non - trivial solution because
b = -a * cos(x+1)/cos(x) = 0 when x = ( pi/2 - 1 ) then you are misguided. Remember that we are talking about a subspace of the space of FUNCTIONS. Thus, your objective is to obtain the 0 function ( the 0 function is the 0 of this vector space ) from a linear combination of the other cosines there ( that is, your resulting linear combination is a function that is 0 for ANY x ). Use the addition formulas to show that this cannot happen

 

[Jamin2112]

Both of them. Apply the formula directly to them.

cos(x+1) = ?
cos(x+2) = ?

Ah, I see. (I think so.)

A = 1 + x
B = 1

----> cos(x) = cos(A - B) = cos(1+x)cos(x) + sin(1+x)sin(x)
cos(1+x) = cos(1+x)
cos(2+x) = cos(A + B) = cos(1+x)cos(x) - sin(1+x)sin(x)Thereupon

a * cos(x) + b * cos(1+x) + c * cos(1+x)

= a * (cos(1+x)cos(x) + sin(1+x)sin(x)) + b * cos(1+x) + c * (cos(1+x)cos(x) - sin(1+x)sin(x))

= ... [URL]http://www.threadbombing.com/data/media/2/maxwell_smart__confused.gif[/URL]I'm still a little confused.

 

[LCKurtz]

Both of them. Apply the formula directly to them.

cos(x+1) = ?
cos(x+2) = ?

Ah, I see. (I think so.)

A = 1 + x
B = 1

...
I'm still a little confused.

I can't argue with that statement. Usually when someone wants to expand cos(1+x) directly using the addition formula for cos(A+B) I would expect them to take A = 1 and B = x. Similarly for cos(2+x). Why are you making such a big problem of it?

 

[Jamin2112]

I can't argue with that statement. Usually when someone wants to expand cos(1+x) directly using the addition formula for cos(A+B) I would expect them to take A = 1 and B = x. Similarly for cos(2+x). Why are you making such a big problem of it?

Sorry, good sir. My mediocre IQ can't figure out the grand simplification that is glaring in front of me.

cos(1+x) = cos(A+B) is A = 1 and B = x.

Resultantly,

a * cos(x) + b * cos(1+X) + c * cos(2+x)

= a * cos(B) + b * cos(A+B) + c * cos(2A+B)

= a * cos(B) + b * ( cos(A)cos(B) - sin(A)sin(B) ) + c * ( cos(2A)cos(B) - sin(2A)sin(B) )

= a * cos(B) + b * ( cos(A)cos(B) - sin(A)sin(B) ) + c * ( (1 - 2sin²(A)cos(B) - 2sin(A)cos(A)sin(B) ) =


...

Am I heading in the right direction?

 

[LCKurtz]

Sorry, good sir. My mediocre IQ can't figure out the grand simplification that is glaring in front of me.

cos(1+x) = cos(A+B) is A = 1 and B = x.

Resultantly,

a * cos(x) + b * cos(1+X) + c * cos(2+x)

= a * cos(B) + b * cos(A+B) + c * cos(2A+B)

= a * cos(B) + b * ( cos(A)cos(B) - sin(A)sin(B) ) + c * ( cos(2A)cos(B) - sin(2A)sin(B) )

= a * cos(B) + b * ( cos(A)cos(B) - sin(A)sin(B) ) + c * ( (1 - 2sin²(A)cos(B) - 2sin(A)cos(A)sin(B) ) =


...

Am I heading in the right direction?

No, you aren't. You are trying to find a basis for the span of {cos(x),cos(x+1),cos(x+2)} You obviously know the formula

cos(A+B) = cos(A)cos(B) - sin(A)sin(B)

If you let A = x and B = 1 what does that formula give you?
If you let A = x and B = 2 what does that formula give you?
If you answer those, it might help you see what you could use for a basis for your span.

 

[Jamin2112]

cos(A+B) = cos(A)cos(B) - sin(A)sin(B)

If you let A = x and B = 1 what does that formula give you?
If you let A = x and B = 2 what does that formula give you?

If A = x and B = 1, then cos(A+B) = cos(A)cos(B) - sin(A)sin(B) = cos(x)cos(1) - sin(x)sin(1).

If A = x and B = 2, then cos(A+B) = cos(x)cos(2) - sin(x)sin(2).

Whereupon a * cos(x) + b * cos(x+1) + c * cos(x+2) = 0
----------> a * cos(x) + b * [ cos(x)cos(1) - sin(x)sin(1) ] + c * [ cos(x)cos(2) - sin(x)sin(2) ]
= cos(x) * [ a + b*cos(1) + c*cos(2) ] + sin(x) * [ -b * sin(1) - c * sin(2) ]

...


I'm sorry! I'm not trolling; I just don't see where this is going.

 

[Mark44]

Notice that a * cos(x) + b * cos(x+1) + c * cos(x+2) can be written as a linear combination of cos(x) and sin(x).

a + b*cos(1) + c*cos(2) and so is -b * sin(1) - c * sin(2). (I didn't check your calculation.)

 

[Jamin2112]

Can you brahs check my work on part (a)?

 

[Jamin2112]

Notice that a * cos(x) + b * cos(x+1) + c * cos(x+2) can be written as a linear combination of cos(x) and sin(x).

a + b*cos(1) + c*cos(2) and so is -b * sin(1) - c * sin(2). (I didn't check your calculation.)

So did you set x = 0?

 

[Mark44]

No, just look at what you wrote.
a * cos(x) + b * cos(x+1) + c * cos(x+2)
= a * cos(x) + b * [ cos(x)cos(1) - sin(x)sin(1) ] + c * [ cos(x)cos(2) - sin(x)sin(2) ]
= cos(x) * [ a + b*cos(1) + c*cos(2) ] + sin(x) * [ -b * sin(1) - c * sin(2) ]


BTW, I had to pick and choose amongst what you wrote, since it didn't make complete sense.

You started by setting a * cos(x) + b * cos(x+1) + c * cos(x+2) = 0, but in your following work, you're really just expanding the left side of the equation above, and you lost the fact that a * cos(x) + b * cos(x+1) + c * cos(x+2) is equal to zero.

 

[Mark44]

Can you brahs check my work on part (a)?

I don't follow what you have here. Presumably you're checking whether e^x, e^x+1, and e^x+2 are linearly independent.

You need to start with the equation ae^x + be^x+1 + ce^x+2 = 0.

That's equivalent to ae^x + bee^x + ce²e^x = 0.

 

[Jamin2112]

I don't follow what you have here. Presumably you're checking whether e^x, e^x+1, and e^x+2 are linearly independent.

You need to start with the equation ae^x + be^x+1 + ce^x+2 = 0.

That's equivalent to ae^x + bee^x + ce²e^x = 0.

Look at part (c) of the original question.

 

[Mark44]

Can you brahs check my work on part (a)?

Look at part (c) of the original question.

Is the work above for part a or part c? Clearly it's part c you're asking about, but mislabeling your question threw me off.

The equation a*1 + b(1 + x) + c(1 + x + x²) = 0 has to be an identity; i.e., for all x, so yes, you can set x = 0, but you don't have to.

Taking the derivative twice gives you the two equations you show, which are
(b + c) + 2cx = 0 (2), and
2c = 0 (3)

(3) implies that c = 0.
From (2), we have b = 0.
From (1) we have a = 0.

Another way to do this doesn't require differentiation.

a + b(1 + x) + c(1 + x + x²) = 0
<==> (a + b + c) + (b + c)x + cx² = 0

Since this equation has to be identically true, the polynomial on the left side must equal the polynomial on the right side, which can be written as 0 + 0x + 0x².

The coefficient of x² on the left side is c, which has to equal to x² coefficient on the right side, implying that c = 0.
Same for the coefficients of x on both sides, implying that b = 0.
Same for the constant terms, implying that a = 0.

Hence the only solution in terms of the constants a, b, and c is a = b = c = 0, so the three functions are linearly independent.

 

[Jamin2112]

Is the work above for part a or part c?

part (c).



But speaking of part (a), isn't {e^x} the basis? Because all the factorization showed us that to span {1, e^1+x,e^2+x} we only need a constant multiplied by e^x.

 

[Mark44]

part (c).



But speaking of part (a), isn't {e^x} the basis? Because all the factorization showed us that to span {1, e^1+x,e^2+x} we only need a constant multiplied by e^x.

Yes.

The set was {e^x, e^{x + 1}, e^{x + 2}}.
Each function is a constant multiple of e^x.

 

[HallsofIvy]

Can you brahs check my work on part (a)?

What you attach is part (c)! You are given 1, 1+ x, and 1+ x+ x^2. Yes, you want to look at a(1)+ b(1+ x)+ c(1+ x+ x^2)= 0 which is the same as a+ b+ bx+ c+ cx+ cx^2= (a+ b+ c)+ (b+ c)x+ cx^2= 0. Obviously if that is the constant 0, its derivative is 0: b+ c+ 2cx= 0. And it that is a constant, its derivative is 0: 2c= 0. From that, c= 0. And from b+ c+ 2cx= b+ 0+ 0(2x) b= 0 (you don't need to set x= 0 for that). Then a+ b+ c= a+ 0+ 0= 0. That is, a= b= c= 0 so they are independent.

Another way to do that is to choose 3 different values for x: if x= 0, a+ b+ c+ (b+ c)(0)+ c(0^2)= a+ b+ c= 0. If x= 1, a+ b+ c+ (b+ c)(1)+ c(1^2)= a+ 2b+ 3c= 0. If x= -1, a+ b+ c+ (b+c)(-1)+ c((-1)^1)= a+ b+ c- b- c+ c= a+ c= 0. We have three equations,
a+ b+ c= 0, a+ 2b+ 3c= 0, and a+ c= 0. Solve those to see that a= b= c= 0 is the only solution.

A third way is to use that 1, x, and x^2 are independent: the only way a quadratic polynomial (or any polynomial) can be 0 for all x is if all of its coefficients are 0. That means that you must have a+ b+ c= 0, b+ c= 0, c= 0. Again, solve those to see that a= b= c= 0.

For (a) which actually has e^x, e^{x+1}, and e^{x+2}, just observe that e^{x+ 1}= e(e^x) and e^{x+2}= (e^2)e^x- that is, both are multiples of e^x.

 

[Jamin2112]

What you attach is part (c)! You are given 1, 1+ x, and 1+ x+ x^2. Yes, you want to look at a(1)+ b(1+ x)+ c(1+ x+ x^2)= 0 which is the same as a+ b+ bx+ c+ cx+ cx^2= (a+ b+ c)+ (b+ c)x+ cx^2= 0. Obviously if that is the constant 0, its derivative is 0: b+ c+ 2cx= 0. And it that is a constant, its derivative is 0: 2c= 0. From that, c= 0. And from b+ c+ 2cx= b+ 0+ 0(2x) b= 0 (you don't need to set x= 0 for that). Then a+ b+ c= a+ 0+ 0= 0. That is, a= b= c= 0 so they are independent.

Another way to do that is to choose 3 different values for x: if x= 0, a+ b+ c+ (b+ c)(0)+ c(0^2)= a+ b+ c= 0. If x= 1, a+ b+ c+ (b+ c)(1)+ c(1^2)= a+ 2b+ 3c= 0. If x= -1, a+ b+ c+ (b+c)(-1)+ c((-1)^1)= a+ b+ c- b- c+ c= a+ c= 0. We have three equations,
a+ b+ c= 0, a+ 2b+ 3c= 0, and a+ c= 0. Solve those to see that a= b= c= 0 is the only solution.

A third way is to use that 1, x, and x^2 are independent: the only way a quadratic polynomial (or any polynomial) can be 0 for all x is if all of its coefficients are 0. That means that you must have a+ b+ c= 0, b+ c= 0, c= 0. Again, solve those to see that a= b= c= 0.

For (a) which actually has e^x, e^{x+1}, and e^{x+2}, just observe that e^{x+ 1}= e(e^x) and e^{x+2}= (e^2)e^x- that is, both are multiples of e^x.

Ah, I see!

Because we have a "for all x" we can plug in 0, or any value, for x.

Back to the sine and cosine thingy.


We can simplify A cos(x) + B cos(x+1) + C cos(x+2):

cos(x) (A + B cos(1) + C cos(2)) + sin(x) (-B sin(1) - C sin(2)) = 0,

or if we want,

tan(x) = (A + B cos(1) + C cos(2)) / (B sin(1) + C sin(2)).

Take x = π / 4. Then

0 = (A + B cos(1) + C cos(2)) / (B sin(1) + C sin(2))

----> (A + B cos(1) + C cos(2)) = 0

----> Infinite solutions.

(?)

 

[Mark44]

Ah, I see!

Because we have a "for all x" we can plug in 0, or any value, for x.

Back to the sine and cosine thingy.


We can simplify A cos(x) + B cos(x+1) + C cos(x+2):

cos(x) (A + B cos(1) + C cos(2)) + sin(x) (-B sin(1) - C sin(2)) = 0,

Stop here. Since this has to be identically true for all x, it must be that A + Bcos(1) + Ccos(2) = 0 and Bsin(1) + Csin(2) = 0.

Since this is a system of two equations in three variables, it is underdetermined, hence there are an infinite number of solutions, hence the solution A = B = C = 0 is not the only solution.

Therefore, the functions cos(x), cos(x + 1), and cos(x + 2) are linearly dependent.

or if we want,

tan(x) = (A + B cos(1) + C cos(2)) / (B sin(1) + C sin(2)).

Take x = π / 4. Then

0 = (A + B cos(1) + C cos(2)) / (B sin(1) + C sin(2))

----> (A + B cos(1) + C cos(2)) = 0

----> Infinite solutions.

(?)

 

[Jamin2112]

Stop here. Since this has to be identically true for all x, it must be that A + Bcos(1) + Ccos(2) = 0 and Bsin(1) + Csin(2) = 0.

Since this is a system of two equations in three variables, it is underdetermined, hence there are an infinite number of solutions, hence the solution A = B = C = 0 is not the only solution.

Therefore, the functions cos(x), cos(x + 1), and cos(x + 2) are linearly dependent.

A basis would be {sin(x), cos(x)}. Right?