Is This a Correct Approach to Determining Linear Independence?

[Jamin2112]

Homework Statement

Critique my understanding.

Homework Equations

From the omniscient Wikipedia:

In linear algebra, a family of vectors is linearly independent if none of them can be written as a linear combination of finitely many other vectors in the collection.

The Attempt at a Solution

So if I had (0 0 1)^T, (0 2 -2)^T, (1 -2 1)^T, and (4 2 3)^T, then I'd check whether at least one of them can be written as a linear combination of the others by looking at

k₁ (0 0 1)^T + k₂ (0 2 -2)^T + k₃ (1 -2 1)^T + k₄ (4 2 3)^T = 0

and seeing whether there is a nontrivial solution.

Why? --- Because I looked to see whether the first one can be written as a linear combination of the other three, I'd have

C (0 0 1)^T = C₂ (0 2 -2)^T + C₃ (1 -2 1)^T + C₄ (4 2 3)^T

----> 0 = - C (0 0 1)^T + C₂ (0 2 -2)^T + C₃ (1 -2 1)^T + C₄ (4 2 3)^T ,

which is essentially the same thing. Ditto when I check so see if the other three vectors can be written as a linear combination of each of their other three. Sometimes my profesor throws functions at us. For example, he asked whether the functions f(x) = cos(x) and g(x) = sin(x) (both R --> R) were linearly independent.

I look at

A cos(x) + B sin(x) = 0.

Since the domain is all real numbers x, I can just choose a convenient real number x and to confirm my sneaking suspicion of linear independence. Let x = 0. Then A = 0 and B = B. Let x = π / 2. Then B = 0 and A = A. … I'm a little confused. Can someone help me with the logic leading to the conclusion that my functions are linearly independent?

 

[bp_psy]

Homework Statement

Critique my understanding.

Homework Equations

From the omniscient Wikipedia:

The Attempt at a Solution

So if I had (0 0 1)^T, (0 2 -2)^T, (1 -2 1)^T, and (4 2 3)^T, then I'd check whether at least one of them can be written as a linear combination of the others by looking at

k₁ (0 0 1)^T + k₂ (0 2 -2)^T + k₃ (1 -2 1)^T + k₄ (4 2 3)^T = 0

and seeing whether there is a nontrivial solution.

Why? --- Because I looked to see whether the first one can be written as a linear combination of the other three, I'd have

C (0 0 1)^T = C₂ (0 2 -2)^T + C₃ (1 -2 1)^T + C₄ (4 2 3)^T

----> 0 = - C (0 0 1)^T + C₂ (0 2 -2)^T + C₃ (1 -2 1)^T + C₄ (4 2 3)^T ,

which is essentially the same thing. Ditto when I check so see if the other three vectors can be written as a linear combination of each of their other three.

This part looks good but if you can get the coefficients you are done. A easier way in this case is to notice that that you have a set of 4 vectors and the dimension is 3 which guaranties that they are dependent since you can't have a set of 4 independents vectors in R3

Sometimes my profesor throws functions at us. For example, he asked whether the functions f(x) = cos(x) and g(x) = sin(x) (both R --> R) were linearly independent.
I look at
A cos(x) + B sin(x) = 0.

Since the domain is all real numbers x, I can just choose a convenient real number x and to confirm my sneaking suspicion of linear independence. Let x = 0. Then A = 0 and B = B. Let x = π / 2. Then B = 0 and A = A. … I'm a little confused. Can someone help me with the logic leading to the conclusion that my functions are linearly independent?

Starting from
A cos(x) + B sin(x) = 0. you can get:cos(x)=K sin(x) where K=-A/B .Is there any K except 0 that satisfies the relation?

 

[Office_Shredder]

Since the domain is all real numbers x, I can just choose a convenient real number x and to confirm my sneaking suspicion of linear independence. Let x = 0. Then A = 0 and B = B. Let x = π / 2. Then B = 0 and A = A. … I'm a little confused. Can someone help me with the logic leading to the conclusion that my functions are linearly independent?

This is a good way to start. There are more high level ways of checking whether functions are linearly independent, for example using the Wronskian, but those are for another class. As bp_psy points out, there are easier ways to do it in the case of two functions. But if there are more, this technique generalizes quite nicely. At the moment we're interested in is whether there exist numbers A and B such that, for every possible choice of x, we have
$$A \cos(x) + B \sin(x) = 0$$
If it's true for all x, then in particular it's true if x=0. Plugging in x=0, we know cos(0)=1 and sin(0)=0 so we get the equation A=0. So if $$A \cos(x) + B \sin(x) = 0$$ is true, then A=0 must be true. Similarly, if we plug in $$x=\pi/2$$, we know that $$\cos(\frac{\pi}{2}) =0$$ and $$\sin(\frac{\pi}{2}) = 1$$ so if $$A \cos(x) + B \sin(x) = 0$$ is true for all x, it must be true for $$x=\pi/2$$ and plugging that number in gives us the equation B=0.

So by checking several values of x, we get A=0 and B=0 are necessary for $$A \cos(x) + B \sin(x) = 0$$ to be true for all x (in fact, for just the two values of x that we guessed). Hence cos(x) and sin(x) must be linearly independent.

If someone asks you to prove a bunch of functions are linearly independent, this is always a valid approach to use. Pick a bunch of points, and prove that linear independence fails on those points (pick as many points as you have functions). There are two problems with this approach - you might pick a bad set of points (if you had tried x=0 and $$x=\pi$$ we wouldn't have gotten linear independence), and the functions might not be linearly independent at all. If you're not able to prove the coefficients are all 0 by picking a finite set of points, it might be because the functions are linearly dependent, but it might be that you just picked a bad set of points, and it's hard to tell which is the case

 

[Jamin2112]

This is a good way to start. There are more high level ways of checking whether functions are linearly independent, for example using the Wronskian, but those are for another class. As bp_psy points out, there are easier ways to do it in the case of two functions. But if there are more, this technique generalizes quite nicely. At the moment we're interested in is whether there exist numbers A and B such that, for every possible choice of x, we have
$$A \cos(x) + B \sin(x) = 0$$
If it's true for all x, then in particular it's true if x=0. Plugging in x=0, we know cos(0)=1 and sin(0)=0 so we get the equation A=0. So if $$A \cos(x) + B \sin(x) = 0$$ is true, then A=0 must be true. Similarly, if we plug in $$x=\pi/2$$, we know that $$\cos(\frac{\pi}{2}) =0$$ and $$\sin(\frac{\pi}{2}) = 1$$ so if $$A \cos(x) + B \sin(x) = 0$$ is true for all x, it must be true for $$x=\pi/2$$ and plugging that number in gives us the equation B=0.

So by checking several values of x, we get A=0 and B=0 are necessary for $$A \cos(x) + B \sin(x) = 0$$ to be true for all x (in fact, for just the two values of x that we guessed). Hence cos(x) and sin(x) must be linearly independent.

If someone asks you to prove a bunch of functions are linearly independent, this is always a valid approach to use. Pick a bunch of points, and prove that linear independence fails on those points (pick as many points as you have functions). There are two problems with this approach - you might pick a bad set of points (if you had tried x=0 and $$x=\pi$$ we wouldn't have gotten linear independence), and the functions might not be linearly independent at all. If you're not able to prove the coefficients are all 0 by picking a finite set of points, it might be because the functions are linearly dependent, but it might be that you just picked a bad set of points, and it's hard to tell which is the case

I see. So it's sort of like the following?

Proof. By Contradiction. Assume that for all x there exist real numbers A and B such that A ≠ 0, B ≠ 0, and A sin(x) + B cos(x) = 0. Call that assumption P.

Let x = 0. Then B = 0. Thus P implies B = 0.

Now let x = π / 2. Then A = 0. Thus P implies A = 0.

We have that P implies A = B = 0, a contradiction. Thus there exists an x such that for all real numbers A and B, A = B = 0 and A sin(x) + B cos(x) = 0. The set of functions {sin(x), cos(x)} is therefore a linearly dependent set.

 

[Jamin2112]

Is there any K except 0 that satisfies the relation?

No, but how do I show that?

 

[Mark44]

Starting from
A cos(x) + B sin(x) = 0. you can get:cos(x)=K sin(x) where K=-A/B .Is there any K except 0 that satisfies the relation?

Well, for x = pi/4, K = 1, but this might not have been what bp_psy had in mind. Since the equation A cos(x) + B sin(x) = 0 has to be identically true for any value of x, my solution of K = 1 does NOT show that cos(x) and sin(x) are linearly independent functions.

No, but how do I show that?

Office_shredder showed one way, and alluded to another way that uses the Wronskian of the functions involved.

Here's a third way:
A cos(x) + B sin(x) = 0

-A sin(x) + B cos(x) = 0 (taking the derivative of both sides)

Solving this system of two equations in two unknowns (A and B) yields A = B = 0 as the only solutions. Hence cos(x) and sin(x) are linearly independent functions.

 

[Jamin2112]

Well, for x = pi/4, K = 1, but this might not have been what bp_psy had in mind. Since the equation A cos(x) + B sin(x) = 0 has to be identically true for any value of x, my solution of K = 1 does NOT show that cos(x) and sin(x) are linearly independent functions.



Office_shredder showed one way, and alluded to another way that uses the Wronskian of the functions involved.

Here's a third way:
A cos(x) + B sin(x) = 0

-A sin(x) + B cos(x) = 0 (taking the derivative of both sides)

Solving this system of two equations in two unknowns (A and B) yields A = B = 0 as the only solutions. Hence cos(x) and sin(x) are linearly independent functions.

Grrr ... The logic of it is still a little confusing for me.

cos(x) and sin(x) are linearly dependent if for all x there exists A and B s.t. ~(A = B = 0) and A * sin(x) + B * cos(x) = 0.

Is that right?

 

[Mark44]

Grrr ... The logic of it is still a little confusing for me.

cos(x) and sin(x) are linearly dependent if for all x there exists A and B s.t. ~(A = B = 0) and A * sin(x) + B * cos(x) = 0.

Is that right?

No. Assuming that ~(A = B = 0) means "it's not true that A = 0 and B = 0"

Here are some examples.
1) cos(x) and 2cos(x) are obviously linearly dependent.
The equation A cos(x) + B*2cos(x) = 0 has a solution A = B = 0, independent of the value of x. There are also many other solutions for A and B; namely, A = 1, B = -1/2, A = 2, B = -1, and so on.
2) cos(x) and sin(x) are linearly independent (which is what part of this thread establishes). The equation A cos(x) + B sin(x) = 0 has exactly one solution for A and B - A = B = 0, and this solution is independent of the value of x.

The difference between linear independence and linear dependence is strictly the difference between how many solutions for the constants there are in the equation c₁ f₁(x) + c₂ f₂(x) + ... + c_n f_n(x) = 0. For a set of n linearly independent functions, there will be exactly one solution for the constants. For a set of n linearly dependent functions, there will be an infinite number of solutions, one of which is c₁ = c₂ = ... = c_n = 0.

 

[Jamin2112]

I think that I'm understanding this a little better.

A sin(x) + B cos(x) = 0 ----> tan(x) = - A / B

So the question is: Does there exist an x such that A sin(x) + B cos(x) = 0 and ~(A = B = 0). If so, then sin(x) and cos(x) and linearly dependent functions. If not, then sin(x) and cos(x) are linearly independent functions.

 

[Mark44]

I think that I'm understanding this a little better.

A sin(x) + B cos(x) = 0 ----> tan(x) = - A / B

But to get to the second equation, you are tacitly assuming that B is not 0.

So the question is: Does there exist an x such that A sin(x) + B cos(x) = 0 and ~(A = B = 0). If so, then sin(x) and cos(x) and linearly dependent functions. If not, then sin(x) and cos(x) are linearly independent functions.

 

[Jamin2112]

But to get to the second equation, you are tacitly assuming that B is not 0.

Is this the following valid reasoning?

A sin(x) + B cos(x) = 0
A cos(x) - B sin(x) = 0 ^[1]
A sin(x) + B (B sin(x) / A) = 0 ^[2]
sin(x) (A + B²/A) = 0
sin(x) ( (A²+B²) / A) = 0
A²+B²=0 ^[3]
A = B = 0


^[1] took the derivative of both sides
^[2] combined lines 1 and 2
^[3] since sin(x) isn't always zero, the numerator of A²+B²) / A must always be zero

 

[Mark44]

Is this the following valid reasoning?

A sin(x) + B cos(x) = 0
A cos(x) - B sin(x) = 0 ^[1]
A sin(x) + B (B sin(x) / A) = 0 ^[2]

In the line above, you replaced cos(x) by Bsin(x)/A. You got this by solving for cos(x) in the second equation, and substituting for cos(x) in the first equation.
Acos(x) = Bsin(x) ==> cos(x) = Bsin(x)/A

To isolate cos(x) you are tacitly assuming that A != 0. You can't then later come along and say that A = 0.

sin(x) (A + B²/A) = 0
sin(x) ( (A²+B²) / A) = 0
A²+B²=0 ^[3]
A = B = 0


^[1] took the derivative of both sides
^[2] combined lines 1 and 2
^[3] since sin(x) isn't always zero, the numerator of A²+B²) / A must always be zero

Instead, what you should do to solve the system
Acos(x) + Bsin(x) = 0
-Asin(x) + Bcos(x) = 0

is add sin(x) times the first equation to cos(x) times the second. That causes the terms in A to drop out and yields an equation in B alone; namely, B(sin²(x) + cos²(x)) = 0 ==> B = 0. Then go back to either of the original equations to show that A must also be zero.

 

[Jamin2112]

To isolate cos(x) you are tacitly assuming that A != 0. You can't then later come along and say that A = 0.

What's the quick fix to that problem?

 

[Mark44]

I added more to my post, showing how it should be done.

 

[Jamin2112]

Instead, what you should do to solve the system
Acos(x) + Bsin(x) = 0
-Asin(x) + Bcos(x) = 0

is add sin(x) times the first equation to cos(x) times the second. That causes the terms in A to drop out and yields an equation in B alone; namely, B(sin²(x) + cos²(x)) = 0 ==> B = 0. Then go back to either of the original equations to show that A must also be zero.

Thanks! I probably wouldn't have figured that out myself. Other sets of functions seem easier.

So, proceeding as before, this time with {e^x, e^-x}:

Ae^x + Be^-x = 0
Ae^x - Be^-x = 0
---> 2Be^-x= 0 ---> B = 0 ---> Ae^x = 0 -----> A = 0

 

[Mark44]

Yes, perfect!

 

[Jamin2112]

Yes, perfect!

http://www.layoutcodez.net/personalized/google/success_baby70989908.jpg





Another question:

(My exam begins 8 minutes from now)

When we look for a basis of a matrix A, we're basically (pun intended) simplifying Ax.

For example, if we have A = [2 -1 3; 1 0 1; 0 -2 1; 1 1 4], well, Ax row reduced is

[1 0 0; 0 1 0; 0 0 1; 0 0 0], so Ax is really just the span of e₁, e₂, e₃.

 

[HallsofIvy]

Frankly, it doesn't even make sense to talk about a "basis" for a matrix. Only vector spaces have "bases". What you are doing is finding a basis for the range of A. Of course, your matrix has non zero determinant so its range is all of R³.
If, for instance, your matrix were
$$A= \begin{bmatrix}2 & -2 & 3 \\ 1 & 0 & 1 \\ 2 & 0 & 2\end{bmatrix}$$
then "row reduction" gives
$$\begin{bmatrix}1 & 0 & 1 \\ 0 & -1 & 1 \\ 0 & 0 & 0\end{bmatrix}$$

Of course, any vector, (x, y, z), in the range of A must satisfy A(a, b, c)= (2a- b+ 3c, a+ b, 2a+ 2b)= (x, y, z), for some (a, b, c) so y= a+ b, z= 2a+ 2b= 2y. A basis for that space is {(1, 0, 0), (0, 1, 2)}. Can you get that from the row reduction of A?