# Linear Independence of functions

1. Jun 4, 2012

### srfriggen

1. The problem statement, all variables and given/known data

This is from Serge Lang's "Linear Algebra, 3rd Edition", page 15.

Consider the vector space of all functions of a variable t. Show that the following pairs of functions are linearly independent:

(a) 1,t

(b) t, t2

(c) t, 4

2. Relevant equations

3. The attempt at a solution

I understand how to DO the problems and attain the correct results, but I don't understand WHY it works. Looking for some insight please.

For example, for part (b) my answer would be to set up an equation with two numbers a and b:

at + bt2=0.

I would first set t = 1 which shows a+b=0.

Then I would set t =-1, showing a=b, therefore a=b=0, showing the two functions cannot be written as linear combinations of one another.

Thanks in advance. Trying to learn this on my own so don't have a teacher to reach out to.

2. Jun 4, 2012

### Staff: Mentor

What I think you might be missing is that the equation with the linear combinations of the functions has to be identically true. For your example, the equation at + bt2 = 0 has to be true for all values of t. The equation is not a conditional equation that is true only for certain values of t.

3. Jun 4, 2012

### srfriggen

I do believe that is what I was missing...

So that is why I am allowed to choose any value of t to manipulate the equation?

If it was helpful to the solution, would I be allowed to take the derivative of that equation?

For example:

If I took the 2nd derivative of at+bt2=0 I would obtain b=0.

Then, choosing t=-1 obtains a=b=0, hence Linearly Independent?

What was in the problem statement where you knew the equation I set up would not be dependent on t?

4. Jun 4, 2012

### Staff: Mentor

Yes.
Sure. If two quantities are equal, then so are their derivatives.
Yes.
The fact that you were checking linear independence of a pair of functions.

5. Jun 4, 2012

### jollyredgiant

A set of vectors is linearly independent if for any vector, $v_i$, in the set of vectors, $\{ v_1, v_2, ..., v_n \}$, it cannot be written as a linear combination of the other vectors. So what this means is that if

$$v_i = c_1 v_1 + ... c_{i-1} v_{i-1} + c_{i+1} v_{i+1} + ... + c_n v_n$$

(where all of the c's are simple numbers, ie scalars) then the set of vectors, $\{ v_1, v_2, ..., v_n \}$, is linearly DEPENDENT. A quick example. Given the set of vectors $\{ (1), (1+t), (2t) \}$, then we can see that set is linearly dependent because

$$1 = 1*(1+t) + (-\frac{1}{2})*(2t)$$

ie, we can write one of the vectors in terms of all the OTHER vectors. This is what it means to be linearly dependent. If we cannot do this, the set is said to be linearly INDEPENDENT.

6. Jun 4, 2012

### Ray Vickson

You almost had the concept, but maybe you did not recognize it in full strength; after all, you DID first put t = 1 then put t = -1 to conclude that a = b = 0. When you did that you _were_ recognizing that you were allowed to use different values of t.

As to the intuition: if you want to show, for example, that 1, t, t^2 are linearly independent you need to show that the only coefficients a, b and c that give f(t) = a*t^2 + b*t + c = 0 _identically_ are a = b = c = 0. Think of it this way: if you think of drawing the graph y = f(t) = a*t^2 + b*t + c, you want to get a straight line along the t-axis (that is, y = 0 identically). As you said, that means that the first and second derivatives are also 0 identically. Putting t = 0 in f(t) = 0 gives you c = 0; putting t = 0 in f'(t) = 2*a*t + b = 0 gives b = 0; then f ''(t) = 2*a = 0 gives a = 0.

RGV

7. Jun 4, 2012

### vela

Staff Emeritus
Let me pile on.

Remember you're working in the vector space of functions. The linear combination $at+bt^2$ is another element of the vector space. That is, it's another function. When you wrote $at+bt^2=0$, the zero on the righthand side can't simply be the scalar 0 because the scalar 0 isn't an element of the vector space. Rather, the 0 on the righthand side actually denotes the function 0 that maps all values of t to 0, e.g. 0(t)=0. This is the reason why the equation has to hold for any value of t.