Thread: Linear Algebra

Need a little help getting started. Thanks!

If given F(R,R) the collection of all functions f:R->R.

1. Then how can I show F(R,R) is a vector space over R, the field of real numbers.
2. If L(f)=f(sqrt(2)), then how can I show that L:F(R,R)->R, is a linear transformation, what is the rank of L?

would the rank be 2 since the dim(F(R,R)) is 2? and does L(f)=f(sqrt(2)) imply f(sqrt(2))=2. So, L(af)=aL(f)=a(f(sqrt(2)))=a(sqrt(2)). Also, L(f+g)=L(f)+L(g)=f(sqrt(2))+g(sqrt(2))=sqrt(2)+sqr t(2)=2sqrt(2)? Therefore its linear.

Originally Posted by GreenandGold

1. Then how can I show F(R,R) is a vector space over R, the field of real numbers.

Define $\displaystyle (f+g)(x) = f(x)+g(x)$ and $\displaystyle k(f)(x) = k(f(x))$ for all $\displaystyle f,g\in F(\mathbb{R},\mathbb{R}),k\in \mathbb{R}$.

If L(f)=f(sqrt(2)), then how can I show that L:F(R,R)->R, is a linear transformation, what is the rank of L?

Show that $\displaystyle L(f_1+f_2) = L(f_1)+L(f_2)$ and $\displaystyle L(kf_1) = kL(f_1)$.

The image of $\displaystyle L$ is $\displaystyle \mathbb{R}$ so the rank is 1, since the dimension of $\displaystyle \mathbb{R}$ over $\displaystyle \mathbb{R}$ is 1.

Since it defined now i prove all the axioms hold?
I then do the same to show that it is linear?

Originally Posted by GreenandGold

Since it defined now i prove all the axioms hold?

Show that how I defined F(R,R) satisfies the definitions (you call them axioms but I do not like to call them that way) of a vector space.

I then do the same to show that it is linear?

I already answered this. Just show that is satisfies the definitiion of a linear transformation (look at my first response).

So, just for the future I first need to define the transformation/function then attack the problem, because that lost me until u defined it?

Also I have few more questions stemming from the previous:

1. Would I use a similar approach to finding L(g), if g = f - f(sqrt(2)), as the previous problem of proving a linear transformation? So does L(g) = 0?

2. How can I show that for any f:R->R I can write f = g + lamda, where g(sqrt(2)) = 0 and lamda is in R? Is the decomposition unique?

Here are some new ideas:

If g = f - f[sqrt(2)] then, L(g) = (f - f[sqrt(2)])(sqrt) = f[sqrt(2)] - f[sqrt(2)]sqrt[2] = f[sqrt(2)] - sqrt(2)f[sqrt(2)] = L(f) - sqrt(2)L(f)

can that be simplified anymore?

Can you possibly explain this decompostion thing to me? Thanks!!!

Originally Posted by GreenandGold

Need a little help getting started. Thanks!

If given F(R,R) the collection of all functions f:R->R.

1. Then how can I show F(R,R) is a vector space over R, the field of real numbers.

You do that by showing it satisfies the definition of "vector space": that there exist a sum and scalar product satifying all the "rules" for a vector space. Here the obvious sum is that if f and g are functions, f+ g is defined by (f+g)(x)= f(x)+ g(x). And, for any real number a, af is the function such that (af)(x)= af(x).

2. If L(f)=f(sqrt(2)), then how can I show that L:F(R,R)->R, is a linear transformation, what is the rank of L?

Again, by showing that it satisfies the definition of "linear transformation". If f and g are two functions, a, b are numbers, what is L(af+ bg)? The rank of L is the dimension of its "image": that is the set of all values of L(f). Do you see that for any function, f, L(f) is a number?

Well my post #6 is stemming from the original conditions in post #1... However after solving the problem I didn't get a number i got
L(f+g) = (f + g)(sqrt(2)) = f(sqrt(2)) + g(sqrt(2)) = L(f) +L(g) therefore additive...

Did i not do that right?