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.
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?
I already answered this. Just show that is satisfies the definitiion of a linear transformation (look at my first response).I then do the same to show that it is linear?
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 = 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!!!
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?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?