Existence of a linear transformation

**arbolis** · March 18th 2009, 12:33 PM

Hi,
This is a part of an exercise I got on the final exam I failed. I couldn't answer well this question : Let $V=\mathbb {R}^4$ and $W=P_3$ the space of polynomials of degree $\leq 3$ .
a)Show that $V$ is isomorph to $W$ . (Done)
b)Does there exist 2 distinct isomorphisms? If so, give them. (Done. The answer was yes if it might help for the next question)
c)Tell whether there exist a linear transformation $T:V\to W$ such that $T(1,0,0,0)=x$ , $T(0,0,0,2)=x$ , $T(1,1,1,1)=x$ and $T(1,0,0,-2)=x$ .
My attempt :It doesn't exist because the vectors $(1,0,0,0)$ , $(0,0,0,2)$ , $(1,1,1,1)$ and $(1,0,0,-2)$ are linear independent. And so $T(1,0,0,0)$ , $T(0,0,0,2)$ , $T(1,1,1,1)$ and $T(1,0,0,-2)$ should... I stopped right there. I'm not sure of how to answer the question.

**clic-clac** · March 18th 2009, 01:30 PM

Hi

Originally Posted by arbolis

It doesn't exist because the vectors $(1,0,0,0)$ , $(0,0,0,2)$ , $(1,1,1,1)$ and $(1,0,0,-2)$ are linear independent.

Four linearly independant vectors in $\mathbb{R}^4$ form a basis. Then any linear map will be completely determined by the images of such vectors. Any different possibility will give different linear maps, but there is no problem.

But here, look closely to the first, second and last vectors. Then, if $T$ is linear, any contradiction?

**arbolis** · March 18th 2009, 02:50 PM

Oops! They are linear dependent. Thus T cannot be a linear transformation?

**clic-clac** · March 19th 2009, 12:07 AM

If $T$ can't be a linear transformation, that's not only because the vectors are dependant: for example, take $(1,1)$ and $(2,2)$ in $\mathbb{R}^2,$ there are linear transformation $T:\mathbb{R}^2\rightarrow\mathbb{R}^2$ such that $T(1,1)=(5,5)$ and $T(2,2)=(10,10)$ , ( $T=5.Id$ works), but $(1,1)$ and $(2,2)$ are dependant.

But here, compare $T(1,0,0,-2)$ and $T((1,0,0,0)-(0,0,0,2)).$

**arbolis** · March 19th 2009, 07:45 AM

But here, compare

and

That's pretty clear now for me.

=x and

=0 which means its not a linear transformation.

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