Question

Give an example of a vector space V and a linear transformation T: V→V such that T is one-to-one but range(V)≠V. Give an example of a vector space V and a linear transformation T: V→V such that T is one-to-one but range(V)≠V. @Mathematics

Answer 1

Well, it's clear that V can't be finite dimensional, because if it were then dim(T(V)) = dim(V) => T(V) = V.

So we need to look for an infinite dimensional vector space.

Think about it for a minute and let me where you land.

Answer 2

That's where I've been trying to think... I've come up with a couple examples in my head but none of them worked perfectly. Like for example, I thought perhaps let V be the vector space of all linear functions f(x)=ax+b where a,b are constants, and then let T be the differentiation operator. T(f(x))=a, which is also a linear function, but of course isn't one-to-one as I realized. It seems that if T were a differentiation operator, we'd always run into that snag, and same for integration. I need a nudge to get thinking in the right direction. Any hints?

Answer 3

Let V = P[x], the vector space of polynomials over the real numbers. There's a natural infinite dimensional basis here, {1, x, x^2, ... }.
What's a linear function V --> V does land on the span of that basis?

Answer 4

sorry, "does NOT land on the span of that basis."

Answer 5

I'm confused...how could it be V-->V if it's not in the span of the basis? For it to be V-->V, wouldn't the output have to be in V, and thus be in the span of the basis?

Answer 6

No. Is there a linear function in P[x] that maps
\[ 1 \mapsto x \]
\[ x \mapsto x^2 \]
etc.?

So we're shifting along the basis.

Answer 7

T(P[x])=xP[x]

Answer 8

AH I see...

Answer 9

Yes. Now the span of 1 is not in the image of T. I.e., there are no constant polynomials in T(V)

Answer 10

It completely didn't occur to me to use x as a scaling factor... arbitrary constants like T(P[x])=cP[x] came to my mind but of course that didn't fix the problem. Interesting. Thank you for your help!

Answer 11

Note that you haven't quite given the right definition of T, as you've written down vector spaces.

Answer 12

Is it sufficient to say \[T:V\rightarrow V\]where\[V=\left\{ a_0+a_1x+a_2x^2+\cdots | a_0,a_1,a_2 \in \mathbb R \right\}\]and\[T(f)=xf \text{ where }f\in V\]

Answer 13

Right. The usual notation is to say a member of P[x] is written as p(x).

Hence the definition of T : P[x] --> P[x] is given by T(p(x)) = x.p(x)

Answer 14

Ah okie dokie.

Answer 15

You'll think about this a little bit more, but let me just point out that multiplying p(x) by x isn't "scaling" p(x) at all, but it's analogous to the linear transformation in R^3 given by
i -> j
j -> k
k -> i

where we 'shift' coefficients among basis vectors. The difference here of course is that the basis is infinite and we don't loop back as we do in R^3 to the first basis vector.

Answer 16

Anyway, nice problem.

Answer 17

Ooh that's interesting! Thanks again for your help!