# Thread: Image of linear operator on finite vector space

1. ## Image of linear operator on finite vector space

Suppose that T is a linear operator on the vector space V (over the ﬁeld
F ). Then for any natural number m, $\displaystyle Im(T^{m+1}) \leq Im(T^m)$.
If V is ﬁnite-dimensional, then $\displaystyle Im(T^{m+1}) = Im(T^m)$ must happen for some natural number m.
Can you give me a counterexample for the case where V is inﬁnite-dimensional? (that is the image is decreasing in size for ever)
Thanks.

2. I just think of one ... polynomials with T = d/dx.
Any more inventive ideas?

3. Originally Posted by vincisonfire
I just think of one ... polynomials with T = d/dx.
Why does this work?

Any more inventive ideas?
I came up with a one that is a little strange.
Consider, $\displaystyle \mathbb{R}^{\infty} = \mathbb{R}\times \mathbb{R}\times ...$ - an infinite tuple.
Okay, I am abusing notation but I think you have an idea of what I mean.
Now, $\displaystyle \mathbb{R}^{\infty}$ is a vector space over $\displaystyle \mathbb{R}$ in a natural way.

Define, $\displaystyle T: \mathbb{R}^{\infty} \to \mathbb{R}^{\infty}$ by $\displaystyle T(a_1,a_2,a_3,...) = (0,a_1,a_2,a_3,...)$.
Notice that $\displaystyle T(\bold{a}+\bold{b}) = T(\bold{a}) + T(\bold{b})$ and $\displaystyle T(a\bold{a}) = aT(\bold{a})$.
Thus, $\displaystyle T$ is a linear transformation.

Define, $\displaystyle \mathbb{R}^{\infty}_1 = \{ (0,r_1,r_2,r_3,...)| r_j \in \mathbb{R}\}$.
Define, $\displaystyle \mathbb{R}^{\infty}_2 = \{(0,0,r_1,r_2,r_3,...)|r_j \in \mathbb{R}\}$.
Define, $\displaystyle \mathbb{R}^{\infty}_3 = \{(0,0,0,r_1,r_2,...)|r_j \in \mathbb{R}\}$.
And in general, $\displaystyle \mathbb{R}^{\infty}_n$ for $\displaystyle n\geq 1$.

Notice that $\displaystyle \mathbb{R}^{\infty}_{n+1} \subset \mathbb{R}^{\infty}_n$.
However, $\displaystyle \text{im}(T^k) = \mathbb{R}^{\infty}_k$.
Thus, $\displaystyle \text{im}(T^{k+1}) \subset \text{im}(T^k)$.

4. Originally Posted by ThePerfectHacker
Why does this work?
No sorry, I was thinking of the increasing Kernel.

5. Originally Posted by ThePerfectHacker
Consider, $\displaystyle \mathbb{R}^{\infty} = \mathbb{R}\times \mathbb{R}\times ...$ - an infinite tuple.
Okay, I am abusing notation but I think you have an idea of what I mean.
Now, $\displaystyle \mathbb{R}^{\infty}$ is a vector space over $\displaystyle \mathbb{R}$ in a natural way.

Define, $\displaystyle T: \mathbb{R}^{\infty} \to \mathbb{R}^{\infty}$ by $\displaystyle T(a_1,a_2,a_3,...) = (0,a_1,a_2,a_3,...)$.
Notice that $\displaystyle T(\bold{a}+\bold{b}) = T(\bold{a}) + T(\bold{b})$ and $\displaystyle T(a\bold{a}) = aT(\bold{a})$.
Thus, $\displaystyle T$ is a linear transformation.
In operator theory, this is called the unilateral shift. It is the device used by the manager of the Hilbert Hotel to ensure that an empty room can always be guaranteed, no matter how many guests arrive.