# Example of linear transformation on infinite dimensional vector space

I haven't had much experience with infinite dimensional vector spaces, and I was working on a problem that asks to prove that for a finite dimensional vector space $V$, and linear transformation $T:V\to V$, $V=imT + ker T \implies V=imT \bigoplus ker T$ . I think I've done this correctly by using the rank-nullity theorem to show $dim(imT \cap kerT)=0$. Next I'm asked to find a counterexample to the assertion for an infinite dimensional $V$ and $T$.

I'm not exactly sure what I'm looking for here. It seems like it should be a linear transformation that maps one or more non-zero elements of $V$ to its own kernel (since their intersection is has to be non-trivial), but its image and kernel still span $V$ somehow.

• Suppose $V$ is the space of all real-valued sequences. Consider a shift operator on this space. – Bungo May 22 '18 at 2:17

## 2 Answers

Consider, $P$ be the set of all polynomials with rational coefficient. Indeed it is a vector space(http://people.math.carleton.ca/~kcheung/math/notes/MATH1107/wk08/08_infinite_dimension_example.html). Define the derivative map from $P$ to $P$ is a linear map.

• Thank you. So would the kernel be all constant polynomials and the image be equal to $P$ ? – Alon Gelber May 22 '18 at 2:32
• Yes, Kernel will be the collection of all constant polynomials. But what do you mean by "and the image be equal to $P$?" – Sujit Bhattacharyya May 22 '18 at 2:36
• @AlonGelber indeed, the image is equal to $P$; take any polynomial $f(x)$ with rational coefficients, can you find a polynomial $g(x)$ with rational coefficients such that $g'(x)=f(x)$? Note: this is an example of a surjective endomorphism which is not injective, which is impossible on finite dimensional spaces. – Dave May 22 '18 at 2:45

Consider the left shift operator which takes one-sided sequences $(x_1, x_2, x_3,\dots)$ to $(x_2, x_3, x_4,\dots)$. It has nonzero kernel and is onto...

(We could consider sequences of elements of some field $F$, such as $\mathbb R$ or $\mathbb C$, say...)