This is a question I encountered in my linear algebra and functional analysis class.

Let $ T : U \rightarrow U $ be a linear transformation where the vector spaces are not necessarily finite-dimensional. I know that we have the isomorphism $U/ \ker(T) \cong \text{ran}(T)$ so that if the dimensions of the cokernel and range are finite, one has equality of dimensions. Can I also deduce $\ker(T) \cong U/\text{ran}(T)$? I basically only need equality of dimension of kernel and codimension of range when these are finite.

Can anyone show me how to prove $\dim(\ker(T)) = \text{codim} \; \text{ran}(T)$ when these are finite but $U$ may be infinite-dimensional? I thank all helpers.


1 Answer 1


This is false in the infinite-dimensional case. For example, take $T_L : \mathbb{R}^{\mathbb{N}} \to \mathbb{R}^{\mathbb{N}}$ to be the left shift

$$T_L(a_1, a_2, \dots) = (a_2, a_3, \dots).$$

Then $\dim \text{ker}(T_L) = 1$ but $\text{codim } \text{im}(T_L) = 0$.

In general the left shift and the right shift

$$T_R(a_1, a_2, \dots) = (0, a_1, a_2, \dots)$$

are the first counterexamples you should look at in infinite-dimensional linear algebra. For example the right shift $T_R$ satisfies $\dim \text{ker}(T_R) = 0$ but $\text{codim } \text{im}(T_R) = 1$. It also famously has no eigenvalues (unlike the left shift).

  • $\begingroup$ thank you so much. Here is a question I have asked and got an answer on math.stackexchange.com/questions/3850619/… $\endgroup$
    – Croc2Alpha
    Oct 5, 2020 at 0:42
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    $\begingroup$ @kroner: the author uses the fact that $I - T$ is a Fredholm operator of Fredholm index zero (because $I$ has Fredholm index $0$ and $I - T$ is a compact perturbation of it which preserves the Fredholm index): en.wikipedia.org/wiki/Fredholm_operator $\endgroup$ Oct 5, 2020 at 0:48
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    $\begingroup$ @kroner: it's a general fact about Fredholm operators that adding a compact operator doesn't change the index, so $I - T$ has the same index as $I$, which is $0$. Although admittedly I don't know how to prove this off the top of my head. I would just argue directly in this case, probably. $\endgroup$ Oct 5, 2020 at 0:51
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    $\begingroup$ @kroner: you can work it out in coordinates by writing $T = \sum \sigma_i u_i \otimes v_i^{\ast}$ but it's not particularly enlightening. I don't know off the top of my head if there's a nicer way. $\endgroup$ Oct 5, 2020 at 2:43
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    $\begingroup$ @kroner: maybe you can make it work but I don't know how to. If you work out in coordinates what it means for a vector to be in $\text{ker}(I - T)$ and again what it means for a vector to be in $\text{ker}(I - T^{\ast})$ you'll see that the problem reduces to the finite-dimensional case, basically. Probably there's a nicer way to do it but I don't know it. $\endgroup$ Oct 5, 2020 at 2:51

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