If $T$ is a linear map from $V$ to $W$ such that $\ker(T)$ and $\operatorname{im}(T)$ are both finite dimensional. Then $V$ is also finite dimensional.

Since the proof of dimension theorem requires $V$ to be finite dimensional, we cannot directly use it here.

Sketch of my proof: Since I havent been introduced with other tools about infinite dimensional spaces, I rely on the fact that if $V$ is infinite dimensional, then there exists an infinite sequence of vectors $\vec v_1,\vec v_2,\dots \in V$ such that $(\vec v_1,\dots ,\vec v_n)$ is linearly independent for any positive integer $n$.

First I construct a basis $(\vec u_1,\dots ,\vec u_m)$ for $\ker(T)$, then proceed to prove that all the lists in the form: $(\vec u_1,\dots, \vec u_m, \vec v_1,\dots , \vec v_n)$ is linearly independent if $\vec v_j $ are not in $\ker(T)$.

Then we can prove that $(T\vec v_1,\dots , T\vec v_n)$ is linearly independent for any value of $n$. Which is then equivalent to $\operatorname{im}(T)$ being infinite dimensional. So we have arrived at a contradiction.

Also, the question looks so obvious, is there a way to show this with a more clever method?

  • $\begingroup$ Why don't you repeat the proof of the dimension theorem? As I recall, the proof takes bases of $\mathrm{Im}T$ and $\ker T$ and produces a base of $V$, which is precisely what you need. $\endgroup$
    – Guy
    Aug 12, 2016 at 15:18
  • $\begingroup$ There is a more general version of the dimension theorem which goes like this: there is a vector space isomorphism $V \approx \text{kernel}(T) \oplus \text{image}(T)$. $\endgroup$
    – Lee Mosher
    Aug 12, 2016 at 15:18
  • $\begingroup$ @Guy Maybe the version that I read was a little bit different, the author extends a basis from $\ker(T)$ to $V$, and show that the basis vectors not in $\ker(T)$ is a space that has dimension equal to $\operatorname{im}(T)$. But I see your idea, it would have worked too $\endgroup$
    – lEm
    Aug 12, 2016 at 15:23
  • $\begingroup$ @LeeMosher Sorry, but I am not sure how $\ker(T) \oplus \operatorname{im}(T)$ would work? Isn't that vectors from $\ker(T)$ and $\operatorname{im}(T)$ of different vector spaces? So how does direct sum make sense with different vector spaces? $\endgroup$
    – lEm
    Aug 12, 2016 at 15:29
  • $\begingroup$ @Bubububu Yes, you're correct. The result LeeMosher described is $V=\ker(T)\oplus\mathrm{im}(T^*)$ where $T^*:W\rightarrow V$ is the adjoint of $T$. In general, however, we have that the dimensions add: $\dim(\ker(T))+\dim(\mathrm{im}(T))=\dim(V)$. $\endgroup$ Aug 12, 2016 at 15:55

1 Answer 1


Here is a quick sketch of a proof, leaving out all the actual linear algebra details and concentrating on the construction of the appropriate bases.

Choose a basis $\{a_i\}_{i \in I}$ for $\text{kernel}(T)$.

Choose a basis $\{b_j\}_{j \in J}$ for $\text{image}(T)$ (make sure your index sets $I,J$ are disjoint).

Choose $b'_j \in T^{-1}(b_j)$, for each $j \in J$.

Note that $a_i \ne b'_j$ for all $i \in I$ and $j \in J$, because $a_i \in \text{kernel}(T)$ but $b_j \ne \vec 0$ and therefore $b'_j \not\in \text{kernel}(T)$.

Finally, $\{a_i\}_{i \in I} \cup \{b'_j\}_{j \in J}$ is a basis for $V$ (This is where all the linear algebra details are hidden).

So far there is no assumption on finite dimension.

But if the kernel is finite dimensional then $I$ is a finite set, and if the image is also finite dimensional then $J$ is a finite set, and therefore $V$ has a finite basis indexed by the finite set $I \cup J$.

By the way, one can also use this argument to construct an isomorphism between $V$ and $\text{kernel}(T) \oplus \text{image}(T)$, taking the basis $\{a_i\}_{i \in I} \cup \{b'_j\}_{j \in J}$ bijectively to the basis $\{a_i\}_{i \in I} \cup \{b_j\}_{j \in J}$. This works without any assumption on cardinality of bases.

  • $\begingroup$ Thanks for the answer, this is simple and clear. Last question: does this specific construction shows that the dimension theorem holds for infinite dimensional spaces? (Assuming axiom of choice) $\endgroup$
    – lEm
    Aug 16, 2016 at 14:41
  • $\begingroup$ Yes, in the sense of cardinal arithmetic. $\endgroup$
    – Lee Mosher
    Aug 16, 2016 at 15:00

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