Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've found in my solution book multiple references to this "dimension theorem" but have been unlucky looking for a more extensive description of this. When does this hold? Are there any other hypothesis needed?


share|cite|improve this question
@Arturo: Thanks for the edit. I was thinking the title looked awful and was looking for the LaTeX code to make it nicer. – Abel Feb 21 '11 at 21:21
No problem. – Arturo Magidin Feb 21 '11 at 21:27
up vote 7 down vote accepted

The Dimension Theorem is often called the "Rank Nullity Theorem", so you may have more luck finding it under that name.

The result holds for any linear transformation and vector spaces; it's not even restricted to finite dimensional spaces, so long as your sum is taken to be a sum of cardinals and you assume the Axiom of Choice (so that all vector spaces have bases).

The Dimension Theorem, aka Rank Nullity Theorem, states:

Dimension Theorem. If $\mathbf{V}$ and $\mathbf{W}$ are any vector spaces (over the same field), and $T$ is any linear transformation from $\mathbf{V}$ to $\mathbf{W}$, then $\dim(\mathrm{ker} T) + \dim(\mathrm{Im} T) = \dim(\mathbf{V})$ where the sum is a sum of cardinals.

The idea is to start with a basis $\beta$ for $\mathrm{ker} T$, and then extend it to a basis $\beta\cup\gamma$ for all of $\mathbf{V}$. Then you show that the image of $\gamma$ under $T$ is linearly independent: if $\mathbf{v}_1,\ldots,\mathbf{v}_n\in\gamma$ are such that $$\alpha_1T(\mathbf{v}_1)+\cdots\alpha_nT(\mathbf{v}_n) = \mathbf{0}$$ in $\mathbf{W}$, then you can rewrite this as $$T(\alpha_1\mathbf{v}_1+\cdots+\alpha_n\mathbf{v}_n) = \mathbf{0}$$ hence $\alpha_1\mathbf{v}_1+\cdots+\alpha_n\mathbf{v}_n\in\mathrm{ker}(T)$. That means that this linear combination of vectors of $\gamma$ lies in the span of $\beta$, but since $\beta\cup\gamma$ is linearly independent, the only way this can happen is if $\alpha_1\mathbf{v}_1+\cdots+\alpha_n\mathbf{v}_n=\mathbf{0}$; this is a linear combination of vectors in $\gamma$, which is linearly independent, so $\alpha_1=\cdots=\alpha_n=0$. This proves that $$T(\gamma) = \{T(\mathbf{v})\mid \mathbf{v}\in\gamma\}$$ is a linearly independent subset of $\mathbf{W}$. It is now easy to show that $\mathrm{Im}(T) = \mathrm{span}T(\beta\cup\gamma) = \mathrm{span}(T(\gamma))$, so that $\dim(\mathrm{Im}(T)) = |\gamma|$. So you have that $$\dim\mathbf{V} = |\beta\cup\gamma| = |\beta|+|\gamma| = \dim(\mathrm{ker}\; T) + \dim(\mathrm{Im}\; T),$$ as desired.

share|cite|improve this answer

No, there is no other hypothesis needed. In particular, the dimension of $V$ may be infinite.


share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.