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

Let $f \colon \mathbb{C}^5 \rightarrow \mathbb{C}^7$ a linear function, $f(2 i e_1 + e_3) = f(e_2)$ and $\mathbb{C}^7=X \oplus Im(f)$. What dimension has $X$?

share|cite|improve this question
The solution is $3 \leq \dim X \leq 7$. Why? – Katy23 May 24 '11 at 18:23
You're not French by any chance, are you? The French word "application" means function/mapping/transformation in English, but in English an application is something else. – Hans Lundmark May 24 '11 at 19:45
I'm not French. Thank you for the correction – Katy23 May 24 '11 at 20:34
up vote 1 down vote accepted

Hint: Apply the rank-nullity theorem. Given that $f$ satisfies at least the relation that you listed, what are the possible dimensions of the kernel?

Second hint: Take a basis $v_1, \ldots v_5$ of $\mathbb{C}^5$ such that $v_1=2ie_1-e_2+3_3$

share|cite|improve this answer
From the rank-nullity theorem we have that $\mathbb{C^5} = Ker(f) \oplus Im(f)$ but no information about $\mathbb{C^7}$ – Katy23 May 24 '11 at 18:27
But you know $\dim X + \dim \operatorname{im}(f)$, and you know $\dim \operatorname{im}(f)+\dim \ker(f)$. – Aaron May 24 '11 at 18:29
Hence $\dim X + \dim Im(f) = 7$ and $\dim Ker(f) + \dim Im(f) = 5$ implies that $\dim X - \dim Ker(f) = 2$. Hence $\dim X = 2 + \dim Ker(f)$. But now? – Katy23 May 24 '11 at 18:37
@Katy23 Yes, so you just need to figure out the possible values of $\dim \ker(f)$. Here's a small hint: the information you've given shows that $\dim \ker(f)\neq 0$. – Aaron May 24 '11 at 18:38
Because $f(2ie_1+e_3-e_2)=0$ we have that $\dim Ker f \geq 1$. Hence $\dim X \geq 3$ and $\dim X \leq 7$. Is this correct? – Katy23 May 24 '11 at 18:41

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.