# Dimension problem

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 $X$?

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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

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 $v_1=2ie_1-e_2+3_3$
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