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 $X$?
Dimension problem
2
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linear-algebra
complex-numbers
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0The solution is $3 \leq \dim X \leq 7$. Why? – 2011-05-24
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0You'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. – 2011-05-24
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0I'm not French. Thank you for the correction – 2011-05-24
1 Answers
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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 $v_1=2ie_1-e_2+3_3$
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0From the rank-nullity theorem we have that $\mathbb{C^5} = Ker(f) \oplus Im(f)$ but no information about $\mathbb{C^7}$ – 2011-05-24
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0But you know $\dim X + \dim \operatorname{im}(f)$, and you know $\dim \operatorname{im}(f)+\dim \ker(f)$. – 2011-05-24
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0Hence $\dim X + \dim Im(f) = 7$ 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? – 2011-05-24
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0@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 $\dim \ker(f)\neq 0$. – 2011-05-24
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0Because $f(2ie_1+e_3-e_2)=0$ we have that $\dim Ker f \geq 1$. Hence $\dim X \geq 3$ and $\dim X \leq 7$. Is this correct? – 2011-05-24
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0@Katy23: You are missing a bit. Your reasoning is right for showing that $\dim X \geq 3$, but and clearly $\dim X \leq 7$ because $X\subset \mathbb{C}^7$, but we need to be more careful than that. What are the possible dimensions of $\dim \ker(f)$? It is at least $1$, and at most $5$ (which would occur if $f$ was the zero map, as $f$ would still satisfy the relation that you listed). But what about the dimensions in between? If you don't have an intuition about it, you might want to try to construct specific linear maps. – 2011-05-24
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0@Katy23: Oh, and can the dimension of the kernel be exactly $1$? Why or why not? – 2011-05-24
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0I think that all dimensions between 1 and 5 are possible. Furthemore $\dim Ker(f)$ can be exactly 1. – 2011-05-24
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0@Katy23 They are, but to explain why, you should look into the following strengthening of the rank-nullity theorem. If $V'\subset V$ and $W'\subset W$ are vector spaces with $\dim V'+\dim W'=\dim V$, then there is a linear map $f:V\to W$ with $\ker f = V'$ and $\operatorname{im} f = W'$, which is the composition of $V\to V/V' \to W' \to W$, where the first map is the natural quotient, the last is the natural inclusion, and the middle map is an isomorphism. It is easy to see this by taking a basis in the right way. – 2011-05-24