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Let $V$ be a proper linear subspace of $\mathbb{C}^{n}$ (viewed as a vector space over $\mathbb{C}$). I want to show that, then there exists $\alpha_{1},\dots,\alpha_{n}\in\mathbb{C}$ (not all zero), such that, for all $(\lambda_{1},\dots,\lambda_{n})\in V$, $$ \alpha_{1}\lambda_{1}+\dots+\alpha_{n}\lambda_{n}=0. $$

I suppose that, If a take the statement to be false, I can arrive somehow to the fact that $V=\mathbb{C}^{n}$, so if I suppose that, for all n-tuple of complex numbers not all zero ($\alpha_{1},\dots,\alpha_{n}\in\mathbb{C}$), we have that there exists an element $(\lambda_{1},\dots,\lambda_{n})$ in $V$ such that $\alpha_{1}\lambda_{1}+\dots+\alpha_{n}\lambda_{n}\neq 0$, I could take $v\in\mathbb{C}^{n}$ and prove that $v\in V$ (by using the previous hypothesis). Nevertheless, I don't know if this would be the right way to think it, and even if it is, I am stuck on the proof completely...

Can someone provide just some guidance so I can prove it ? Thanks in advanced!

rubikman23
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    If you take a proper subspace, can you build its orthogonal? – blamethelag Feb 24 '23 at 23:57
  • @blamethelag , yes! If $v_{1},\dots,v_{k}$ is a basis of $V$ (with $k=\dim(V)<n$), by taking, $v_{k+1},\dots,v_{n}$ linearly independent vectors (among them) that are not in $V$, I would have a basis of the orthogonal, and $v_{1},\dots,v_{n}$ would be in fact a basis of $\mathbb{C}^{n}$! Where does this leads me? Thanks for answering! – rubikman23 Feb 25 '23 at 00:03
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    I don't buy "I would have a basis of the orthogonal". The best you would have with your construction is a direct sum which may not be perpendicular. A drawing of two lines in $\mathbb R^2$ can show you why. To correct your construction: start with an orthogonal basis of $V$ (Graham Schmidt for instance). Then iterate Graham Schmidt as long as you can, this will give you an orthogonal splitting $\mathbb C^n = V \oplus W$. Can you take $w \in W$ non null? Then for sure $V \subset { w }^\perp$. – blamethelag Feb 25 '23 at 08:31

2 Answers2

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For $a,b\in\mathbb C^n$, with $a=(a_1,a_2,\ldots,a_n)$ and $b=(b_1,b_2,\ldots,b_n)$, define $a\cdot b:=a_1b_1+a_2b_2+\ldots+a_nb_n\in\mathbb C$. It is easy to see that the map $\cdot:\mathbb C^n\times\mathbb C^n\to\mathbb C$ is bilinear.

Now, let $(v_1,v_2,\ldots,v_k)$ be a basis of $V$, where $k<n$ i.e. $k\le n-1$.

Further, define a map $L:\mathbb C^n\to \mathbb C^k$ given by $L(a):=(a\cdot v_1,a\cdot v_2,\ldots,a\cdot v_k)$ for $a\in\mathbb C^n$. It is easy to prove that $L$ is linear. As per Rank-Nullity theorem, $\dim\ker(L)=n-\dim\operatorname{im}(L)$. As $\operatorname{im}(L)\subset \mathbb C^k$, we have $\dim\operatorname{im}(L)\le k\le n-1$. Thus, $\dim\ker(L)\ge n-(n-1)=1$. So, $\ker L$ is nontrivial, i.e. you can find $a\in\ker L, a\ne 0$.

However, this means $a\cdot v_1=a\cdot v_2=\ldots=a\cdot v_k=0$, which then implies that $a\cdot v=0$ for all $v\in V$.

Note that the same proof applies to finite-dimensional vector spaces over any field, not necessarily $\mathbb C$.

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Hint: Let $e_1,e_2,\dots,e_k$ be an orthonormal basis for $V$. Take any $w$ in the complement of $V$.

Analyze the vector

$$ w^{'} = w - \sum_{i=1}^{k} \langle w,e_i \rangle e_i $$

(an $n$ tuple of complex numbers)

Note that for $a,b \in \Bbb C^n$,

$$ \langle a, b \rangle =a_1 \bar{b_1}+a_2 \bar{b_2}+\ldots +a_n \bar{b_n}$$

CopyPasteIt
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  • I would not necessarily invoke the orthonormality as the dot product is a sesquilinear form, whereas what the OP has is a bilinear form. For one thing, $\langle v,v\rangle=0$ does not imply $v=0$ (E.g. $v=(1,i)\in\mathbb C^2$). One needs to be careful about what is still valid here. That is why I avoided mentioning dot products in my answer. –  Feb 25 '23 at 01:06
  • @StinkingBishop I'm using inner product notation. – CopyPasteIt Feb 25 '23 at 01:07
  • Maybe, but what you have here is not an inner product. (Conjugate symmetry is not satisfied.) –  Feb 25 '23 at 01:21
  • @StinkingBishop Conjugate symmetry is certainly satisfied - my answer depends on it. – CopyPasteIt Feb 25 '23 at 01:25
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    Okay, I think I understand what you mean to say. You use the inner product as-is. However, once the OP calculates $w'$, they need to conjugate every component of it and what they get will then work for them. Sorry for being a bit slow, maybe worth mentioning that in the answer. –  Feb 25 '23 at 01:29
  • @StinkingBishop I upvoted your comment - you got me to write out the inner product in my answer, and after reading the OP's question again, well, there it is... – CopyPasteIt Feb 25 '23 at 01:48
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    Thanks! My overall issue here is that the "product" used in the OP's question does not involve conjugation at all, so it is not immediately clear from your answer how you use the ordinary inner product (which does involve conjugation) to infer something about that other product. (In fact I first assumed wrongly that you used $\langle\cdot,\cdot\rangle$ for that other product, thanks for clarifying.) –  Feb 25 '23 at 01:56