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Suppose we have 2 subspaces $\mathcal{A} \subseteq \mathbf{R}^n$ and $\mathcal{B}\subseteq \mathbf{R}^n$ that have the same dimension, say dim $\mathcal{A}=$ dim $\mathcal{B}=l>0$. Is it true that they have a non null intersection?


Ignore this part and focus on the top question answered correctly by @paf


I'm reading a section of Meyer's book, Matrix Analysis and Applied Linear Algebra, on the Courant–Fischer Theorem. I'm following the proof of pages 550-551, but I don't seem to understand why $\tilde{\mathcal{V}} \cap \mathcal{F} \neq \emptyset$ . . . This two subspaces have the same dimension so I figure it might be a general property that it's being used. Is this the case? If not, how would your show that $\tilde{\mathcal{V}} \cap \mathcal{F} \neq \emptyset$, following the reasoning of Meyer?

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  • $\begingroup$ I'm guessing you're missing some hypothesis, something along the lines of $\cal\dim A+\dim B>n$. $\endgroup$ – Asaf Karagila Jun 5 '18 at 17:32
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    $\begingroup$ Two lines with a single point in common... $\endgroup$ – Saucy O'Path Jun 5 '18 at 17:54
  • $\begingroup$ I don't think we have that type of hypothesis. The only general property I can think off is that dim$ \mathcal{A} \cup \mathcal{B}\geq l$, but this will not help. @AsafKaragila $\endgroup$ – J.D. Jun 5 '18 at 17:56
  • $\begingroup$ @Saucy: I'm confused. What's the dimension of $\{0\}$? $\endgroup$ – Asaf Karagila Jun 5 '18 at 17:59
  • $\begingroup$ @AsafKaragila just edit the question such that l>0. If l=0 then one could easily arrive at a contradiction. Simply choose $\mathcal{A}=\mathcal{B}=\emptyset$ and conclude that $l=0$ and $l>0$. $\endgroup$ – J.D. Jun 5 '18 at 18:20
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I'm not sure of your question, hence I'll give two answers.

1. Two vector subspaces of $\Bbb R^n$ always have non-empty ($\ne \varnothing$) intersection. Indeed, by definition, every vector subspace must contain the zero vector.

2. But their intersection can be exactly $\{0\}$ (thus having dimension $0$). Simply take two distinct vector subspaces of dimension 1 in $\Bbb R^2$ (i.e. 2 lines through 0)! To have an example valid for all $n$, if $V = \{a_1x_1 + \dots + a_nx_n = 0\}\subset \Bbb R^n$, then the line $W$ generated by $(a_1,\dots,a_n)$ intersects $V$ only at the zero vector.

So don't confuse $\varnothing$ and $\{0\}$...

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  • $\begingroup$ @pad Hum I see. I was not confusing $\{0\}$ and $\emptyset$. I simply did not think off $\{0\}$. But I see your point. In reality, for the theorem I was trying to prove, one actually needs that this vector [that belongs to the intersection] has unit norm . . . In that case, I don't see how Meyer's is finding such a vector to complete it's prove. Nonetheless your answer was correct and I thank you for your response. Do you have any suggestions? $\endgroup$ – J.D. Jun 5 '18 at 19:07
  • $\begingroup$ I assume that $(\mathbf{e}_1,...,\mathbf{e}_n)$ is the canonical basis of $\Bbb R^n$ with the canonical inner product. Since $\mathcal F= \text{span}(\mathbf{e}_1,...,\mathbf{e}_i)$ and $\tilde{\mathcal V} = (\mathbf{e}_1,...,\mathbf{e}_{i-1})^{\perp}$ and since $\mathbf{e}_i\in \tilde{\mathcal V}$, we obtain that $\tilde{\mathcal V}\cap \mathcal F$ contains $\mathbf{e}_i$ (and in fact, this intersection is exactly $\text{span}(\mathbf{e}_i)$). $\endgroup$ – paf Jun 5 '18 at 19:23
  • $\begingroup$ Of course. What a dumb question. Thank you for checking the prove and answering my question. I confused the definition of the sets and was doing the intersection of $\mathcal{V}$ with $\{e_i,\dots,e_n\}$,which is the orthogonal complement of $\tilde{\mathcal{V}}$. Nevermind, you are absolutely correct. In fact my question is incorrect. The sets I mentioned in the question do not have the same dimension . . . I think it's better to delete the question not to confuse anyone . . . $\endgroup$ – J.D. Jun 5 '18 at 20:03

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