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Let $A_i, A'_i, i=1\cdots k$ and $I=\operatorname{span}\{v_0\}$ be subspaces of $V$ such that $\operatorname{dim }A_i=\operatorname{dim} A'_i, \forall i$ and $\operatorname{dim} \bigcap_{1}^kA_i=\operatorname{dim} \bigcap_{1}^kA'_i $. The following is true or false :$$\operatorname{dim} \bigcap_{1}^k(A_i\oplus I)=\operatorname{dim} \bigcap_{1}^k(A'_i\oplus I) $$


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up vote 3 down vote accepted

In general, this isn't true. My original answer was wrong, here is the updated version:

Take $V=\mathbb{R}^3$, $k=2$. Set $v_0=(1, 0, 0),\, I=span\{v_0\}$.

Take $A_1=span\{(1, -1, 0)\},\,A_2=span\{(1, 1, 0)\}$.

Take $A'_1=span\{(0, 1, 0)\},\,A'_2=span\{(0, 0, 1)\}$.

All the dimensions of $A_i$ and $A'_i$ are equal to 1. The dimensions of $\bigcap A_i$ and $\bigcap A'_i$ are 0. On the LHS we have $\dim \bigcap(A_i \oplus I) = 2$, and on the RHS we have $\dim \bigcap(A'_i \oplus I) = 1$.

UPDATE: here are my thoughts on the additional question from the comments: what conditions would be enough to guarantee the equality from the original question?

I can only think of one thing: we can require that $I$ has a trivial intersection with $\sum_1^k A_i$ and with $\sum_1^k A'_i$. And we can drop the requirement that $\dim A_i = \dim A'_i$. Here is the proof.

Lemma Let $A, B, C$ be subspaces of $V$ such that $C \cap (A+B) = \{0\}$. Then $$ C \oplus (A \cap B) = (C \oplus A) \cap (C \oplus B). $$ Proof. It is quite clear that $C \oplus (A \cap B) \subset (C \oplus A) \cap (C \oplus B)$, so we only need to prove that $C \oplus (A \cap B) \supset (C \oplus A) \cap (C \oplus B)$.

Suppose that a vector $v$ belongs to both subspaces $C \oplus A$ and $C \oplus B$. Then there are vectors $c_1,c_2 \in C$, $a \in A$ and $b \in B$ such that $v=c_1+a=c_2+b$. Subtracting one from another, we get $c_1-c_2=a-b$. The LHS belongs to $C$ and the RHS belongs to $A+B$. Therefore, $c_1=c_2$ and $a=b$, and so $a \in A\cap B$. But then $v=c_1+a \in C \oplus (A \cap B)$, QED.

Corollary Let $A_i,\,i=1,\,\ldots,\,k$ and $C$ be subspaces of $V$, and $C \cap \sum_1^k A_i = \{0\}$. Then $$ C\oplus \bigcap_1^k A_i = \bigcap_1^k (C \oplus A_i). $$ Proof. Just use induction and the Lemma.

From this corollary we get this statement: if $A_i,A'_i$ and $I$ are finite-dimensional subspaces of $V$ such that $\dim \bigcap_1^k A_i = \dim \bigcap_1^k A'_i$ and $I$ has a trivial intersection with $\sum_1^k A_i$ and with $\sum_1^k A'_i$, then $\dim \bigcap_1^k (A_i \oplus I) = \dim \bigcap_1^k (A'_i \oplus I)$.

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No, i assume that $v_0$ not belong to any $A_i$ and $A'_i$. – Firsttime Nov 2 '12 at 13:13
Well, you should probably put that in the question statement, this condition makes a huge difference. – Dan Shved Nov 2 '12 at 13:16
That's why direct sum in here – Firsttime Nov 2 '12 at 13:18
Oops, my bad. Well, I'll update my answer then. – Dan Shved Nov 2 '12 at 13:18
There. I hope there are no stupid mistakes this time ) – Dan Shved Nov 2 '12 at 13:27

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