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Let $V$ be a vector space. If $W_1,W_2$ subspaces of $V$ then $(W_1\cap W_2)^0=W_1^0+W_2^0$. One way is very clear, and it's $W_1^0+W_2^0\subseteq (W_1\cap W_2)^0$. But I cant quite understand how to prove the other way? I tried writing a base for the intersection and expanding it to each of the other spaces, then operate $t\in (W_1\cap W_2)^0$ on a linear combination of a generic vector from $W_1^0+W_2^0$ which I wrote a a linear combination of the base I expanded. Got stuck. Is that even the right way?

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The main thing you need to use here is that a linear form defined on a subspace can be extended to a linear form on the whole space. To do this, just complete any basis of the subspace to a basis of the whole space, and define the linear form to be $0$ on the added basis vectors.

Now given a linear form $\alpha$ that vanishes on $W_1\cap W_2$, we wish to write it as the sum of a linear form $\alpha_1$ that vanishes on $W_1$ and a linear form $\alpha_2$ that vanishes on $W_2$. Consider the restriction $\beta_1$ of $\alpha$ to $W_1$; it needs to be zero (only) on $W_1\cap W_2$. We first wish to extend $\beta_1$ to a linear form $\beta_2$ on $W=W_1+W_2$ that vanishes on $W_2$, which can be done as follows. By the second isomorphism theorem one has $W/W_2\cong W_1/(W_1\cap W_2)$, and $\beta_1$ defines a linear form on $W_1/(W_1\cap W_2)$; through the isomorphism it gives a linear form on $W/W_2$ which lifts to our linear form $\beta_2$ on $W$ that vanishes on $W_2$. Now further extend $\beta_2$ to a linear form $\beta$ on all of $V$. By construction $\beta$ coincides with $\alpha$ on $W_1$ and vanishes on $W_2$. One can therefore take $\alpha_1=\alpha-\beta$ and $\alpha_2=\beta$.

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  • $\begingroup$ I find two complementary subspaces, say $U_1$ and $U_2$ such that $W_1 = U_1 \oplus W_1\cap W_2$ and $W_2 = U_2 \oplus W_1\cap W_2$, and then we can find a third subspace $U_3$ such that $V = U_3 \oplus U_1 \oplus W_1\cap W_2 \oplus U_2$. Now by taking a basis for each of these subspaces, I can define a linear functional that annihilates $W_1\cap W_2$ as a sum of two linear functionals. Do I understand this right, is this what your solution means ? $\endgroup$ – me10240 Feb 6 '16 at 19:50
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We know that if U and V are two subspaces of a finite-dimensional vector space: $\dim (U+V)=\dim U + \dim V -\dim (U\cap V)$. To prove the equality requested if suffice to show that $\dim (W_1^0+W_2^0)=\dim (W_1\cap W_2)^0$. This is a direct consequent from the previous equality and the fact that $W_1^0\cap W_2^0=(W_1+ W_2)^0$ and $\dim U^0=\dim V-\dim U$ for all subspace $U$ of $V$.

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