The annihilator of an intersection is the sum of annihilators

Question

Given a subset $X$ of a vector space $V$, let $X^\circ$ be the annihilator of $X$, that is $X^\circ = \{y\in V^* | \; y(x)=0, \;\forall\; x\in X\}$, where $V^*$ is the dual space of $V$.

Question: If $M$ and $N$ are subspaces of an $n$-dimensional vector space $V$, then $(M\cap N)^\circ =M^\circ +N^\circ $.

It's easy to prove this result by using that $(M+N)^\circ =M^\circ \cap N^\circ $, which is easy to prove too.

My doubt is: Is there a "direct way" to prove that $(M\cap N)^\circ =M^\circ +N^\circ $ (without make use of other equality)?

Thanks!

Answer 1

There is a reason why the first relation $$(M+N)^\circ =M^\circ \cap N^\circ \tag1$$ is easier to prove than $$(M\cap N)^\circ =M^\circ +N^\circ\tag2$$ Indeed, (1) is true for arbitrary vector spaces. (The proof is short: a linear function annihilates $M+N$ if and only if it annihilates both $M$ and $N$.)

But (2) fails for general vector spaces (it holds for finite dimensional ones). More specifically, the inclusion $(M\cap N)^\circ \supseteq M^\circ +N^\circ$ is true in general vector spaces, but not the reverse inclusion. So, any proof of (2) must use the finiteness of dimension somewhere. It cannot be a twin of the proof of (1).

(A gerw said, (2) can be made true in full generality by replacing $M^\circ +N^\circ$ with $\overline{M^\circ +N^\circ}$.)

Answer 2

The equality $(M\cap N)^o=M^o+N^o$ does hold in the general infinite-dimensional case. Here is an elegant proof:

first, recall that if $T:X\to Y$ is a linear operator, then $(\mathrm{Ker}(T))^o=\mathrm{Im}(T^*)$. To prove the nontrivial inclusion, pick $\alpha$ in $(\mathrm{Ker}(T))^o$ and define a linear functional $\beta:\mathrm{Im}(T)\to K$ by passing $\alpha$ to the quotient through $T$, i.e., set $\beta(y)=\alpha(x)$, with $x\in X$ chosen with $T(x)=y$. Now check that $\beta$ is well-defined and linear. Pick a linear extension $\bar\beta$ of $\beta$ to the whole space $Y$ and note that $T^*(\bar\beta)=\alpha$.

Now, to prove the equality $(M\cap N)^o=M^o+N^o$ for subspaces $M$ and $N$ of $X$, consider the linear map $T:X\to X/M\times X/N$ given by $T(x)=(x+M,x+N)$ and use the equality $(\mathrm{Ker}(T))^o=\mathrm{Im}(T^*)$. The kernel of $T$ is obviously equal to $M\cap N$ and the map $T^*$ is identified with the sum map $S:M^o\times N^o\to X^*$, $S(\alpha,\beta)=\alpha+\beta$, with the dual space of a product identified with the product of the dual spaces and the dual of a quotient $X/M$ identified with the annihilator $M^o$ in the canonical way.

Now, if $X$ is a Banach space and $M$, $N$ are closed subspaces of $X$, the equality $(M\cap N)^o=M^o+N^o$ does not hold in general if the dual spaces are understood in the topological sense (i.e., just continuous linear functionals). What holds in general is that $(M\cap N)^o$ is the closure of $M^o+N^o$ in the weak*-topology. This follows with the same proof as above, noting that for a bounded operator $T:X\to Y$ between Banach spaces, we have that $(\mathrm{Ker}(T))^o$ is equal to the weak*-closure of $\mathrm{Im}(T^*)$. Moreover, it holds that:

$(\mathrm{Ker}(T))^o=\mathrm{Im}(T^*)\Longleftrightarrow\mathrm{Im}(T)\ \text{is norm-closed}\Longleftrightarrow\mathrm{Im}(T^*)\ \text{is norm-closed}.$

Thus $(M\cap N)^o=M^o+N^o$ if and only if the sum $M^o+N^o$ is norm-closed. However, it is not true in general that $(M\cap N)^o$ equals the norm-closure of $M^o+N^o$ (as it is not true in general that $(\mathrm{Ker}(T))^o$ equals the norm-closure of $\mathrm{Im}(T^*)$). Here is a simple counterexample: set $X=\ell_1\times\ell_2$, $M=\ell_1\times\{0\}$ and $N=\big\{(x,x):x\in\ell_1\big\}$. We have $M\cap N=\{0\}$, so that $(M\cap N)^o=X^*$. However, identifying $X^*$ with $\ell_\infty\times\ell_2$ in the natural way, we have $M^o+N^o=\ell_2\times\ell_2$ and the norm closure of $M^o+N^o$ is $c_0\times\ell_2$.

Answer 3

You may work with orthonormal bases: Let $\{u_i\} \cup \{v_i\} \cup \{w_i\} \cup \{x_i\}$ a basis in $X$, such that $\{u_i\}$ is a basis of $M\cap N$, $\{u_i\} \cup \{v_i\}$ of $M$ and $\{u_i\} \cup \{w_i\}$ of $N$.

Then, $M^\circ$ is the linear hull of $\{w_i\} \cup \{x_i\}$, $N^\circ$ is the hull of $\{v_i\} \cup \{x_i\}$ and $(M \cap N)^\circ$ is the hull of $\{v_i\} \cup \{w_i\} \cup \{x_i\}$ (if I have done anything right).

Anyway, this is a mess. The easy proof you mentioned has also the advantage, that it works in infinite dimensions (in this case you have that the closure of $M^\circ + N^\circ$ is equal to $(M\cap N)^\circ$).