If complementary subspaces are almost orthogonal, is the same true for their orthogonal complements?

Question

Suppose that $W_1,W_2$ are complementary subspaces in a finite-dimensional vector space $V$ (so $W_1 + W_2 = V$ and $W_1 \cap W_2 = \{0\}$). Fix an inner product $\langle \cdot,\cdot\rangle$ on $V$ and further suppose that $W_1$ and $W_2$ are `almost orthogonal' with respect to this inner product i.e. if $w_1 \in W_1$ and $w_2 \in W_2$ are unit vectors, then $|\langle w_1, w_2 \rangle| \le \epsilon$.

It is true that $W_1^\perp$ and $W_2^\perp$ are complemented, but are they almost orthogonal? More precisely, can we estimate the quantity $$ \sup_{v_i \in W_i^\perp, ||v_i|| = 1} |\langle v_1, v_2 \rangle| $$ and does it vanish as $\epsilon \to 0$.

Answer 1

The head line question is answered with a plain yes.
And this yes remains true if $V$ is an infinite-dimensional Hilbert space.

It is assumed that $V=W_1\oplus W_2$, and the two complementary subspaces are necessarily closed (this merits special mention in the case $\dim V=\infty$).
Let $P_j$ denote the orthogonal projector (= idempotent and self-adjoint) onto $W_j$: $$\sup_{w_j\in W_j\\ \|w_j\| = 1}\big|\langle w_1, w_2 \rangle\big| \;=\;\sup_{v_j\in V\\ \|v_j\| = 1}\big|\langle P_1v_1, P_2v_2 \rangle\big| \;=\;\sup_{v_j\in V\\ \|v_j\| = 1}\big|\langle v_1, P_1P_2v_2 \rangle\big| \:=\:\|P_1P_2\|\:=\:\epsilon\,<\,1$$

The last estimate is a non-obvious fact, cf Norm estimate for a product of two orthogonal projectors . Only if $W_1$ and $W_2$ are (completely) orthogonal one has $\epsilon=0\,$.

Look at the corresponding quantity for the direct sum $V=W_2{}^\perp\oplus W_1{}^\perp\,$: $$\sup_{w_j\in W_j{}^\perp\\ \|w_j\| = 1} \big|\langle w_2, w_1 \rangle\big| \;=\; \sup_{v_j\in V\\ \|v_j\| = 1} \big|\langle (\mathbb 1-P_2)v_2, (\mathbb 1-P_1)v_1 \rangle\big| \:=\: \|(\mathbb 1-P_2)(\mathbb 1-P_1)\|$$

Because of $V=W_1\oplus W_2 = W_2{}^\perp\oplus W_2 = W_2{}^\perp\oplus W_1{}^\perp$ one can find unitaries $U_1:W_1\to W_2{}^\perp$ and $U_2:W_2\to W_1{}^\perp$, and thus define on $V$ the unitary operator $$U: W_1\oplus W_2\xrightarrow{U_1\oplus\,U_2}W_2{}^\perp\oplus W_1{}^\perp$$ which respects the direct sums. Then $\mathbb 1-P_2=UP_1U^*$ and vice versa, hence $$\|(\mathbb 1-P_2)(\mathbb 1-P_1)\|\;=\;\|UP_1U^*UP_2U^*\|\;=\;\|P_1P_2\| = \epsilon\,.$$

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Remark$\:\;\epsilon\,$ can be written as $\cos\gamma$, and $\gamma$ is interpreted as angle between the subspaces. This was the motivation for the post A "Crookedness criterion" for a pair of orthogonal projectors? .

Answer 2

Let $A_1 = {W_1}^\perp$ and $A_2 = {W_2}^\perp$. Notice that $\dim A_1 = \dim W_2$ and that $\dim A_2 = \dim W_1$ so in order for them to be complementary we need only show that $A_1 \cap A_2 = \{0\}$.

Now, any $v\in V$ can be uniquely written as $v = w_1^v + w_2^v$, where $w_1^v \in W_1$ and $w_2^v\in W_2$. If $a \in$ $A_1 \cap A_2$ then $a \perp v = w_1^v + w_2^v$ for all $v \in V$, and hence $a = 0$ as we sought to show.

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Now, let $a \in A_1$ with $\lVert a\rVert = 1$ and write $a = w_1^a + w_2^a$. Then

\begin{align}1 &= \langle a, a \rangle \\&= \underbrace{\langle a, w_1^a \rangle}_0 + \langle a, w_2^a \rangle \\&= \langle w_1^a , w_2^a \rangle + \langle w_2^a , w_2^a \rangle\\{} \end{align} \begin{align} &\implies 1 - \langle w_2^a , w_2^a \rangle = \langle w_1^a , w_2^a \rangle\tag{$*$} \\&\implies \big|1 - \langle w_2^a , w_2^a \rangle\big| = \big|\langle w_1^a , w_2^a \rangle\big| \leqslant \epsilon, \end{align}

so that most of $\lVert a \rVert$ is concentrated on the $W_2$ component. Now, of course, we can write

\begin{align}1 &= \langle a, a \rangle \\&= \langle w_1^a+w_2^a, w_1^a+w_2^a \rangle \\&= \langle w_1^a, w_1^a\rangle + 2 \langle w_1^a, w_2^a\rangle + \langle w_2^a, w_2^a\rangle \\&= \langle w_1^a, w_1^a\rangle + 2 \Big(1 - \langle w_2^a , w_2^a \rangle\Big) + \langle w_2^a, w_2^a\rangle\tag{$**$} \\&= \langle w_1^a, w_1^a\rangle + 2 - \langle w_2^a, w_2^a\rangle \end{align} \begin{align} &\implies \langle w_1^a, w_1^a\rangle = \langle w_2^a, w_2^a\rangle - 1, \end{align}

where we substituted $(*)$ into $(**)$. This allows us to conclude that $\big|1 - \langle w_2^a , w_2^a \rangle\big| = \langle w_2^a, w_2^a\rangle - 1$, and hence

$$\left\{\begin{array}{} 0\leqslant \langle w_1^a, w_1^a\rangle \leqslant \epsilon\\ -\epsilon \leqslant \langle w_1^a , w_2^a \rangle \leqslant 0\\ 1 \leqslant \langle w_2^a , w_2^a \rangle \leqslant 1+\epsilon \end{array}\right.\tag{A}$$

Similarly, if $b\in A_2$ has $\lVert b \rVert = 1$, most of $\lVert b \rVert$ is concentrated on the $W_1$ component, with

$$\left\{\begin{array}{} 0\leqslant \langle w_2^b, w_2^b\rangle \leqslant \epsilon\\ -\epsilon \leqslant \langle w_1^b , w_2^b \rangle \leqslant 0\\ 1 \leqslant \langle w_1^b , w_1^b \rangle \leqslant 1+\epsilon \end{array}\right.\tag{B}$$

Finally, we can estimate $\langle a, b\rangle$ with

\begin{align} \langle a, b\rangle &= \langle a, w_1^b + w_2^b\rangle \\&= \underbrace{\langle a, w_1^b\rangle}_{0} + \langle a, w_2^b\rangle \\&= \langle w_1^a, w_2^b\rangle + \langle w_2^a, w_2^b\rangle, \end{align}

so that

\begin{align} |\langle a, b\rangle| &\leqslant |\langle w_1^a, w_2^b\rangle| + |\langle w_2^a, w_2^b\rangle| \\&\leqslant \epsilon + |\langle w_2^a, w_2^b\rangle| \tag{by almost orthogonality} \\&\leqslant \epsilon + \sqrt{\langle w_2^a, w_2^a\rangle\cdot\langle w_2^b, w_2^b\rangle}\quad\quad\quad\quad \tag{by Cauchy-Schwarz} \\&\leqslant \epsilon + \sqrt{(1+ \epsilon) \cdot \epsilon} \tag{by $(A)$ and $(B)$} \end{align}

It follows that

$$ \sup_{v_i \,\in \,W_i^\perp\,\cap\, \partial B(0; 1)} |\langle v_1, v_2 \rangle| \leqslant \epsilon + \sqrt{(1+ \epsilon) \cdot \epsilon}\, ,$$

so that it vanishes as $\epsilon \to 0$. I'm not sure if the estimate is sharp, and moreover $V$ being finite dimensional did not come into play here. We merely used the fact that $V = W_1 \oplus W_2$ and the 'almost orthogonality'.