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 $\epsilon \to 0$.


2 Answers 2


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 $$\|(\mathbb 1-P_2)(\mathbb 1-P_1)\|\;=\;\|UP_1U^*UP_2U^*\|\;=\;\|P_1P_2\| = \epsilon\,.$$

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? .

  • $\begingroup$ Dear Hanno, I think the argument is quite clear, although I don't know if I agree with your claim that the same is true for infinite-dimensional Hilbert spaces (at least with the given proof). I don't think it is clear that the unitary maps U_i exist when one considers an infinite-dimensional, non-seperable Hilbert space. $\endgroup$ Jul 29, 2019 at 1:27
  • $\begingroup$ Thanks for feedback. Shall consider your pt more closely begin of Sept when turning active again. $\endgroup$
    – Hanno
    Jul 31, 2019 at 10:09

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.

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'.

  • $\begingroup$ There is a mistake in the first bound (*) - it may not be true that w_1^a and w_2^a have norm 1. The argument is easily fixed, although it becomes slightly more complicated. $\endgroup$ Jul 29, 2019 at 1:22
  • $\begingroup$ That is not assumed, only that $a$ has norm $1$. $\endgroup$ Jul 29, 2019 at 3:15
  • $\begingroup$ Maybe I should have phrased my criticism as a question - sorry about that. I do not see how you can conclude that |<w_1^a, w_2^a>| \le \epsilon, since the definition of 'almost-orthogonality' only applies to unit vectors, and it is not obvious to me why w_1^a and w_2^a are unit vectors. $\endgroup$ Jul 29, 2019 at 3:31
  • $\begingroup$ Oh, you are absolutely right. My bad. $\endgroup$ Jul 29, 2019 at 3:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.