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I was thinking about the converse of the Pythagorean theorem:

$\lVert x + y\rVert^2 = \lVert x\rVert^2 + \lVert y\rVert^2 \implies x \perp y$

Does this hold if the inner product $\langle \cdot,\cdot\rangle$ is complex-valued and it induces the norm $\lVert \cdot \rVert$ ?

Here's what I've done:

Assuming an inner product that has linearity in the first argument, i.e. $\langle ax,y\rangle = a\langle x,y\rangle$, and conjugate linearity in the second argument: $\langle x,ay\rangle = a^*\langle x,y\rangle$, we have:

$\lVert x+y\rVert^2 = \langle x+y,x+y\rangle = \lVert x \rVert^2 + \lVert y \rVert^2 + \langle x,y\rangle + \langle y,x\rangle$ .

Using the assumption that the Pythagorean theorem holds and Hermitian symmetry, i.e. $\langle x,y\rangle^* = \langle y,x\rangle$, then we must have:

$\langle x,y\rangle + \langle y,x\rangle = 0 \Leftrightarrow \langle x,y\rangle = -\langle x,y\rangle^*$

Since the inner product is complex, let $\langle x,y\rangle = a + bi$.

Then, $\langle x,y\rangle = -\langle x,y\rangle^* \Leftrightarrow a + bi = -a + bi$ .

Equating real and imaginary parts, $a=0$ and $b=b$, i.e $\operatorname{Re}[\langle x,y\rangle] = 0$ .

Thus, we can have two vectors x and y that aren't orthogonal, i.e. $\langle x,y\rangle \ne0$, but do satisfy the Pythagorean Theorem.

Is this correct? The logic seems right, but I still have some uncertainty.

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  • $\begingroup$ If you want to formulate the Pythagorean theorem with vectors, the equation should be $\|x+y\|^2 = \|x\|^2 + \|y\|^2$, not $\|x+y\|^2 = \|x\| + \|y\|$. $\endgroup$ – Henning Makholm May 22 '16 at 1:29
  • $\begingroup$ just take $y=ix$ and do the math $\endgroup$ – Filburt Nov 1 '17 at 15:26
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You are right.

And it has to be that way, because the norm that the Hermitian product induces on $\mathbb C^n$ is the same as the Euclidean norm on $\mathbb R^{2n}$.

Therefore, for a fixed $x\in\mathbb C^n$, the set $$ \{ y\in\mathbb C^n \mid \|x+y\|^2 = \|x\|^2+\|y\|^2 \} $$ has dimension $2n-1$ over $\mathbb R$ -- and therefore it cannot be a linear subspace over $\mathbb C$. In particular it cannot equal $\{y\in\mathbb C^n \mid \langle x,y\rangle = 0 \}$.

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  • $\begingroup$ I'm confused by this question - your answer is clearly right - can you give a concrete example of two vectors $x,y$ for which the condition holds but the converse of the Pythagorean theorem fails? For some reason I keep on thinking they would have to be pure imaginary, but that makes just as little sense as if they were real. $\endgroup$ – Chill2Macht May 22 '16 at 1:43
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    $\begingroup$ @William: The OP's computation shows that this happens when $\langle x,y\rangle$ is pure imaginary. For example, this would be the case in $\mathbb C^1$ for the vectors $(1)$ and $(i)$. $\endgroup$ – Henning Makholm May 22 '16 at 1:46
  • $\begingroup$ OK I think I see it now, and this is a counterexample because in $\mathbb{C}^1$ those two vectors are linearly dependent, even though their values in the standard isomorphism with $\mathbb{R}^2$ would be linearly independent? Thank you so much -- this makes sense now -- considered as vectors in $\mathbb{R}^{2n}$, they would have to be linearly independent (perpendicular), but as vectors in $\mathbb{C}^n$ that no longer has to be the case. $\endgroup$ – Chill2Macht May 22 '16 at 1:53
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    $\begingroup$ @William: Correct -- when you switch from viewing the space as a vector space over $\mathbb C$ instead of a vector space over $\mathbb R$, you get more linear combinations (because you have more scalars to make them with), and therefore it becomes more difficult for a set to be linearly independent. [But note that "linearly independent" and "perpendicular" are quite different concepts even in the real case!] $\endgroup$ – Henning Makholm May 22 '16 at 2:06
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Pythagorean theorem is valid for non- orthogonal set of vectors as well. There's is good discussion on this at https://arxiv.org/abs/1509.08667 For the Pythagorean theorem, pair wise orthogonality not required, just sum of the cross terms must be zero. For example, there is set of LINOEP vectors for which square of the norm is preserved.

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    $\begingroup$ Could you elaborate? What is a LINOEP vector? $\endgroup$ – Robert Shore Feb 13 at 3:54
  • $\begingroup$ Linearly independent no orthogonal yet energy preserving (LINOEP) set of vectors. These vectors are useful in some applications e.g. energy preserving empirical mode decomposition. In signal processing, energy is same as square of the norm. $\endgroup$ – Pushpendra Singh Feb 13 at 16:54
  • $\begingroup$ Please refer the following link for more details about LINOEP vectors: arxiv.org/abs/1409.5710 $\endgroup$ – Pushpendra Singh Feb 13 at 17:01

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