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Thanks for reading.

In short, my question is this:

What's the geometric interpretation of inner products when we allow our vectors to have complex components?


Now, to give a little context...

When I think of vectors that allow for complex components, I think of it as we're attaching a new orthogonal axis to each axis of the cartesian plane in which we put that corresponding complex component.

So $2$ dimensional vectors that allow for complex components are really $4$ dimensional, since both their $x$ component and their $y$ component have an additional "complex axis" attached to them.

I'm not sure if this is a good way to think about vectors with complex components, but it's the way I've been thinking about them thus far.

The dot product between two real-valued vectors $\vec{a}\cdot\vec{b}$ can be interpreted as multiplying the magnitude of the projection of $\vec{a}$ onto $\vec{b}$ by the magnitude of $\vec{b}$.

If we allow $\vec{a}$ and $\vec{b}$ to be complex, then the above only seems to be true when the components of $\vec{a}$ and the components of $\vec{b}$ point in the same "direction" on their corresponding complex planes.

An example:

Let $\vec{a}=\begin{bmatrix} i\\ 1 \end{bmatrix}$ and $\vec{b}=\begin{bmatrix} i\\ -3 \end{bmatrix}$

Then, $\vec{a}\cdot\vec{b}=(i)(-i)+(1)(-3)=(1)-3=-2$

Which can indeed be interpreted as the magnitude of the projection of $\vec{a}$ onto $\vec{b}$ multiplied by the magnitude of $\vec{b}$.

Another example:

Let $\vec{a}=\begin{bmatrix} 2+i\\ 1 \end{bmatrix}$ and $\vec{b}=\begin{bmatrix} 4+2i\\ -3 \end{bmatrix}$

Then, $\vec{a}\cdot\vec{b}=(2+i)(4-2i)+(1)(-3)=(8+2)-3=7$

Once again, this can be interpreted as the magnitude of the projection of $\vec{a}$ onto $\vec{b}$ multiplied by the magnitude of $\vec{b}$.

Note that $(2+i)$ and $(4+2i)$, the two $x$ components of our vectors, point in the same direction on the $x$ complex plane.


However, now let:

$\vec{a}=\begin{bmatrix} i\\ 1 \end{bmatrix}$

...and...

$\vec{b}=\begin{bmatrix} 2+3i\\ 1+i \end{bmatrix}$

...then $\vec{a}\cdot\vec{b}=4+i$.

We got a complex number as a result...how am I supposed to geometrically interpret that?

Thanks!!!

joshuaronis
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2 Answers2

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Think of vectors $z,w \in\Bbb C^n$ (in your case, $n=2$). If you write $z=a+bi$ and $w=c+di$, with $a,b,c,d\in\Bbb R^n$, then you can identify $z$ with the vector $\tilde z=(a,b)\in\Bbb R^{2n}$, and so on. When you compute the hermitian inner product $\langle z,w\rangle$ (which you wrote with dot), you end up with $$\langle z,w\rangle = (a\cdot c + b\cdot d) + i(b\cdot c - a\cdot d).$$ The real part is the real dot product $\tilde z\cdot\tilde w$. The imaginary part is skew-symmetric (meaning that you get the negative when you switch the order of $z$ and $w$); indeed, it can be written naturally as an alternating $2$-tensor, called the Kähler form of $\Bbb C^n$. It is naturally computing (perhaps up to a constant factor) the signed area of the parallelogram in $\Bbb R^{2n}$ spanned by $\tilde z$ and $\tilde w$.

Ted Shifrin
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[This is a repetition of the answer here]

Consider ${ \mathbb{C} ^n }.$

Distance: Let ${ z = (z _1, \ldots, z _n) \in \mathbb{C} ^n }.$ Viewing it as ${ (\text{Re}(z _1) + i \text{Im}(z _1), \ldots , \text{Re}(z _n) + i \text{Im}(z _n) ) }$ suggests looking at ${ \lVert z \rVert \overset{\text{def}}{=} \sqrt{ \vert z _1 \vert ^2 + \ldots + \vert z _n \vert ^2} }$ (which we expect to behave like length).

Real Orthogonality: Let ${ x, y \in \mathbb{C} ^n }$ be nonzero. For ${ n = 1 ,}$ we know ${ (x,y \text{ are perpendicular}) }$ ${ \iff (\text{point } x \text{ is equidistant from points } y, (-y)) }$ ${ \iff (\lVert x - y \rVert = \rVert x + y \rVert) }.$ This suggests looking at the notion ${ (x,y \text{ are real orthogonal}) \overset{\text{def}}{\iff} (\lVert x - y \rVert = \rVert x + y \rVert ). }$

Complex Orthogonality: The condition ${ \lVert x - y \rVert = \lVert x + y \rVert }$ can be rewritten as ${ \sum _{j=1} ^{n} \vert x _j - y _j \vert ^2 = \sum _{j=1} ^{n} \vert x _j + y _j \vert ^2 }$ that is ${ \sum (\overline{x _j} - \overline{y _j})(x _j - y _j) = \sum (\overline{x _j} + \overline{y _j})(x _j + y _j) }$ that is ${ \text{Re}(\sum _{j=1} ^{n} \overline{x _j} y _j) = 0 }.$
This suggests defining, for algebraic convenience, a stronger notion ${ (x,y \text{ are complex orthogonal}) \overset{\text{def}}{\iff} ( \sum _{j=1} ^{n} \overline{x _j} y _j = 0 ) }.$

Inner Product: Note ${ \lVert z \rVert = \sqrt{\sum _{j=1} ^{n} \overline{z _j} z _j} }$ and ${ (x,y \text{ are complex orthogonal}) \overset{\text{def}}{\iff} ( \sum _{j=1} ^{n} \overline{x _j} y _j = 0 ) }.$ This suggests looking at ${ \langle x, y \rangle \overset{\text{def}}{=} \sum _{j=1} ^{n} \overline{x _j} y _j. }$
Now all the above notions are ${ \lVert z \rVert = \sqrt{\langle z,z \rangle} },$ and ${ (x,y \text{ are real orthogonal}) \overset{\text{def}}{\iff} (\text{Re}(\langle x,y \rangle) = 0) },$ and ${ (x,y \text{ are complex orthogonal}) \overset{\text{def}}{\iff} (\langle x,y \rangle = 0 ) }.$
[${ \langle \cdot, \cdot \rangle }$ also has pleasant algebraic properties, like ${ \langle \alpha x, y \rangle = \overline{\alpha}\langle x, y \rangle },$ ${ \langle x, \beta y \rangle = \beta \langle x, y \rangle }$ and ${ \overline{\langle x, y \rangle} = \langle y, x \rangle}$ amd ${ \langle x + y, z \rangle = \langle x, z \rangle + \langle y, z \rangle },$ ${ \langle x, y + z \rangle = \langle x, y \rangle + \langle x, z \rangle }$].


TLDR: The constraint ${ \text{Re}(\sum _{j=1} ^{n} \overline{x _j} y _j) = 0 }$ has geometric significance, and considering ${ \sum _{j=1} ^{n} \overline{x _j} y _j }$ is for algebraic convenience.


Parallel vectors: As in vector spaces, ${ x, y \in \mathbb{C} ^{n} }$ nonzero are (real/complex) parallel if and only if one is a (real/complex) scalar multiple of the other respectively.

Decomposing into orthogonal and parallel vectors: Consider nonzero ${ x, y \in \mathbb{C} ^n }.$ We’ll try decomposing ${ x }$ as ${ x = x _{\parallel} + x _{\perp} ,}$ so that ${ x _{\parallel} }$ is (complex) parallel to ${ y }$ and ${ x _{\perp} }$ is (complex) orthogonal to ${ y }$.
This amounts to writing ${ x = \lambda y + (x - \lambda y) }$ where ${ \langle x - \lambda y, y \rangle = 0 .}$ So ${ \lambda = \frac{\langle y, x \rangle}{\langle y, y \rangle}, }$ that is ${ x = \frac{\langle y, x \rangle}{\langle y, y \rangle} y + \left( x - \frac{\langle y, x \rangle}{\langle y, y \rangle} y \right) }$ is the required decomposition.

Cauchy-Schwarz: In above setup ${ \lVert x \rVert = \sqrt{\langle x _{\parallel} + x _{\perp}, x _{\parallel} + x _{\perp} \rangle} },$ which on using ${ \langle x _{\parallel}, x _{\perp} \rangle = 0 }$ becomes ${ \lVert x \rVert = \sqrt{ \lVert x _{\parallel} \rVert ^2 + \lVert x _{\perp} \rVert ^2} }.$
Especially ${ \lVert x \rVert }$ ${ \geq \lVert x _{\parallel} \rVert }$ ${ = \frac{\vert \langle y, x \rangle \vert}{\lVert y \rVert} }.$ Equality occurs if and only if ${ x _{\perp} = 0 }$ that is ${ x,y }$ are parallel.