Under the standard definition of the inner product as:

$⟨(w_1,...,w_n),(z_1,...,z_n)⟩=w_1 \bar z_1 +···+w_n\bar z_n$

What does it mean when $⟨(w_1,...,w_n),(z_1,...,z_n)⟩=0$? For real vectors it means that there is a right angle between the two vectors in space they are in. On the complex place, however, there is a different interpretation of this as (1,0) can be multiplied by $i$ to get to (0,1). So we have a rotation operation that can be linearly multiplied. For two complex 1-d vectors to be orthogonal: $(a+bi)(c-di)=0$


$ac+bd=0$ and $ad-bc=0$

We have real parts($a$) interacting with complex(like $adi$) and vice versa. So what does orthogonality mean geometrically here? I'm sure its different that the real case.

Here is a quote from an answer I found very interesting:

If ${\bf x}={\bf a}+{\bf b}i$, ${\bf y}={\bf c}+{\bf d}i$ in $C^n$, define vectors ${\bf u}=({\bf a},{\bf b})$, ${\bf v}=({\bf c},{\bf d})$, ${\bf w}=(-{\bf d},{\bf c})$ in $R^{2n}$. Then $$ {\bf x}\cdot {\bf y} = {\bf u}\cdot {\bf v}+i{\bf u}\cdot {\bf w} $$ so the real and imaginary parts of ${\bf x}\cdot {\bf y}$ have exactly the geometric interpretations you are familiar with, as applied to ${\bf u}$, ${\bf v}$ and ${\bf w}$, where ${\bf w}$ is a rotation of ${\bf v}$.

Source:interpretation of dot product of complex vectors

Why does $w$ need to be such?

Bonus Questions The inner product can also be complex itself. What does this mean in terms of projections? Is it the real projection + the complex projection ? If two complex vectors have a real inner product what does this say about them?

  • $\begingroup$ $w$ is $i\mathbf{v}$; because you need to figure out how to deal with complex scalars, but an arbitrary complex scalar $\alpha + i\beta$ can be decomposed as "multiply by the real scalar $\alpha$, then multiply by the real scalar $\beta$ separately, and then this by $i$, and then add the results", the only thing that stands between the real inner product and extending it to complex numbers is to figure out what multiplication by $i$ does. And by conjugate symmetry, it is enough to figure out what $\langle u,iv\rangle$ is to figure out arbitrary scalars. $\endgroup$ Commented Aug 2, 2019 at 18:24

1 Answer 1


This is the "obvious" extension of the usual real inner product on $\mathbb{R}^{2n}$, identified with $\mathbb{C}^n$, to allow for complex scalars.

Explicitly, you can view $\mathbb{C}^n$ as a $2n$-dimensional real vector space by restricting the scalars you allow to be real numbers (just like the complex plane, which is a $1$-dimensional complex vector space, can also be viewed as a $2$-dimensional real vector space if we only allow real scalars, so that now $1$ and $i$ become linearly independent).

Now, if you look at $\mathbb{C}^n$ as a real vector space, then it has a natural real inner product stucture. Namely, you have an inner product $[\cdot\,,\cdot]$ given by $$\Bigl[(a_1+ib_1,\ldots,a_n+ib_n),(c_1+id_1,\ldots,c_n+id_n)\Bigr] = \sum_{j=1}^n (a_jc_j + b_jd_j),\qquad a_k,b_k,c_k,d_k\in\mathbb{R}.$$ With this real inner product (so it is symmetric, not conjugate symmetric, and it is only homogeneous relative to real scalars, not arbitrary complex scalars, we have that for any vector $\mathbf{v}=(a_1+ib_1,\ldots,a_n+ib_n)$, we have $$[\mathbf{v},i\mathbf{v}] = [(a_1+ib_1,\ldots,a_n+ib_n), (-b_1+ia_1,\ldots,-b_n+ia_n)] = \sum_{j=1}^n(-a_jb_j+b_ja_j) = 0.$$ So that multiplication by $i$ does result in a vector that is orthogonal to the original one, just as we might expect from looking at the effect of multiplication by $i$ in the complex plane.

However, this inner product is not a complex inner product, because it is not homogeneous relative to complex scalars and is not conjugate symmetric. However, we can use this inner product to define a complex inner product on $\mathbb{C}^{n}$, $\langle\cdot\,,\cdot\rangle$, as follows: given vectors $\mathbf{x},\mathbf{y}\in\mathbb{C}^n$, define $$\langle \mathbf{x},\mathbf{y}\rangle = [\mathbf{x},\mathbf{y}] + i[\mathbf{x},i\mathbf{y}].$$ It is a good exercise to verify that this is indeed a complex inner product: linear in the first component, positive definite, and conjugate symmetric. It is the natural way to extend the real inner product we obviously have to a complex inner product.

What will this inner product give us in $\mathbb{C}^n$? Well, we get $$\langle (a_1+ib_1,\ldots,a_n+ib_n),(c_1+id_1,\ldots,c_n+id_n)\rangle\qquad\qquad\qquad\qquad\qquad\qquad\\ \begin{align*} \qquad&= [(a_1+ib_1,\ldots,a_n+ib_n),(c_1+id_1,\ldots,c_n+id_n)]\\&\qquad \mathop{+} i[(a_1+ib_1,\ldots,a_n+ib_n),i(c_1+id_1,\ldots,c_n+id_n)]\\ &= \sum_{j=1}^n(a_jc_j + b_jd_j) + i[(a_1+ib_1,\ldots,a_n+ib_n),(-d_1+ic_1,\ldots,-d_n+ic_n)]\\ &= \sum_{j=1}^n(a_jc_j+b_jd_j) + i\sum_{j=1}^n(-a_jd_j + c_jb_j)\\ &= \sum_{j=1}^n\Bigl(a_jc_j + b_jd_j + i(c_jb_j - a_jd_j)\Bigr) \\ &= \sum_{j=1}^n\Bigl( (a_j+ib_j)(c_j-id_j)\Bigr)\\ &= \sum_{j=1}^n (a_j+ib_j)\overline{(c_j+id_j)}, \end{align*}$$ that is, the standard inner product on $\mathbb{C}^n$.

Summary: If you identify the complex plane with $\mathbb{R}^2$ with its usual inner product, which respects our usual notions of "orthogonality" between complex numbers viewed as vectors on the plane, and want to extend that inner product to allow complex scalars so that you get an inner product for the complex vector space, then you are naturally led to the definition of the standard complex inner product. So this inner product does capture our "usual" geometric understanding of "orthogonal"; it's just that allowing complex scalars skews it somewhat.

From the bonus question: again, think of a vector in $\mathbb{C}^n$ as corresponding to a real vector in $\mathbb{R}^{2n}$ in the obvious way. The real part of $\langle \mathbf{x},\mathbf{y}\rangle$ is just their inner product as vectors in $\mathbb{R}^{2n}$, and the imaginary part of $\langle \mathbf{x},\mathbf{y}\rangle$ is the inner product of $\mathbf{x}$ and $i\mathbf{y}$.

The inner product is real if $\mathbf{x}$ is orthogonal to $i\mathbf{y}$, when viewed as vectors in $\mathbb{R}^{2n}$.

  • $\begingroup$ I have updated my question $\endgroup$ Commented Aug 2, 2019 at 18:22

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