Why the Slot Convention Matters
Over \(\mathbb{R}\), the familiar dot product does not force us to think about which argument is linear: there is no difference between a scalar and its complex conjugate. Over \(\mathbb{C}\), that distinction matters. An inner product must conjugate one of its arguments, and authors choose either the first or the second argument to be conjugate-linear. Both conventions are consistent, but scalar factors must be handled according to the convention in use.
In this tutorial, inner products on complex vector spaces are linear in the first argument. The second argument is therefore conjugate-linear. We will state the rules carefully and use them for vectors in \(\mathbb{C}^n\). A vector in \(\mathbb{C}^n\) is a column, and its coordinates may be complex numbers.
Here \(\overline{z}\) denotes the complex conjugate of \(z\). Property (1) says that sums and scalars can be pulled out of the first slot without alteration. Property (2) is conjugate symmetry. Together, these two properties determine how sums and scalars behave in the second slot; that behavior is not an additional choice.
Proof. Apply conjugate symmetry, then linearity in the first argument, then conjugate symmetry to each resulting inner product:
Taking \(b=0\) gives \(\langle\mathbf{u},a\mathbf{v}\rangle=\overline{a}\langle\mathbf{u},\mathbf{v}\rangle\). Then linearity in the first argument gives \(\langle a\mathbf{u},b\mathbf{v}\rangle=a\langle\mathbf{u},b\mathbf{v}\rangle=a\overline{b}\langle\mathbf{u},\mathbf{v}\rangle\). This proves both formulas. \(\square\)
The Standard Inner Product on \(\mathbb{C}^n\)
For column vectors \(\mathbf{u}=(u_1,\dots,u_n)^{\top}\) and \(\mathbf{v}=(v_1,\dots,v_n)^{\top}\), the standard inner product under our linear-first convention is:
The conjugate is on the coordinates of the second vector. This formula satisfies the inner product definition. It is linear in \(\mathbf{u}\) because each coordinate \(u_j\) occurs without conjugation. Swapping the vectors and conjugating gives the same sum, so it has conjugate symmetry. Finally, setting the two vectors equal gives \(\sum_{j=1}^n |u_j|^2\), a nonnegative real number that is zero precisely when every coordinate is zero.
The norm does not depend on which of the two consistent conventions is chosen: conjugating the whole inner product does not change \(\langle\mathbf{u},\mathbf{u}\rangle\). Orthogonality is also unchanged, since a complex number is zero exactly when its conjugate is zero. The actual inner product value, however, may be conjugated when the convention changes.
Worked Example: Computing an Inner Product and Norms
Let \(\mathbf{u}=(1+i,\,2-i)^{\top}\) and \(\mathbf{v}=(2-i,\,1+i)^{\top}\). Compute \(\langle\mathbf{u},\mathbf{v}\rangle\), \(\|\mathbf{u}\|\), and \(\|\mathbf{v}\|\).
Conjugate the coordinates of the second vector before multiplying coordinate by coordinate:
For the norms, square the magnitude of each coordinate and add. Since \(|1+i|^2=1^2+1^2=2\) and \(|2-i|^2=2^2+(-1)^2=5\), we get \(\|\mathbf{u}\|^2=2+5=7\), so \(\|\mathbf{u}\|=\sqrt{7}\). The coordinates of \(\mathbf{v}\) have the same magnitudes in the opposite order, so \(\|\mathbf{v}\|^2=5+2=7\), giving \(\|\mathbf{v}\|=\sqrt{7}\).
Check: \(\langle\mathbf{v},\mathbf{u}\rangle\) is the conjugate of \(\langle\mathbf{u},\mathbf{v}\rangle=2\), so it is also \(2\). The norm calculations are sums of squared magnitudes and are nonnegative real numbers, as required.
Consequences for Length and Orthogonality
The axioms give useful checks on computations. The scalar rule for a vector paired with itself is \(\langle a\mathbf{u},a\mathbf{u}\rangle=|a|^2\langle\mathbf{u},\mathbf{u}\rangle\), not \(a^2\langle\mathbf{u},\mathbf{u}\rangle\). Consequently, multiplying a vector by a complex scalar changes its norm by the scalar's magnitude.
Proof. By the two-slot scalar rule, \(\langle a\mathbf{u},a\mathbf{u}\rangle=a\overline{a}\langle\mathbf{u},\mathbf{u}\rangle=|a|^2\|\mathbf{u}\|^2\). Both sides are nonnegative real numbers. Taking their nonnegative square roots gives \(\|a\mathbf{u}\|=|a|\|\mathbf{u}\|\). For orthogonality, conjugate symmetry gives \(\langle\mathbf{v},\mathbf{u}\rangle=\overline{\langle\mathbf{u},\mathbf{v}\rangle}\). Thus if either inner product is zero, its conjugate is zero as well, proving both directions. \(\square\)
Worked Example: Checking Orthogonality in \(\mathbb{C}^2\)
Determine whether \(\mathbf{u}=(1,i)^{\top}\) and \(\mathbf{v}=(i,1)^{\top}\) are orthogonal, and find their norms.
Conjugating the second vector's coordinates gives \((-i,1)^{\top}\). Therefore
Hence \(\mathbf{u}\perp\mathbf{v}\). For each vector, the squared magnitudes of its coordinates are \(1\) and \(1\), so \(\|\mathbf{u}\|^2=2\), \(\|\mathbf{v}\|^2=2\), and both norms equal \(\sqrt{2}\).
Check: Reversing the order gives \(\langle\mathbf{v},\mathbf{u}\rangle=i\cdot1+1\cdot(-i)=i-i=0\). This agrees with the theorem that orthogonality is symmetric.
Comparing the Two Common Conventions
Some texts define the standard complex inner product to be conjugate-linear in the first argument and linear in the second. Under that convention the coordinate formula is \(\sum_{j=1}^n\overline{u_j}v_j\). It is the conjugate of the linear-first value \(\sum_{j=1}^n u_j\overline{v_j}\). This difference does not affect norms or the question of whether an inner product is zero, but it does affect the result of a general computation and the scalar rules.
| Feature | Linear in the first argument | Conjugate-linear in the first argument |
|---|---|---|
| Coordinate formula | \(\sum u_j\overline{v_j}\) | \(\sum\overline{u_j}v_j\) |
| Scalar in first slot | \(a\langle\mathbf{u},\mathbf{v}\rangle\) | \(\overline{a}\langle\mathbf{u},\mathbf{v}\rangle\) |
| Scalar in second slot | \(\overline{a}\langle\mathbf{u},\mathbf{v}\rangle\) | \(a\langle\mathbf{u},\mathbf{v}\rangle\) |
| Norm and orthogonality | Unchanged | Unchanged |
Worked Example: Keeping Track of Scalar Placement
Use the linear-first convention with \(\mathbf{u}=(1+i)^{\top}\), \(\mathbf{v}=(2-i)^{\top}\), \(a=1+i\), and \(b=-i\). Compute \(\langle a\mathbf{u},b\mathbf{v}\rangle\) directly and by the scalar rule.
First, \(a\mathbf{u}=(1+i)(1+i)=2i\), and \(b\mathbf{v}=(-i)(2-i)=-2i+i^2=-1-2i\). Direct calculation gives
Now \(\langle\mathbf{u},\mathbf{v}\rangle=(1+i)\overline{(2-i)}=(1+i)(2+i)=1+3i\), while \(\overline{b}=i\). The scalar rule gives
Check: the direct value and the scalar-rule value are both \(-4-2i\). The scalar \(a\) comes out unchanged because the inner product is linear in its first slot, and the scalar \(b\) comes out conjugated, as \(\overline{b}\), because the inner product is conjugate-linear in its second slot.
Common Pitfall: Forgetting the Conjugate
A frequent error is to multiply matching coordinates without conjugating the second vector. For example, using \(\mathbf{u}=(1,i)^{\top}\) and \(\mathbf{v}=(i,1)^{\top}\), the incorrect calculation \(1\cdot i+i\cdot1=2i\) omits the conjugates. The correct calculation is \(1\cdot\overline{i}+i\cdot\overline{1}=-i+i=0\). The wrong result would incorrectly say the vectors are not orthogonal.
The same omission can even produce an impossible squared length. If one incorrectly computes \(\langle\mathbf{u},\mathbf{u}\rangle\) for \(\mathbf{u}=(1,i)^{\top}\) without conjugation, the result is \(1^2+i^2=0\), despite \(\mathbf{u}\ne\mathbf{0}\). The correct value is \(1\cdot1+i\cdot\overline{i}=1+1=2\), so \(\|\mathbf{u}\|=\sqrt{2}\). A squared norm must be a nonnegative real number and can be zero only for the zero vector; use these facts as checks.
When working with a complex inner product, first identify which argument is linear. Then conjugate the appropriate coordinates, apply scalar factors according to their slot, and only afterward simplify. Under the convention used here, the reliable pattern is: conjugate the second argument, multiply corresponding coordinates, and add.
Check Your Understanding
Use the linear-first convention throughout. Show enough intermediate work to make the conjugations and scalar placement explicit.
- Compute \(\langle(1-i,\,2)^{\top},(i,\,1+i)^{\top}\rangle\).
- Find the norm of \((2i,\,1-i,\,3)^{\top}\).
- Determine whether \((1,\,i)^{\top}\) and \((2,\,2i)^{\top}\) are orthogonal.
- For arbitrary \(\mathbf{u},\mathbf{v}\) and \(a,b\in\mathbb{C}\), write the correct formula for \(\langle a\mathbf{u},b\mathbf{v}\rangle\) under the linear-first convention.
- Explain why switching to the conjugate-first convention changes an inner product value only by complex conjugation, and why it leaves norms and orthogonality unchanged.
