Consequences Of The Inner Product Axioms

What the Inner Product Axioms Imply

An inner product gives a scalar measurement of how two vectors relate. Its axioms impose strong restrictions on that measurement, beyond the rules for adding vectors and scaling them. In particular, the zero vector contributes no inner product, no nonzero vector can be orthogonal to itself, and reversing the order of two complex vectors conjugates the result. These facts are useful checks when computing with inner products on vector, matrix, or function spaces.

We work over \(\mathbb{C}\) and use the convention that the inner product is linear in its first argument and conjugate-linear in its second. Write \(\overline{c}\) for the complex conjugate of a scalar \(c\). A form \(\langle\cdot,\cdot\rangle:V\times V\to\mathbb{C}\) is sesquilinear under this convention if it is additive in each argument and satisfies \(\langle c\mathbf{u},\mathbf{v}\rangle=c\langle\mathbf{u},\mathbf{v}\rangle\) and \(\langle\mathbf{u},c\mathbf{v}\rangle=\overline{c}\langle\mathbf{u},\mathbf{v}\rangle\).

We will derive conjugate symmetry rather than assume it as a separate axiom. This is an equivalent way to present the complex inner product axioms: in the common presentation that includes conjugate symmetry as an axiom, sesquilinearity and positive definiteness are also required, and the diagonal values are then real and positive for nonzero vectors. Here, the reality of the diagonal values lets us prove the symmetry property directly.

The Zero Vector Has Zero Inner Product

Additivity already constrains what happens when an argument is zero. For the first argument, \(\mathbf{0}+\mathbf{0}=\mathbf{0}\), so additivity gives \(\langle\mathbf{0},\mathbf{v}\rangle=\langle\mathbf{0},\mathbf{v}\rangle+\langle\mathbf{0},\mathbf{v}\rangle\). Subtracting the same scalar from both sides shows that this value is zero. The same reasoning in the second argument proves the corresponding result there.

Proof. Additivity in the first argument gives \(\langle\mathbf{0},\mathbf{v}\rangle =\langle\mathbf{0}+\mathbf{0},\mathbf{v}\rangle =\langle\mathbf{0},\mathbf{v}\rangle+\langle\mathbf{0},\mathbf{v}\rangle\). Subtracting \(\langle\mathbf{0},\mathbf{v}\rangle\) yields \(\langle\mathbf{0},\mathbf{v}\rangle=0\). In the second argument, \(\langle\mathbf{u},\mathbf{0}\rangle =\langle\mathbf{u},\mathbf{0}+\mathbf{0}\rangle =\langle\mathbf{u},\mathbf{0}\rangle+\langle\mathbf{u},\mathbf{0}\rangle\). Subtracting the repeated term gives \(\langle\mathbf{u},\mathbf{0}\rangle=0\). \(\square\)

This proof uses additivity, not positive definiteness. In particular, it applies to any sesquilinear form, even one that is not an inner product. Once conjugate symmetry has been established, the two zero-vector identities also follow from one another, but proving both directly makes clear that each follows from the relevant additivity axiom.

Self-Orthogonality Forces the Zero Vector

Two vectors are orthogonal when their inner product is zero. For complex inner products, conjugate symmetry will show that reversing the arguments preserves this condition. Before using that symmetry, positive definiteness already gives an especially important consequence: a vector cannot have zero inner product with itself unless it is the zero vector.

Proof. Positive definiteness states that \(\langle\mathbf{v},\mathbf{v}\rangle=0\) if and only if \(\mathbf{v}=\mathbf{0}\). Applying the forward implication to the given equality gives \(\mathbf{v}=\mathbf{0}\). Conversely, if \(\mathbf{v}=\mathbf{0}\), the zero-vector proposition gives \(\langle\mathbf{v},\mathbf{v}\rangle=\langle\mathbf{0},\mathbf{0}\rangle=0\). Thus the equality holds exactly for the zero vector. \(\square\)

The converse is included because it identifies the full set of vectors whose self-inner product is zero. The strict positivity condition for every nonzero vector is essential: without it, a sesquilinear form might assign zero to the self-inner product of a nonzero vector.

Deriving Conjugate Symmetry

Conjugate symmetry says that the order of the arguments may be reversed at the cost of conjugating the scalar. It is not the same as ordinary symmetry: complex values need not remain unchanged when conjugated. To derive the property, compare the diagonal inner products of \(\mathbf{u}+\mathbf{v}\) and \(\mathbf{u}+i\mathbf{v}\). Both diagonal values must be real.

Proof. Fix \(\mathbf{u},\mathbf{v}\in V\), and set \(a=\langle\mathbf{u},\mathbf{v}\rangle\) and \(b=\langle\mathbf{v},\mathbf{u}\rangle\). By sesquilinearity, expansion of the self-inner product of the sum gives \[ \langle\mathbf{u}+\mathbf{v},\mathbf{u}+\mathbf{v}\rangle =\langle\mathbf{u},\mathbf{u}\rangle+a+b+\langle\mathbf{v},\mathbf{v}\rangle. \] The left side and the two diagonal terms are real. Therefore \(a+b\) is real.

Next expand the self-inner product of \(\mathbf{u}+i\mathbf{v}\). Linearity in the first argument gives the factor \(i\) in \(\langle i\mathbf{v},\mathbf{u}\rangle\), while conjugate-linearity in the second gives \(\overline{i}=-i\) in \(\langle\mathbf{u},i\mathbf{v}\rangle\). Also, \(\langle i\mathbf{v},i\mathbf{v}\rangle=i\overline{i}\langle\mathbf{v},\mathbf{v}\rangle=\langle\mathbf{v},\mathbf{v}\rangle\). Hence \[ \langle\mathbf{u}+i\mathbf{v},\mathbf{u}+i\mathbf{v}\rangle =\langle\mathbf{u},\mathbf{u}\rangle+\langle\mathbf{v},\mathbf{v}\rangle+i(b-a). \] The diagonal terms and the left side are real, so \(i(b-a)\) is real.

Write \(a=x+iy\) and \(b=s+it\), with \(x,y,s,t\in\mathbb{R}\). Since \(a+b\) is real, its imaginary part gives \(y+t=0\). Further, \[ i(b-a)=i\big((s-x)+i(t-y)\big)=-(t-y)+i(s-x). \] Since this is real, its imaginary part gives \(s-x=0\). Thus \(s=x\) and \(t=-y\), so \(b=x-iy=\overline{a}\). By the definitions of \(a\) and \(b\), this is exactly \(\langle\mathbf{v},\mathbf{u}\rangle=\overline{\langle\mathbf{u},\mathbf{v}\rangle}\). \(\square\)

This proof accounts for both real and imaginary parts: the sum \(\mathbf{u}+\mathbf{v}\) determines the real-part relation, and the sum \(\mathbf{u}+i\mathbf{v}\) determines the imaginary-part relation. Positivity is used through the axiom that every diagonal inner product is real; no assumption that the off-diagonal inner products are real is made.

A Common Mistake: Scaling the Second Argument

The most frequent error under the linear-in-the-first-slot convention is to treat the second argument as linear rather than conjugate-linear. The wrong calculation is \(\langle\mathbf{1},i\mathbf{1}\rangle=i\langle\mathbf{1},\mathbf{1}\rangle=i\), where \(\mathbf{1}\) denotes the scalar \(1\) in \(\mathbb{C}\) with the standard inner product \(\langle z,w\rangle=z\overline{w}\). The correction uses the conjugate of the scalar in the second slot:

Direct substitution verifies the correction, and \(\langle i,1\rangle=i\overline{1}=i=\overline{-i}\) checks conjugate symmetry in this case. The sign is not a cosmetic detail: using \(i\) instead of \(-i\) can invalidate calculations involving complex cross terms, including the proof of conjugate symmetry above.

The three consequences provide quick consistency checks. A calculation with a zero argument must return zero; a positive-definite inner product cannot make a nonzero vector self-orthogonal; and reversing the arguments must conjugate the result. These checks apply to inner products on complex coordinate spaces and to the matrix and function settings introduced earlier, provided the stated inner product axioms hold.