Why Induced Norms Are Norms

From an Inner Product to a Length

An inner product assigns a scalar to a pair of vectors. When both inputs are the same vector, the result can be interpreted as the square of its length. This interpretation is not automatic: it relies on the positivity and scalar rules in the inner product axioms. In particular, for a complex vector space the scalar \(\langle\mathbf{v},\mathbf{v}\rangle\) must be real and nonnegative, even though inner products of two different vectors can be complex.

We work over \(\mathbb{F}=\mathbb{R}\) or \(\mathbb{F}=\mathbb{C}\). An inner product on a vector space \(V\) over \(\mathbb{F}\) is a function \(\langle\cdot,\cdot\rangle:V\times V\to\mathbb{F}\) that is linear in its first slot, conjugate-symmetric, and positive definite. Linearity in the first slot means, in particular, \(\langle a\mathbf{u},\mathbf{v}\rangle=a\langle\mathbf{u},\mathbf{v}\rangle\) and \(\langle\mathbf{u}+\mathbf{w},\mathbf{v}\rangle=\langle\mathbf{u},\mathbf{v}\rangle+\langle\mathbf{w},\mathbf{v}\rangle\). Conjugate symmetry means \(\langle\mathbf{u},\mathbf{v}\rangle=\overline{\langle\mathbf{v},\mathbf{u}\rangle}\); over \(\mathbb{R}\), this is ordinary symmetry. Positive definiteness means \(\langle\mathbf{v},\mathbf{v}\rangle\) is real and nonnegative for every \(\mathbf{v}\), and equals zero exactly when \(\mathbf{v}=\mathbf{0}\). For a real scalar, its modulus is its absolute value; for a complex scalar \(a\), \(|a|=\sqrt{a\overline a}\).

The definition of an inner product already tells us that the quantity under the square root is a nonnegative real number. The other properties need attention too. In particular, definiteness uses the “exactly when” part of positive definiteness, while absolute homogeneity uses both slots of the inner product: scaling the first slot contributes \(a\), and scaling the second contributes \(\overline a\).

The Three Properties from the Axioms

Proof. Fix \(\mathbf{v}\in V\). By positive definiteness, \(\langle\mathbf{v},\mathbf{v}\rangle\) is a real number greater than or equal to zero. Its nonnegative square root is therefore real and nonnegative, so \(\|\mathbf{v}\|\geq0\). Also, a nonnegative real number has square root zero exactly when it is zero. Thus \(\|\mathbf{v}\|=0\) if and only if \(\langle\mathbf{v},\mathbf{v}\rangle=0\), which by positive definiteness holds if and only if \(\mathbf{v}=\mathbf{0}\).

Now fix \(a\in\mathbb{F}\). Conjugate symmetry and linearity in the first slot imply conjugate-linearity in the second slot: \(\langle\mathbf{u},a\mathbf{w}\rangle=\overline a\langle\mathbf{u},\mathbf{w}\rangle\). Indeed, \(\langle\mathbf{u},a\mathbf{w}\rangle=\overline{\langle a\mathbf{w},\mathbf{u}\rangle}=\overline{a\langle\mathbf{w},\mathbf{u}\rangle}=\overline a\langle\mathbf{u},\mathbf{w}\rangle\). Consequently, \(\langle a\mathbf{v},a\mathbf{v}\rangle=a\overline a\langle\mathbf{v},\mathbf{v}\rangle=|a|^2\langle\mathbf{v},\mathbf{v}\rangle\). Both sides are nonnegative real numbers. Taking their nonnegative square roots gives \(\|a\mathbf{v}\|=|a|\|\mathbf{v}\|\), since \(|a|\geq0\) and \(\|\mathbf{v}\|\geq0\). This also covers \(a=0\): both sides are zero. \(\square\)

The proof shows why the scalar modulus, rather than the scalar itself, appears in the homogeneity rule. A length cannot acquire a negative or nonreal factor when a vector is rescaled. The product \(a\overline a=|a|^2\) is what ensures that the squared length is multiplied by a nonnegative real number.

For completeness, the zero-vector case also follows directly from the axioms. Linearity gives \(\langle\mathbf{0},\mathbf{0}\rangle=\langle\mathbf{0}+\mathbf{0},\mathbf{0}\rangle=2\langle\mathbf{0},\mathbf{0}\rangle\), so \(\langle\mathbf{0},\mathbf{0}\rangle=0\), and hence \(\|\mathbf{0}\|=0\). This agrees with the theorem’s definiteness statement.

Worked Examples in Real and Complex Spaces

A Common Mistake with Complex Scalars

Pitfall: multiplying a complex vector’s norm by the scalar itself. Let \(\mathbf{z}=(1,0)^\top\in\mathbb{C}^2\), with the standard complex inner product, and take \(a=i\). Since \(\|\mathbf{z}\|=1\), an incorrect use of homogeneity might claim \(\|i\mathbf{z}\|=i\|\mathbf{z}\|=i\). This cannot be a norm value: it is not even real.

The correction is to use the modulus. Here \(i\mathbf{z}=(i,0)^\top\), and its squared norm is \(i\overline{i}+0=1\), so \(\|i\mathbf{z}\|=1\). Also, \(|i|=\sqrt{i\overline{i}}=1\), and the correct formula gives \(|i|\|\mathbf{z}\|=1\cdot1=1\). The inner product calculation explains the correction: scaling both slots produces \(i\overline i\), not \(i\) alone.

A second error is to infer definiteness from nonnegativity alone. A nonnegative function could still be zero at a nonzero vector. For the induced quantity, the “zero only at \(\mathbf{0}\)” conclusion comes specifically from positive definiteness of the inner product. Both parts of that axiom are needed: nonnegativity makes the square root meaningful, and its zero condition establishes definiteness.

What Has Been Established

The calculations in the examples illustrate the general proof, but the result does not depend on coordinates or on a particular formula for an inner product. It applies to every inner product space over \(\mathbb{R}\) or \(\mathbb{C}\), whether finite-dimensional or not. Only the inner product axioms were used: positivity for nonnegativity and definiteness, and linearity together with conjugate symmetry for scalar homogeneity.

These three properties justify treating \(\sqrt{\langle\mathbf{v},\mathbf{v}\rangle}\) as a candidate length. They do not, by themselves, establish every norm axiom: the triangle inequality is a separate requirement. Its proof for induced norms uses a bound on inner products and is developed separately. Keeping these logical steps distinct prevents the mistake of treating a verified subset of the norm properties as a complete proof.