The Standard Inner Product On C^n

Complex Coordinates Require Conjugation

For real vectors, the standard dot product multiplies corresponding coordinates and adds the results. Applying that same rule without change to complex coordinates does not give a suitable measure of squared length: for instance, multiplying \(i\) by \(i\) gives \(-1\), not a nonnegative number. Complex conjugation fixes this problem. It also determines how scalars behave in each argument, so it is important to keep the convention consistent throughout a calculation.

We use the course convention that an inner product on a complex vector space is linear in its first argument and conjugate-linear in its second. For \(z=a+bi\), where \(a,b\in\mathbb{R}\), its complex conjugate is \(\overline{z}=a-bi\), and its squared modulus is \(|z|^2=z\overline{z}=a^2+b^2\). Vectors in \(\mathbb{C}^n\) are columns, and scalar multiplication and addition are coordinatewise.

With this convention, it is the entries of the second vector that are conjugated in \(\langle\mathbf{u},\mathbf{v}\rangle\). In particular, the scalar rules are \(\langle a\mathbf{u}+b\mathbf{w},\mathbf{v}\rangle=a\langle\mathbf{u},\mathbf{v}\rangle+b\langle\mathbf{w},\mathbf{v}\rangle\) and \(\langle\mathbf{u},a\mathbf{v}+b\mathbf{w}\rangle=\overline{a}\langle\mathbf{u},\mathbf{v}\rangle+\overline{b}\langle\mathbf{u},\mathbf{w}\rangle\), for \(a,b\in\mathbb{C}\). Some texts instead define the conjugate-linear argument first; that is a different convention. Here, using conjugation in the first argument would not match the linear-first convention.

Checking the Inner Product Axioms

To verify that the formula really defines an inner product, check its scalar rules, conjugate symmetry, and positive definiteness. Positive definiteness means that \(\langle\mathbf{u},\mathbf{u}\rangle\) is a nonnegative real number and equals zero exactly when \(\mathbf{u}=\mathbf{0}\).

Proof. Applying the definition coordinate by coordinate gives \(\langle a\mathbf{u}+b\mathbf{w},\mathbf{v}\rangle=\sum_{j=1}^n(au_j+bw_j)\overline{v_j}=a\sum_{j=1}^n u_j\overline{v_j}+b\sum_{j=1}^n w_j\overline{v_j}=a\langle\mathbf{u},\mathbf{v}\rangle+b\langle\mathbf{w},\mathbf{v}\rangle\). For the second argument, \(\overline{av_j+bw_j}=\overline{a}\,\overline{v_j}+\overline{b}\,\overline{w_j}\). Therefore \(\langle\mathbf{u},a\mathbf{v}+b\mathbf{w}\rangle=\overline{a}\langle\mathbf{u},\mathbf{v}\rangle+\overline{b}\langle\mathbf{u},\mathbf{w}\rangle\). Also, \(\overline{\langle\mathbf{v},\mathbf{u}\rangle}=\overline{\sum_{j=1}^n v_j\overline{u_j}}=\sum_{j=1}^n\overline{v_j}u_j=\sum_{j=1}^n u_j\overline{v_j}=\langle\mathbf{u},\mathbf{v}\rangle\). Finally, \(\langle\mathbf{u},\mathbf{u}\rangle=\sum_{j=1}^n u_j\overline{u_j}=\sum_{j=1}^n |u_j|^2\). Each term is a nonnegative real number. If the sum is zero, every \(|u_j|^2=0\), so every \(u_j=0\) and \(\mathbf{u}=\mathbf{0}\). If \(\mathbf{u}=\mathbf{0}\), the sum is zero. This proves all the required properties. \(\square\)

The theorem gives a useful calculation rule: a squared length is found by multiplying each coordinate by its own conjugate and adding. If \(u_j=a_j+b_ji\), its contribution is \(u_j\overline{u_j}=a_j^2+b_j^2\). Thus the result is real and cannot be negative, even though a general inner product \(\langle\mathbf{u},\mathbf{v}\rangle\) may be complex.

Computing Inner Products and Squared Lengths

It is useful to keep the scalar rules visible when a complex scalar is moved into an argument. The next calculation checks both rules on the same vectors, rather than relying on a verbal shortcut.

The Standard Coordinate Vectors

The standard coordinate vectors \(\mathbf{e}_1,\dots,\mathbf{e}_n\) are the columns whose coordinates are all zero except for a \(1\) in the indicated position. The standard inner product makes each of these vectors have squared length \(1\), and distinct coordinate vectors have inner product \(0\).

Proof. In the standard inner product, the \(r\)th coordinate of \(\mathbf{e}_j\) is \(1\) when \(r=j\) and \(0\) otherwise. Thus the sum \(\langle\mathbf{e}_j,\mathbf{e}_k\rangle=\sum_{r=1}^n(\mathbf{e}_j)_r\overline{(\mathbf{e}_k)_r}\) has a nonzero term exactly when \(r=j=k\), and that term equals \(1\). This proves the first claim. For the uniqueness claim, write arbitrary vectors as \(\mathbf{u}=\sum_{j=1}^n u_j\mathbf{e}_j\) and \(\mathbf{v}=\sum_{k=1}^n v_k\mathbf{e}_k\). Linearity in the first argument and conjugate-linearity in the second give \(\langle\mathbf{u},\mathbf{v}\rangle=\sum_{j=1}^n\sum_{k=1}^n u_j\overline{v_k}\langle\mathbf{e}_j,\mathbf{e}_k\rangle\). All terms with \(j\ne k\) vanish, and each term with \(j=k\) has \(\langle\mathbf{e}_j,\mathbf{e}_j\rangle=1\). The expression is therefore \(\sum_{j=1}^n u_j\overline{v_j}\), which is the standard formula. \(\square\)

This characterization explains why the coordinate formula is natural: once the scalar convention is fixed, requiring the coordinate vectors to have unit squared length and to be mutually orthogonal determines every inner product. It also gives a quick check on calculations: a coordinate vector contributes only in its own coordinate, so mixing two distinct coordinate directions produces zero.

Common Mistakes and a Reliable Routine

Mistake: omit conjugation. For the one-coordinate vector \(\mathbf{u}=(i)^{\top}\), an incorrect calculation without conjugation gives \(i\cdot i=i^2=-1\) as its squared length. That cannot satisfy positive definiteness. The correction is \(\langle\mathbf{u},\mathbf{u}\rangle=i\overline{i}=i(-i)=-i^2=1\).

Mistake: conjugate the wrong argument for this convention. The formula \(\sum_j\overline{u_j}v_j\) is used in the alternative convention that is conjugate-linear in the first argument. It is not the formula used here. For example, with \(\mathbf{u}=(i)^{\top}\) and \(\mathbf{v}=(1)^{\top}\), this tutorial gives \(\langle\mathbf{u},\mathbf{v}\rangle=i\), whereas conjugating the first argument gives \(-i\). Neither convention is inherently invalid, but the formula and scalar rules must be kept together; switching between them mid-calculation changes the answer.

For a direct calculation, write the formula before multiplying, conjugate every coordinate of the second vector, and simplify each product before adding. For a squared length, use the same rule with the vector in both slots; each product then becomes a squared modulus. This makes it straightforward to check that the final value is real and nonnegative.