What Equality Means Over the Complex Numbers
The complex Cauchy–Schwarz inequality says that the magnitude of an inner product is no greater than the product of the norms. Its equality case is more specific: equality occurs precisely when the two vectors point along the same complex one-dimensional direction. Over \(\mathbb{C}\), that may involve multiplication by a nonreal scalar, so equality does not require the vectors to be identical or real multiples of one another.
Let \(V\) be a vector space over \(\mathbb{C}\) with inner product \(\langle\cdot,\cdot\rangle\), linear in the first slot and conjugate-linear in the second. Its induced norm is \(\|\mathbf{x}\|=\sqrt{\langle\mathbf{x},\mathbf{x}\rangle}\). We use the complex Cauchy–Schwarz inequality, established in the earlier material: for \(\mathbf{u},\mathbf{v}\in V\), \(|\langle\mathbf{u},\mathbf{v}\rangle|\leq\|\mathbf{u}\|\|\mathbf{v}\|\). Equality means that the two nonnegative real quantities on the two sides are equal.
The two alternatives in the definition matter when a vector is zero. If \(\mathbf{v}=\mathbf{0}\) and \(\mathbf{u}\ne\mathbf{0}\), then \(\mathbf{u}\) cannot be written as a scalar multiple of \(\mathbf{v}\), but \(\mathbf{v}=0\mathbf{u}\). The symmetric wording captures this case as well as the case where both vectors are nonzero. When both vectors are nonzero, either alternative implies that each is a scalar multiple of the other.
The Remainder After Removing a Multiple
For \(\mathbf{v}\ne\mathbf{0}\), we can choose a particular multiple of \(\mathbf{v}\) that captures the inner product with \(\mathbf{u}\). Define \(c=\langle\mathbf{u},\mathbf{v}\rangle/\|\mathbf{v}\|^2\), and let \(\mathbf{w}=\mathbf{u}-c\mathbf{v}\). The denominator is nonzero because positive definiteness of the inner product gives \(\|\mathbf{v}\|^2>0\). This choice makes the remainder satisfy \(\langle\mathbf{w},\mathbf{v}\rangle=0\).
Proof. By linearity in the first slot and the definition of \(c\), \(\langle\mathbf{w},\mathbf{v}\rangle=\langle\mathbf{u},\mathbf{v}\rangle-c\langle\mathbf{v},\mathbf{v}\rangle=\langle\mathbf{u},\mathbf{v}\rangle-c\|\mathbf{v}\|^2=0\). Since \(\mathbf{u}=c\mathbf{v}+\mathbf{w}\), expand the inner product defining its squared norm. The term \(\langle c\mathbf{v},c\mathbf{v}\rangle\) is \(|c|^2\|\mathbf{v}\|^2\). The cross terms vanish: \(\langle\mathbf{w},c\mathbf{v}\rangle=\overline{c}\langle\mathbf{w},\mathbf{v}\rangle=0\), and \(\langle c\mathbf{v},\mathbf{w}\rangle=c\langle\mathbf{v},\mathbf{w}\rangle=0\), because \(\langle\mathbf{v},\mathbf{w}\rangle=\overline{\langle\mathbf{w},\mathbf{v}\rangle}=0\). The remaining term is \(\|\mathbf{w}\|^2\). Therefore \(\|\mathbf{u}\|^2=|c|^2\|\mathbf{v}\|^2+\|\mathbf{w}\|^2\). \(\square\)
This identity gives a direct way to distinguish equality from strict inequality. The chosen multiple accounts for the part of \(\mathbf{u}\) that contributes to its inner product with \(\mathbf{v}\); any nonzero remainder contributes to \(\|\mathbf{u}\|\) but not to that inner product. The proposition makes this distinction exact without requiring \(V\) to be finite-dimensional.
Proving the Equality Criterion
Proof. First suppose equality holds. If \(\mathbf{v}=\mathbf{0}\), then \(\mathbf{v}=0\mathbf{u}\), as required. If \(\mathbf{v}\ne\mathbf{0}\), define \(c\) and \(\mathbf{w}\) as in the proposition. Then \(|\langle\mathbf{u},\mathbf{v}\rangle|=|c|\|\mathbf{v}\|^2\). Squaring the assumed equality and using the proposition gives \(|c|^2\|\mathbf{v}\|^4=\|\mathbf{u}\|^2\|\mathbf{v}\|^2=(|c|^2\|\mathbf{v}\|^2+\|\mathbf{w}\|^2)\|\mathbf{v}\|^2\). Cancel the common term \(|c|^2\|\mathbf{v}\|^4\). The result is \(\|\mathbf{w}\|^2\|\mathbf{v}\|^2=0\). Since \(\|\mathbf{v}\|^2>0\), we have \(\|\mathbf{w}\|^2=0\); positive definiteness gives \(\mathbf{w}=\mathbf{0}\). Thus \(\mathbf{u}=c\mathbf{v}\). Conversely, suppose first that \(\mathbf{u}=c\mathbf{v}\). Then \(\langle\mathbf{u},\mathbf{v}\rangle=c\|\mathbf{v}\|^2\) and \(\|\mathbf{u}\|=|c|\|\mathbf{v}\|\). Taking magnitudes gives \(|\langle\mathbf{u},\mathbf{v}\rangle|=|c|\|\mathbf{v}\|^2=\|\mathbf{u}\|\|\mathbf{v}\|\), including \(c=0\) or \(\mathbf{v}=\mathbf{0}\). If instead \(\mathbf{v}=c\mathbf{u}\), conjugate-linearity in the second slot gives \(\langle\mathbf{u},\mathbf{v}\rangle=\overline{c}\|\mathbf{u}\|^2\), while \(\|\mathbf{v}\|=|c|\|\mathbf{u}\|\). Hence \(|\langle\mathbf{u},\mathbf{v}\rangle|=|c|\|\mathbf{u}\|^2=\|\mathbf{u}\|\|\mathbf{v}\|\). Both directions are proved. \(\square\)
When both vectors are nonzero and \(\mathbf{u}=c\mathbf{v}\), the scalar \(c\) is nonzero. The ratio \(|c|\) controls their relative lengths, while \(c/|c|\) has magnitude one and controls their complex phase. The next result states this normalized form precisely.
Proof. If equality holds, the theorem gives \(\mathbf{u}=c\mathbf{v}\). Since both vectors are nonzero, \(c\ne0\), and \(\|\mathbf{u}\|=|c|\|\mathbf{v}\|\). Set \(\alpha=c/|c|\). Then \(|\alpha|=1\), and \(\mathbf{u}/\|\mathbf{u}\|=c\mathbf{v}/(|c|\|\mathbf{v}\|)=\alpha\mathbf{v}/\|\mathbf{v}\|\). Conversely, suppose the displayed normalized relation holds. Multiplying by \(\|\mathbf{u}\|\) shows that \(\mathbf{u}=\alpha(\|\mathbf{u}\|/\|\mathbf{v}\|)\mathbf{v}\), so \(\mathbf{u}\) is a complex scalar multiple of \(\mathbf{v}\). The theorem then gives equality. \(\square\)
Worked Examples in \(\mathbb{C}^2\)
Worked Example: Equality with a Nonreal Scalar
Use the standard inner product \(\langle\mathbf{u},\mathbf{v}\rangle=\sum_j u_j\overline{v_j}\) on \(\mathbb{C}^2\). Let \(\mathbf{v}=\begin{bmatrix}1\\i\end{bmatrix}\) and \(\mathbf{u}=\begin{bmatrix}1+i\\-1+i\end{bmatrix}\). Check whether equality holds.
The vectors satisfy \(\mathbf{u}=(1+i)\mathbf{v}\), since \((1+i)i=-1+i\). Directly calculate the inner product, remembering to conjugate the entries of the second vector:
Also, \(\|\mathbf{u}\|^2=|1+i|^2+|-1+i|^2=2+2=4\), so \(\|\mathbf{u}\|=2\). And \(\|\mathbf{v}\|^2=|1|^2+|i|^2=1+1=2\), so \(\|\mathbf{v}\|=\sqrt{2}\). Finally, \(|2+2i|=\sqrt{2^2+2^2}=2\sqrt{2}\), while \(\|\mathbf{u}\|\|\mathbf{v}\|=2\sqrt{2}\). Equality holds, as predicted by the theorem.
Worked Example: A Nonzero Remainder Gives Strict Inequality
Take \(\mathbf{u}=\begin{bmatrix}1\\i\end{bmatrix}\) and \(\mathbf{v}=\begin{bmatrix}1\\1\end{bmatrix}\) in \(\mathbb{C}^2\). Check the inequality and use the remainder from the proposition to account for the strictness.
The inner product and norms are \(\langle\mathbf{u},\mathbf{v}\rangle=1+i\), \(\|\mathbf{u}\|^2=2\), and \(\|\mathbf{v}\|^2=2\). Therefore \(|\langle\mathbf{u},\mathbf{v}\rangle|=\sqrt{2}\), while \(\|\mathbf{u}\|\|\mathbf{v}\|=\sqrt{2}\sqrt{2}=2\), so the inequality is strict.
For the remainder calculation, \(c=(1+i)/2\), and hence \(\mathbf{w}=\mathbf{u}-c\mathbf{v}=\begin{bmatrix}1-(1+i)/2\\i-(1+i)/2\end{bmatrix}=\begin{bmatrix}(1-i)/2\\(i-1)/2\end{bmatrix}\). Each entry has squared magnitude \(2/4=1/2\), so \(\|\mathbf{w}\|^2=1\). The proposition checks numerically: \(|c|^2\|\mathbf{v}\|^2+\|\mathbf{w}\|^2=(1/2)(2)+1=2=\|\mathbf{u}\|^2\). The nonzero remainder is why equality does not hold.
Worked Example: Equality When One Vector Is Zero
Let \(\mathbf{u}=\begin{bmatrix}1\\2i\end{bmatrix}\) and \(\mathbf{v}=\begin{bmatrix}0\\0\end{bmatrix}\). The inner product is \(\langle\mathbf{u},\mathbf{v}\rangle=1\overline{0}+2i\overline{0}=0\). Also, \(\|\mathbf{v}\|=0\), so \(|\langle\mathbf{u},\mathbf{v}\rangle|=0=\|\mathbf{u}\|\|\mathbf{v}\|\). Equality holds, even though \(\mathbf{u}\) is not a scalar multiple of \(\mathbf{v}\). The correct multiple relation is \(\mathbf{v}=0\mathbf{u}\).
Check: both sides of the inequality are zero, and the symmetric scalar-multiple condition is satisfied.
Common Pitfall: Forgetting the Conjugate
For the standard complex inner product, the second vector's entries must be conjugated. Omitting conjugation changes the calculation and can give a false equality conclusion. For example, take \(\mathbf{u}=\begin{bmatrix}i\\1\end{bmatrix}\) and \(\mathbf{v}=\begin{bmatrix}1\\i\end{bmatrix}\). The incorrect bilinear calculation without conjugation gives \(i\cdot1+1\cdot i=2i\), whose magnitude is \(2\). Since both norms are \(\sqrt{2}\), that wrong calculation appears to give equality.
The correct calculation is \(\langle\mathbf{u},\mathbf{v}\rangle=i\overline{1}+1\overline{i}=i-i=0\). Thus the left side is \(0\), whereas the product of the norms is \(\sqrt{2}\sqrt{2}=2\); the inequality is strict. The vectors are not scalar multiples: if \(\mathbf{u}=c\mathbf{v}\), the first coordinates force \(c=i\), but then the second coordinate would be \(i^2=-1\), not \(1\). Correct conjugation and the scalar-multiple test agree.
A second useful check is that equality concerns the magnitude of the inner product, not necessarily the inner product itself. In the first worked example, the inner product is \(2+2i\), a nonreal number, while the product of the norms is the real number \(2\sqrt{2}\). Their magnitudes agree, exactly as the inequality states.
Check Your Understanding
Use the complex inner product convention that is linear in the first slot, and justify each equality decision.
- For \(\mathbf{u}=\begin{bmatrix}2i\\-2\end{bmatrix}\) and \(\mathbf{v}=\begin{bmatrix}1\\i\end{bmatrix}\), identify the complex scalar relating the vectors and verify equality by calculating both sides.
- Let \(\mathbf{u}=\begin{bmatrix}1\\0\end{bmatrix}\) and \(\mathbf{v}=\begin{bmatrix}1\\1\end{bmatrix}\). Calculate the inner product and norms, then decide whether equality holds.
- Explain why the equality criterion must include a separate scalar-multiple direction when \(\mathbf{v}=\mathbf{0}\) and \(\mathbf{u}\ne\mathbf{0}\).
- Assuming both vectors are nonzero and equality holds, show how to obtain a scalar of magnitude one relating their normalized vectors.
- For \(\mathbf{u}=\begin{bmatrix}1\\i\end{bmatrix}\) and \(\mathbf{v}=\begin{bmatrix}1\\-i\end{bmatrix}\), compute the standard complex inner product with the conjugation shown explicitly, then test equality.
