The Complex Inequality and Its Setting
Over the real numbers, the Cauchy–Schwarz inequality compares an inner product with the lengths of its two inputs. Over the complex numbers, the same bound holds, but an inner product need not itself be real: it may have both a magnitude and a phase. The proof must account for that phase before using the real-valued idea of minimizing a quadratic. A suitable unit complex scalar removes the phase, after which nonnegativity of a squared norm supplies the bound.
Let \(V\) be a complex inner product space. We use the convention that the inner product is linear in its first slot and conjugate-linear in its second: for \(\mathbf{u},\mathbf{v},\mathbf{w}\in V\) and \(a,b\in\mathbb{C}\), \(\langle a\mathbf{u}+b\mathbf{v},\mathbf{w}\rangle=a\langle\mathbf{u},\mathbf{w}\rangle+b\langle\mathbf{v},\mathbf{w}\rangle\), while \(\langle\mathbf{u},a\mathbf{v}+b\mathbf{w}\rangle=\overline{a}\langle\mathbf{u},\mathbf{v}\rangle+\overline{b}\langle\mathbf{u},\mathbf{w}\rangle\). It also satisfies conjugate symmetry, \(\langle\mathbf{v},\mathbf{u}\rangle=\overline{\langle\mathbf{u},\mathbf{v}\rangle}\), and positive definiteness: \(\langle\mathbf{u},\mathbf{u}\rangle\) is real and nonnegative, and equals zero exactly when \(\mathbf{u}=\mathbf{0}\). The induced norm is \(\|\mathbf{u}\|=\sqrt{\langle\mathbf{u},\mathbf{u}\rangle}\).
The absolute value is essential: the inequality compares nonnegative real quantities, while \(\langle\mathbf{u},\mathbf{v}\rangle\) may be complex. The proof will use two facts. First, multiplying one vector by a scalar of magnitude one can adjust the phase of the inner product without changing the vector’s norm. Second, every squared norm is nonnegative.
Choosing a Phase
Proof. Since \(z\ne0\), its magnitude \(|z|\) is positive. Thus \(|\eta|=|z|/|z|=1\). Also, \(\overline{\eta}=\overline{z}/|z|\), so \(\overline{\eta}z=\overline{z}z/|z|=|z|^2/|z|=|z|\). \(\square\)
Apply the lemma to \(z=\langle\mathbf{u},\mathbf{v}\rangle\), when this inner product is nonzero, and set \(\mathbf{w}=\eta\mathbf{v}\). Because the inner product is conjugate-linear in its second slot, we get \(\langle\mathbf{u},\mathbf{w}\rangle=\overline{\eta}\langle\mathbf{u},\mathbf{v}\rangle=|z|\). This is a nonnegative real number. The phase adjustment does not change the norm: \(\|\mathbf{w}\|^2=\langle\eta\mathbf{v},\eta\mathbf{v}\rangle=\eta\overline{\eta}\langle\mathbf{v},\mathbf{v}\rangle=\|\mathbf{v}\|^2\).
Minimizing a Squared Norm
For any real number \(t\), expand \(\|\mathbf{u}-t\mathbf{w}\|^2\) using linearity in the first slot and conjugate-linearity in the second. Because \(t\) is real, its conjugate is also \(t\), so both cross terms have the same real value:
Here \(\langle\mathbf{w},\mathbf{u}\rangle=\overline{\langle\mathbf{u},\mathbf{w}\rangle}=|z|\), and \(\|\mathbf{w}\|=\|\mathbf{v}\|\). If \(\mathbf{v}\ne\mathbf{0}\), then \(\|\mathbf{v}\|^2>0\). The quadratic expression in \(t\) is minimized at \(t=|z|/\|\mathbf{v}\|^2\). Substitution into the expression gives \(\|\mathbf{u}\|^2-|z|^2/\|\mathbf{v}\|^2\). Since this is a squared norm, it is nonnegative; multiplying by the positive number \(\|\mathbf{v}\|^2\) yields \(|z|^2\leq\|\mathbf{u}\|^2\|\mathbf{v}\|^2\).
Proof of the theorem. Set \(z=\langle\mathbf{u},\mathbf{v}\rangle\). If \(\mathbf{v}=\mathbf{0}\), then \(z=0\) by linearity, and both sides of the claimed inequality are zero. Suppose \(\mathbf{v}\ne\mathbf{0}\). If \(z=0\), then \(|z|=0\leq\|\mathbf{u}\|\,\|\mathbf{v}\|\), since norms are nonnegative. If \(z\ne0\), choose \(\eta=z/|z|\) and \(\mathbf{w}=\eta\mathbf{v}\) as above. For real \(t\), nonnegativity of the squared norm gives $$ 0\leq\|\mathbf{u}-t\mathbf{w}\|^2 =\|\mathbf{u}\|^2-2t|z|+t^2\|\mathbf{v}\|^2. $$ Take \(t=|z|/\|\mathbf{v}\|^2\), which is defined because \(\mathbf{v}\ne\mathbf{0}\). Then $$ 0\leq\|\mathbf{u}\|^2-\frac{|z|^2}{\|\mathbf{v}\|^2}, \qquad |z|^2\leq\|\mathbf{u}\|^2\|\mathbf{v}\|^2. $$ Both sides of the last inequality are nonnegative. Taking their nonnegative square roots proves \(|z|\leq\|\mathbf{u}\|\,\|\mathbf{v}\|\), as required. \(\square\)
Worked Examples in Complex Spaces
Worked Example: Vectors in \(\mathbb{C}^2\)
Use the standard inner product \(\langle\mathbf{u},\mathbf{v}\rangle=u_1\overline{v_1}+u_2\overline{v_2}\) to verify the inequality for \(\mathbf{u}=(1+i,2)^{\top}\) and \(\mathbf{v}=(i,1-i)^{\top}\). Then carry out the phase and minimization steps.
First compute the inner product, conjugating each entry of the second vector:
Thus \(z=3+i\) and \(|z|=\sqrt{3^2+1^2}=\sqrt{10}\). The squared norms are \(\|\mathbf{u}\|^2=|1+i|^2+|2|^2=2+4=6\) and \(\|\mathbf{v}\|^2=|i|^2+|1-i|^2=1+2=3\). The phase is \(\eta=(3+i)/\sqrt{10}\), so \(|\eta|=1\) and \(\langle\mathbf{u},\eta\mathbf{v}\rangle=\overline{\eta}(3+i)=\sqrt{10}\). The minimizing real parameter is \(t=\sqrt{10}/3\).
For this choice, the squared norm in the proof is $$ \|\mathbf{u}-t\eta\mathbf{v}\|^2 =6-2t\sqrt{10}+3t^2 =6-\frac{20}{3}+\frac{10}{3} =\frac{8}{3}\geq0. $$
Check: The theorem’s bound is \(|\langle\mathbf{u},\mathbf{v}\rangle|=\sqrt{10}\leq\sqrt{6}\sqrt{3}=\sqrt{18}\), since \(10\leq18\). The computed squared norm is nonnegative, consistent with the minimization argument.
Worked Example: When the Inner Product Is Zero
Let \(\mathbf{u}=(1,i)^{\top}\) and \(\mathbf{v}=(1,-i)^{\top}\) in \(\mathbb{C}^2\) with the standard inner product. This example checks the zero-inner-product case, where the phase \(z/|z|\) is not defined.
Conjugating the entries of \(\mathbf{v}\) gives \((1,i)^{\top}\), so $$ \langle\mathbf{u},\mathbf{v}\rangle =1\cdot1+i\cdot i=1-1=0. $$ Also, \(\|\mathbf{u}\|^2=|1|^2+|i|^2=2\) and \(\|\mathbf{v}\|^2=|1|^2+|-i|^2=2\). Therefore \(|\langle\mathbf{u},\mathbf{v}\rangle|=0\leq\sqrt{2}\sqrt{2}=2\).
Check: The inner product calculation gives exactly zero, so there is no phase to adjust. The inequality follows directly from \(0\leq2\), as in the zero case of the proof.
Worked Example: Complex-Valued Functions
On \([0,1]\), take the inner product \(\langle f,g\rangle=\int_0^1 f(x)\overline{g(x)}\,dx\), linear in the first slot. For \(f(x)=1+ix\) and \(g(x)=x\), compute both sides of Cauchy–Schwarz and the minimum from the proof.
Since \(g\) is real-valued, $$ \langle f,g\rangle=\int_0^1 (1+ix)x\,dx =\int_0^1 x\,dx+i\int_0^1 x^2\,dx =\frac12+\frac{i}{3}. $$ Its magnitude is \(\sqrt{(1/2)^2+(1/3)^2}=\sqrt{13}/6\). The squared norms are $$ \|f\|^2=\int_0^1(1+x^2)\,dx=\frac43, \qquad \|g\|^2=\int_0^1x^2\,dx=\frac13. $$
Here \(z=(3+2i)/6\), so \(\eta=(3+2i)/\sqrt{13}\). Then \(\overline{\eta}z=\sqrt{13}/6\), and the minimizing parameter is \(t=(\sqrt{13}/6)/(1/3)=\sqrt{13}/2\). The minimum squared norm is $$ \|f\|^2-\frac{|z|^2}{\|g\|^2} =\frac43-\frac{13/36}{1/3} =\frac43-\frac{13}{12} =\frac14. $$
Check: The inequality reads \(\sqrt{13}/6\leq\sqrt{4/3}\sqrt{1/3}=2/3\). Squaring gives \(13/36\leq16/36\). The computed minimum is \(1/4\), which is nonnegative.
A Common Phase Error
Pitfall: With an inner product linear in the first slot, multiplying the second input by \(\eta\) multiplies the inner product by \(\overline{\eta}\), not by \(\eta\). For example, take \(\mathbf{u}=(1+i,0)^{\top}\), \(\mathbf{v}=(1,0)^{\top}\) in \(\mathbb{C}^2\). Then \(z=\langle\mathbf{u},\mathbf{v}\rangle=1+i\) and \(\eta=z/|z|=(1+i)/\sqrt{2}\). A mistaken use of linearity in the second slot would give \(\langle\mathbf{u},\eta\mathbf{v}\rangle=\eta z=i\sqrt{2}\), which is not real. The correction uses conjugate-linearity: $$ \langle\mathbf{u},\eta\mathbf{v}\rangle =\overline{\eta}z =\frac{1-i}{\sqrt{2}}(1+i) =\sqrt{2}. $$
The conjugate is not a cosmetic detail. The proof needs the two cross terms in the squared-norm expansion to add to a real number, specifically \(2t|z|\). Choosing the phase in accordance with the slot convention ensures this happens. If the convention were linear in the second slot instead, the phase adjustment would have to be made in the first slot.
Why the Argument Is Useful
The proof applies in every complex inner product space, not only in coordinate spaces. Its assumptions are exactly the inner product properties used in the argument: conjugate symmetry identifies the two cross terms, the slot rules justify the expansion, and positive definiteness makes the squared norm nonnegative. That is why the same reasoning works for complex vectors, matrices with a complex inner product, and suitable spaces of complex-valued functions.
The phase step also explains why taking the absolute value in the theorem is natural. A unit scalar can rotate a nonzero complex inner product to the positive real axis while leaving the norm unchanged. Once aligned, the magnitude of that inner product can be compared with the norms by a real quadratic calculation. In practice, the proof’s reusable pattern is to handle the zero case separately, rotate a nonzero inner product to a nonnegative real number, expand a squared norm, and choose the real parameter that minimizes it.
Check Your Understanding
For each question, keep track of which slot is conjugate-linear and state any zero case that matters.
- For \(z\ne0\), verify directly that \(\eta=z/|z|\) has magnitude one and \(\overline{\eta}z=|z|\).
- Expand \(\|\mathbf{u}-t\mathbf{w}\|^2\) for real \(t\), without assuming \(\langle\mathbf{u},\mathbf{w}\rangle\) is real.
- Why must the proof treat \(\mathbf{v}=\mathbf{0}\) separately before dividing by \(\|\mathbf{v}\|^2\)?
- For \(\mathbf{u}=(1+i,1)^{\top}\) and \(\mathbf{v}=(1,2i)^{\top}\) in standard \(\mathbb{C}^2\), compute the inner product and both norms, then check the inequality.
- For a nonzero \(z=\langle\mathbf{u},\mathbf{v}\rangle\), what real value of \(t\) minimizes the squared norm after the phase adjustment, and why is this choice permitted?
