What Equality Says
The real Cauchy–Schwarz inequality compares the size of an inner product with the product of two norms. Its equality case gives more information than the inequality alone: it identifies exactly when the two vectors carry the same direction information, up to a real scalar. The scalar may be positive or negative, so the absolute value in the inequality matters.
Let \(V\) be a real inner product space; it need not be finite-dimensional. Its real inner product is written \(\langle\mathbf{u},\mathbf{v}\rangle\), and its induced norm is \(\|\mathbf{u}\|=\sqrt{\langle\mathbf{u},\mathbf{u}\rangle}\). In particular, positive definiteness says that \(\|\mathbf{u}\|=0\) exactly when \(\mathbf{u}=\mathbf{0}\). The real Cauchy–Schwarz inequality states that \(|\langle\mathbf{u},\mathbf{v}\rangle|\leq\|\mathbf{u}\|\|\mathbf{v}\|\) for every \(\mathbf{u},\mathbf{v}\in V\).
That zero-vector case is important. If either vector is \(\mathbf{0}\), then both sides of Cauchy–Schwarz are zero, so equality holds regardless of the other vector. Any correct equality criterion must include such pairs rather than divide by a norm that could be zero.
The Equality Criterion
Proof. First assume equality. If \(\mathbf{v}=\mathbf{0}\), then the pair is dependent: \(1\cdot\mathbf{v}+0\cdot\mathbf{u}=\mathbf{0}\), and \(\mathbf{v}=0\mathbf{u}\). Now suppose \(\mathbf{v}\ne\mathbf{0}\), so \(\|\mathbf{v}\|^2>0\). Define the real scalar \(\alpha=\langle\mathbf{u},\mathbf{v}\rangle/\|\mathbf{v}\|^2\). Symmetry and bilinearity of the real inner product give
The last equality follows from the assumed equality after squaring it: \(\langle\mathbf{u},\mathbf{v}\rangle^2=\|\mathbf{u}\|^2\|\mathbf{v}\|^2\). Positive definiteness now implies \(\mathbf{u}-\alpha\mathbf{v}=\mathbf{0}\), so \(\mathbf{u}=\alpha\mathbf{v}\), and the pair is dependent.
Conversely, suppose the pair is dependent. If \(\mathbf{v}=\mathbf{0}\), equality holds because both sides are zero. If \(\mathbf{v}\ne\mathbf{0}\), a dependence relation \(a\mathbf{u}+b\mathbf{v}=\mathbf{0}\) with \(a,b\) not both zero must have \(a\ne0\); otherwise \(b\mathbf{v}=\mathbf{0}\) would force \(b=0\), contradicting that the coefficients are not both zero. Thus \(\mathbf{u}=-(b/a)\mathbf{v}\). Write \(\mathbf{u}=c\mathbf{v}\). Then \(\langle\mathbf{u},\mathbf{v}\rangle=c\|\mathbf{v}\|^2\), while \(\|\mathbf{u}\|=|c|\|\mathbf{v}\|\). Therefore
So equality holds. Finally, a scalar-multiple pair is dependent, since if \(\mathbf{u}=c\mathbf{v}\), then \(\mathbf{u}-c\mathbf{v}=\mathbf{0}\); and if \(\mathbf{v}=c\mathbf{u}\), then \(-c\mathbf{u}+\mathbf{v}=\mathbf{0}\). This proves all the stated equivalences. \(\square\)
The theorem is an exact test in both directions. Proportionality guarantees equality, and equality guarantees proportionality. In coordinate spaces, proportionality can be checked by finding one scalar that multiplies every coordinate of one vector to produce the other; checking just one coordinate is not enough.
Coordinate Examples
Worked Example: Equality with a Positive Inner Product
Let \(\mathbf{u}=\begin{bmatrix}2\\1\end{bmatrix}\) and \(\mathbf{v}=\begin{bmatrix}4\\2\end{bmatrix}\) in \(\mathbb{R}^2\) with the standard inner product. Check equality and the scalar-multiple criterion.
First compute the inner product coordinate by coordinate:
The squared norms and norms are
Hence \(|\langle\mathbf{u},\mathbf{v}\rangle|=10\) and \(\|\mathbf{u}\|\|\mathbf{v}\|=\sqrt{5}\,(2\sqrt{5})=10\), so equality holds. The coordinate check agrees: \(\mathbf{v}=2\mathbf{u}\). Check: substituting that relation gives \(\langle\mathbf{u},2\mathbf{u}\rangle=2\|\mathbf{u}\|^2=10\) and \(\|\mathbf{u}\|\|2\mathbf{u}\|=\sqrt{5}\,(2\sqrt{5})=10\).
Worked Example: Equality with a Negative Inner Product
Let \(\mathbf{u}=\begin{bmatrix}1\\2\\0\end{bmatrix}\) and \(\mathbf{v}=\begin{bmatrix}-3\\-6\\0\end{bmatrix}\) in \(\mathbb{R}^3\). Determine whether equality holds, paying attention to the absolute value.
The inner product is
The norms are found from the sums of squared coordinates:
Thus \(|\langle\mathbf{u},\mathbf{v}\rangle|=|-15|=15\), and \(\|\mathbf{u}\|\|\mathbf{v}\|=\sqrt{5}\,(3\sqrt{5})=15\). Equality holds even though the inner product is negative. Indeed, \(\mathbf{v}=-3\mathbf{u}\).
Check: using \(\mathbf{v}=-3\mathbf{u}\) gives \(\langle\mathbf{u},\mathbf{v}\rangle=-3\langle\mathbf{u},\mathbf{u}\rangle=-3(5)=-15\), and \(\|\mathbf{v}\|=|-3|\|\mathbf{u}\|=3\sqrt{5}\). These values reproduce the equality above.
Worked Example: Strict Inequality for Nonproportional Vectors
Let \(\mathbf{u}=\begin{bmatrix}1\\2\end{bmatrix}\) and \(\mathbf{v}=\begin{bmatrix}2\\1\end{bmatrix}\). Compute both sides of Cauchy–Schwarz and check whether either vector is a scalar multiple of the other.
The inner product and norms are
Since \(4<5\), the inequality is strict. The vectors are not scalar multiples: if \(\mathbf{v}=c\mathbf{u}\), the first coordinate forces \(c=2\), but then the second coordinate would have to be \(4\), not \(1\). If \(\mathbf{u}=c\mathbf{v}\), the first coordinate forces \(c=\tfrac12\), but the second would have to be \(\tfrac12\), not \(2\). This confirms the criterion.
Check: the computed values satisfy \(4\leq5\), and they are unequal. The failed coordinate tests establish that the pair is not dependent, as the equality theorem predicts.
Signed Equality and the Scalar’s Sign
The absolute value in Cauchy–Schwarz treats positive and negative inner products alike. If the absolute value is removed, the sign of the proportionality scalar determines which signed equality can occur. The following result includes zero vectors explicitly.
Proof. If \(\mathbf{v}=\mathbf{0}\), then \(\langle\mathbf{u},\mathbf{v}\rangle=0\) and \(\|\mathbf{u}\|\|\mathbf{v}\|=0\), so both signed equalities hold. Now let \(\mathbf{v}\ne\mathbf{0}\). If the positive signed equality holds, set \(c=\langle\mathbf{u},\mathbf{v}\rangle/\|\mathbf{v}\|^2\). The residual computation in the first theorem gives \(\|\mathbf{u}-c\mathbf{v}\|^2=0\), so \(\mathbf{u}=c\mathbf{v}\). Moreover, \(c=\|\mathbf{u}\|/\|\mathbf{v}\|\geq0\), using \(\|\mathbf{v}\|>0\). Conversely, if \(\mathbf{u}=c\mathbf{v}\) with \(c\geq0\), then \(\langle\mathbf{u},\mathbf{v}\rangle=c\|\mathbf{v}\|^2\), and \(\|\mathbf{u}\|\|\mathbf{v}\|=|c|\|\mathbf{v}\|^2=c\|\mathbf{v}\|^2\), proving the positive equality.
For the negative signed equality with \(\mathbf{v}\ne\mathbf{0}\), the same scalar \(c=\langle\mathbf{u},\mathbf{v}\rangle/\|\mathbf{v}\|^2\) gives \(\mathbf{u}=c\mathbf{v}\) by the residual computation, and now \(c=-\|\mathbf{u}\|/\|\mathbf{v}\|\leq0\). Conversely, if \(\mathbf{u}=c\mathbf{v}\) with \(c\leq0\), then \(|c|=-c\), so \(\|\mathbf{u}\|\|\mathbf{v}\|=-c\|\mathbf{v}\|^2=-\langle\mathbf{u},\mathbf{v}\rangle\). Thus the negative equality holds as well. \(\square\)
When \(\mathbf{u}=\mathbf{0}\) and \(\mathbf{v}\ne\mathbf{0}\), the theorem uses \(c=0\), so both signed equalities hold. When \(\mathbf{v}=\mathbf{0}\), both equalities hold without requiring \(\mathbf{u}\) to be zero. These boundary cases explain why the theorem separates \(\mathbf{v}=\mathbf{0}\) rather than making a claim about a scalar obtained by dividing by \(\|\mathbf{v}\|^2\).
Common Pitfall: Dropping the Absolute Value
A common mistake is to replace the equality condition \(|\langle\mathbf{u},\mathbf{v}\rangle|=\|\mathbf{u}\|\|\mathbf{v}\|\) with \(\langle\mathbf{u},\mathbf{v}\rangle=\|\mathbf{u}\|\|\mathbf{v}\|\). That wrongly excludes dependent pairs with a negative scalar. For \(\mathbf{u}=\begin{bmatrix}1\\0\end{bmatrix}\) and \(\mathbf{v}=\begin{bmatrix}-2\\0\end{bmatrix}\), the incorrect calculation stops at \(-2\ne2\) and concludes that equality fails. The correction is to compare the absolute value: \(|\langle\mathbf{u},\mathbf{v}\rangle|=|-2|=2\), while \(\|\mathbf{u}\|=1\) and \(\|\mathbf{v}\|=2\), so the product of norms is \(2\). Equality does hold, because \(\mathbf{v}=-2\mathbf{u}\).
A second error is to claim that equality holds only when the vectors are identical. The first worked example has \(\mathbf{v}=2\mathbf{u}\), and the second has \(\mathbf{v}=-3\mathbf{u}\); both satisfy equality. The correct condition is proportionality, including a negative scalar and the cases involving a zero vector.
Using the Criterion Reliably
For coordinate vectors, a useful check is to calculate both sides and then test the scalar relation. If the computed quantities agree, the theorem guarantees dependence; if the vectors are dependent, the theorem guarantees equality. Doing both checks is valuable in a worked problem because an arithmetic error in a norm or inner product can otherwise lead to a false conclusion.
Multiply corresponding coordinates and add, using the given real inner product.
For the standard inner product on \(\mathbb{R}^n\), square each coordinate, add, and take the nonnegative square root.
Use \(|\langle\mathbf{u},\mathbf{v}\rangle|\), not just \(\langle\mathbf{u},\mathbf{v}\rangle\), when testing equality in Cauchy–Schwarz.
If equality holds, verify a single scalar works in every coordinate. Its sign determines the sign of the inner product when neither vector is zero.
This criterion is useful whenever an estimate involving an inner product becomes exact: equality identifies the dependent case. It also provides a quick diagnostic for computations. A proposed equality example that has no scalar-multiple relation must contain an arithmetic or reasoning error; a proposed strict example with dependent vectors must be checked again.
Check Your Understanding
For each coordinate pair, compute enough to decide whether equality holds, and justify the conclusion using proportionality.
- Do \(\begin{bmatrix}3\\-1\end{bmatrix}\) and \(\begin{bmatrix}-6\\2\end{bmatrix}\) give equality in real Cauchy–Schwarz? What is the scalar relation?
- For \(\mathbf{u}=\begin{bmatrix}0\\0\end{bmatrix}\) and \(\mathbf{v}=\begin{bmatrix}1\\-4\end{bmatrix}\), do both signed equality cases hold?
- Are \(\begin{bmatrix}1\\3\end{bmatrix}\) and \(\begin{bmatrix}3\\1\end{bmatrix}\) proportional? Compute both sides of Cauchy–Schwarz.
- State why the proof using \(\alpha=\langle\mathbf{u},\mathbf{v}\rangle/\|\mathbf{v}\|^2\) needs a separate case when \(\mathbf{v}=\mathbf{0}\).
- If \(\mathbf{v}\ne\mathbf{0}\) and \(\mathbf{u}=c\mathbf{v}\) with \(c<0\), which signed equality holds?
