Dot Products, Norms, and Cauchy-Schwarz

From Dot Products to Length

A dot product combines the components of two vectors into a single number. A particularly important case occurs when both inputs are the same vector. In that case, each component is multiplied by itself, so the result records the sum of the component squares. This connects the dot product directly to the length, or norm, of a vector.

That connection gives a practical way to compute lengths and supports a fundamental inequality: the absolute value of a dot product is no greater than the product of the two vector lengths. This inequality is useful even before we know the angle between the vectors; it gives a reliable numerical bound using only their components.

The Norm and the Self-Dot-Product Identity

For vectors in three-dimensional Euclidean space, write \(\mathbf{v}=(v_1,v_2,v_3)\). The dot product of \(\mathbf{u}=(u_1,u_2,u_3)\) and \(\mathbf{v}\) is \(u_1v_1+u_2v_2+u_3v_3\). The Euclidean norm \(\|\mathbf{v}\|\) is the vector’s nonnegative length.

The definition of the dot product immediately gives the key identity. Set both input vectors equal to \(\mathbf{v}\), and each product is a component squared.

To see the last statement, a sum of squares can equal zero only if each square equals zero. Thus \(v_1=v_2=v_3=0\). Conversely, all three components are zero for \(\mathbf{v}=\mathbf{0}\), so its self-dot-product is zero.

Cauchy–Schwarz Inequality

The self-dot-product identity says that every vector has a nonnegative squared length. Apply that fact not only to a given vector, but to a difference of vectors. The right choice of difference produces a quadratic expression whose nonnegativity bounds the dot product.

The absolute value is essential: a dot product may be negative, but its magnitude is bounded. The proof below first treats the case in which \(\mathbf{v}\ne\mathbf{0}\), so that \(\|\mathbf{v}\|^2\) is positive and division by it is permitted.

Here is the complete calculation. Since a squared norm is nonnegative,

Expand using symmetry and linearity of the dot product, and then use \(\mathbf{v}\cdot\mathbf{v}=\|\mathbf{v}\|^2\):

Let \(d=\mathbf{u}\cdot\mathbf{v}\) and \(q=\|\mathbf{v}\|^2\), where \(q>0\) because \(\mathbf{v}\ne\mathbf{0}\). Substituting \(t=d/q\) gives

Multiplication by \(q>0\) preserves the inequality, so \(d^2\le \|\mathbf{u}\|^2\|\mathbf{v}\|^2\). Both norms are nonnegative. Taking square roots therefore yields \(|d|\le\|\mathbf{u}\|\|\mathbf{v}\|\), as required. If \(\mathbf{v}=\mathbf{0}\), then \(\mathbf{u}\cdot\mathbf{v}=0\) and \(\|\mathbf{v}\|=0\), so the inequality holds as equality. This proves the theorem in all cases.

When Does Equality Occur?

The proof also identifies equality. When \(\mathbf{v}\ne\mathbf{0}\), equality in Cauchy–Schwarz occurs exactly when the vector used in the proof has squared norm zero:

A vector has zero norm exactly when it is the zero vector. Thus equality holds precisely when \(\mathbf{u}\) is a scalar multiple of \(\mathbf{v}\). If \(\mathbf{v}=\mathbf{0}\), equality also holds, since both sides vanish. This is a condition for equality in the bound; it does not say that the dot product must always equal the product of the norms.

Why the Inequality Is Useful

Cauchy–Schwarz converts information about lengths into a bound on a dot product. It can therefore be used when the dot product is unknown, difficult to calculate, or only needed within a range. The theorem also guarantees that calculations involving dot products remain controlled by the sizes of the vectors.

The proof’s method is as important as the inequality itself. The key is to start from a quantity known to be nonnegative, namely a squared norm, and choose its vector so that expansion produces the desired dot product. This approach works here because the dot product is symmetric and linear, and because the self-dot-product is the squared norm.

A Common Mistake: Dropping the Absolute Value

A frequent error is to treat the dot product as necessarily nonnegative. For example, let \(\mathbf{p}=(1,0,0)\) and \(\mathbf{q}=(-2,0,0)\). Then

The norms are \(\|\mathbf{p}\|=\sqrt{1}=1\) and \(\|\mathbf{q}\|=\sqrt{4}=2\). The incorrect claim \(\mathbf{p}\cdot\mathbf{q}\ge0\) would say \(-2\ge0\), which is false. Cauchy–Schwarz instead compares the absolute value with the product of norms:

Another mistake is to add component magnitudes when finding a norm. For \((3,4,0)\), that incorrect procedure gives \(3+4+0=7\). The norm is instead \(\sqrt{3^2+4^2+0^2}=\sqrt{25}=5\). The norm comes from the square root of the sum of component squares, not the sum of component magnitudes.