Independence of Orthogonal Nonzero Vectors

Why Orthogonality Detects Independence

Linear independence is often tested by placing vectors in the columns of a matrix and row reducing. For an orthogonal list, there is a more direct route: take the inner product of a proposed linear relation with one vector at a time. Orthogonality makes all the other terms vanish, leaving only the coefficient of that vector. If the vector is nonzero, its inner product with itself is positive, so that coefficient must be zero.

This argument works in real and complex inner product spaces, and it does not require the space to be finite-dimensional. The list itself must be finite so that its linear combination is a finite sum. In a complex space, we use the convention that the inner product is linear in its first slot and conjugate-linear in its second.

Positive definiteness of an inner product says that \(\langle\mathbf{v},\mathbf{v}\rangle>0\) for every nonzero \(\mathbf{v}\). Equivalently, \(\langle\mathbf{v},\mathbf{v}\rangle=\|\mathbf{v}\|^2\), and this quantity is zero exactly when \(\mathbf{v}=\mathbf{0}\). That positivity is the step that turns an inner-product equation into a conclusion about a coefficient.

The Independence Theorem

Proof. Let \((\mathbf{v}_1,\dots,\mathbf{v}_p)\) be an orthogonal list of nonzero vectors, and suppose \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p=\mathbf{0}\), where each \(c_i\in\mathbb{F}\). If \(p=0\), the list has no coefficients, so the independence condition holds vacuously. Suppose \(p\ge1\), and fix any index \(k\) with \(1\le k\le p\). Take the inner product of the relation with \(\mathbf{v}_k\) in the second slot. Linearity in the first slot gives

The left side is \(\langle\mathbf{0},\mathbf{v}_k\rangle=0\). In the sum on the right, each term with \(i\ne k\) vanishes by orthogonality. Since \(\mathbf{v}_k\ne\mathbf{0}\), positive definiteness gives \(\langle\mathbf{v}_k,\mathbf{v}_k\rangle>0\), so \(c_k=0\). The index \(k\) was arbitrary, and the same argument applies to every index from \(1\) to \(p\). Thus every coefficient is zero, which proves linear independence. \(\square\)

The argument uses the inner product to isolate one coefficient at a time; it does not require solving a system of equations. Over \(\mathbb{C}\), the same calculation is valid because the coefficient \(c_i\) is in the first slot, where the inner product is linear. If the inner product convention were linear in the second slot instead, the slots in this proof would be reversed.

The Nonzero Condition Matters

Proof. If every vector in the orthogonal list is nonzero, the theorem proves that the list is linearly independent. For the reverse direction, suppose the list contains a zero vector, say \(\mathbf{v}_k=\mathbf{0}\). Assign coefficient \(1\) to \(\mathbf{v}_k\) and coefficient \(0\) to every other vector. Then \(\sum_{i=1}^{p}c_i\mathbf{v}_i=\mathbf{0}\), but the coefficients are not all zero. This is a nontrivial linear relation, so the list is dependent. If the list is empty, it has no zero vector and is independent vacuously, consistent with the statement. \(\square\)

This corollary distinguishes two conditions that should not be conflated. Orthogonality only says that distinct vectors have zero inner product; it does not rule out a zero vector, whose inner product with every vector is zero. Nonzero vectors are needed to ensure that the self-inner-product left after taking an inner product is positive.

Using the Test in Other Inner Product Spaces

The theorem is not limited to coordinate vectors. It applies whenever the space has an inner product and the vectors in the list are pairwise orthogonal and nonzero. In a matrix space, for example, one may use the Frobenius inner product: for real matrices \(\mathbf{A}=(a_{ij})\) and \(\mathbf{B}=(b_{ij})\) of the same size, define \(\langle\mathbf{A},\mathbf{B}\rangle=\sum_{i,j}a_{ij}b_{ij}\). Its squared norm is the sum of the squares of all matrix entries, which is positive for a nonzero matrix.

In each example, the independence test has the same structure: verify pairwise orthogonality, verify that no listed vector is zero, and apply the theorem. For a long list, this can be substantially simpler than row reducing a matrix whose columns are the vectors. The test is sufficient when orthogonality is established; it does not say that every independent list must be orthogonal.

Common Pitfall: Forgetting to Check for Zero

A tempting but incorrect statement is “every orthogonal list is linearly independent.” For a concrete counterexample, take \((1,0)\), \((0,1)\), and \((0,0)\) in \(\mathbb{R}^2\). Each pair has dot product zero, so the list is orthogonal under the definition. But it is dependent: \(0(1,0)+0(0,1)+1(0,0)=(0,0)\) is a relation with a nonzero coefficient. The correction is to check that every vector is nonzero before using the theorem.

There is also a useful way to organize a valid test. For a list of \(p\) vectors, check all pairs with distinct indices, not just one or two selected pairs. Then check the squared norm of each vector. For an orthogonal list, these checks establish exactly the hypotheses needed for the independence theorem. If one pair has nonzero inner product, this theorem does not decide independence; another method may be needed.

The essential mechanism is worth remembering: in a proposed relation, inner product with \(\mathbf{v}_k\) removes every term except the one involving \(c_k\). Orthogonality removes the other terms, and nonzero length prevents the remaining factor from vanishing. Together, these facts force each coefficient to be zero.