It will be important to compute the set of all vectors that are orthogonal to a given set of vectors. It turns out that a vector is orthogonal to a set of vectors if and only if it is orthogonal to the span of those vectors, which is a subspace, so we restrict ourselves to the case of subspaces.

Definition

Let be a subspace of Its orthogonal complement is the subspace

The symbol is read “ perp”.

This is the set of all vectors in that are orthogonal to all of the vectors in We will show below that is indeed a subspace.

Note

We now have two similar-looking pieces of notation:

Try not to confuse the two.

Pictures of orthogonal complements

The orthogonal complement of a line through the origin in is the perpendicular line

The orthogonal complement of is since the zero vector is the only vector that is orthogonal to all of the vectors in

For the same reason, we have

Since any subspace is a span, the following proposition gives a recipe for computing the orthogonal complement of any subspace. However, below we will give several shortcuts for computing the orthogonal complements of other common kinds of subspaces.

To justify the first equality, we need to show that a vector is perpendicular to the all of the vectors in if and only if it is perpendicular only to Since the are contained in we really only have to show that if then is perpendicular to every vector in Indeed, any vector in has the form for suitable scalars so

The zero vector is in because the zero vector is orthogonal to every vector in

Let be in so and for every vector in We must verify that for every in Indeed, we have

Let be in so for every in and let be a scalar. We must verify that for every in Indeed, we have

Next we prove the third assertion. Let be a basis for so and let be a basis for so We need to show First we claim that is linearly independent. Suppose that Let and so is in is in and Then is in both and which implies is perpendicular to itself. In particular, so and hence Therefore, all coefficients are equal to zero, because and are linearly independent.

It follows from the previous paragraph that Suppose that Then the matrix

has more columns than rows (it is “wide”), so its null space is nonzero by this note in Section 4.2. Let be a nonzero vector in Then

Finally, we prove the second assertion. Clearly is contained in this says that everything in is perpendicular to the set of all vectors perpendicular to everything in Let By 3, we have so The only -dimensional subspace of is all of so

See these paragraphs for pictures of the second property. As for the third: for example, if is a (-dimensional) plane in then is another (-dimensional) plane. Explicitly, we have

the orthogonal complement of the -plane is the -plane.

Definition

The row space of a matrix is the span of the rows of and is denoted

If is an matrix, then the rows of are vectors with entries, so is a subspace of Equivalently, since the rows of are the columns of the row space of is the column space of

We showed in the above proposition that if has rows then

Taking orthogonal complements of both sides and using the second fact gives

Replacing by and remembering that gives

To summarize:

Recipes: Shortcuts for computing orthogonal complements

In other words, the span of the rows of has the same dimension as the span of the columns of even though the first lives in and the second lives in This fact is often stated as “the row rank equals the column rank.”