- Understand the basic properties of orthogonal complements.
- Recipes: shortcuts for computing the orthogonal complements of common subspaces.
- Picture: orthogonal complements in and
- Vocabulary words: orthogonal complement, row space.
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.
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.
We now have two similar-looking pieces of notation:
Try not to confuse the two.
The orthogonal complement of a line through the origin in is the perpendicular line
The orthogonal complement of a line in is the perpendicular plane
The orthogonal complement of a plane in is the perpendicular line
We see in the above pictures that
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.
Let be vectors in and let Then
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
Therefore, is in
To prove the second equality, we let
By the row-column rule for matrix multiplication in Section 3.3, for any vector in we have
Therefore, is in if and only if is perpendicular to each vector
By the proposition, computing the orthogonal complement of a span means solving a system of linear equations. For example, if
then is the solution set of the homogeneous linear system associated to the matrix
This is the solution set of the system of equations
In order to find shortcuts for computing orthogonal complements, we need the following basic facts.
Let be a subspace of Then:
For the first assertion, we verify the three defining properties of subspaces.
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
by the row-column rule for matrix multiplication in Section 3.3. Since it follows from this proposition that is in and similarly, is in As above, this implies is orthogonal to itself, which contradicts our assumption that is nonzero. Therefore, as desired.
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.
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
For any vectors we have
For any matrix we have
Let be an matrix. By the rank theorem in Section 3.9, we have
On the other hand the third fact says that
which implies Since we have
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.”