Oblique Projections and Idempotent Matrices

Projections Need Not Be Orthogonal

A projection keeps the component of a vector in a chosen subspace and removes the component assigned to a complementary subspace. In an orthogonal projection, the removed component is perpendicular to the chosen subspace. But a projection can instead use a complement that meets the chosen subspace at an angle other than 90 degrees. Such a map is an oblique projection.

The algebraic condition capturing both kinds is simple: applying the map twice has the same effect as applying it once. This condition is called idempotency. To understand what it means geometrically, the key is to identify both the subspace that remains and the subspace that is removed.

Definition: Let \(U\) and \(V\) be subspaces of \(\mathbb{R}^n\) such that \(\mathbb{R}^n=U\oplus V\). This means every \(x\in\mathbb{R}^n\) has a unique representation \(x=u+v\), with \(u\in U\) and \(v\in V\). The projection onto \(U\) along \(V\) is the linear map \(P\) defined by \(P(u+v)=u\). Its range is \(U\), and its null space is \(V\). If \(U\) and \(V\) are orthogonal, this is an orthogonal projection; otherwise, it is oblique.

The phrase “along \(V\)” specifies which part of the input is discarded. Thus a projection is determined not just by its range, but also by its null space. A given subspace generally has many possible complements, and choosing a different complement gives a different projection onto that same subspace.

Idempotency Characterizes Projections

For a square matrix \(P\), idempotency means \(P^2=P\). The equality says that once a vector has been mapped to its projected value, another application does not change it. The next result shows that this algebraic property is equivalent to being a projection onto the range along the null space.

Theorem: A real \(n\times n\) matrix \(P\) is idempotent if and only if \(\mathbb{R}^n=\mathcal{R}(P)\oplus\mathcal{N}(P)\) and \(P\) is the projection onto \(\mathcal{R}(P)\) along \(\mathcal{N}(P)\). Here \(\mathcal{R}(P)\) denotes the range and \(\mathcal{N}(P)\) the null space.

Proof. First suppose \(P^2=P\). For any \(x\in\mathbb{R}^n\), write

$$ x=Px+(I-P)x. $$

The first term belongs to \(\mathcal{R}(P)\). For the second,

$$ P(I-P)x=(P-P^2)x=0, $$

so \((I-P)x\in\mathcal{N}(P)\). Therefore every vector is a sum of a vector in the range and a vector in the null space. To show the sum is direct, suppose \(z\in\mathcal{R}(P)\cap\mathcal{N}(P)\). Since \(z\in\mathcal{R}(P)\), there is an \(x\) with \(z=Px\). Then

$$ z=P x=P^2x=Pz=0, $$

where \(Pz=0\) because \(z\in\mathcal{N}(P)\). Thus the intersection contains only zero, and the sum is direct. Moreover, \(P\) leaves every range vector fixed: if \(u=Px\), then \(Pu=P^2x=Px=u\). It sends every null-space vector to zero. Hence it is precisely the projection onto \(\mathcal{R}(P)\) along \(\mathcal{N}(P)\).

Conversely, suppose \(P\) is the projection onto \(U\) along \(V\). For any \(x=u+v\), with \(u\in U\) and \(v\in V\), the definition gives \(Px=u\). Applying \(P\) again gives \(P(Px)=Pu=u=Px\), because \(P\) keeps every vector in \(U\) fixed. This holds for every \(x\), so \(P^2=P\). \(\square\)

The proof also gives a practical decomposition that does not require choosing bases for the two subspaces: for every \(x\), its projected component is \(Px\), and its discarded component is \((I-P)x\). These components lie in the range and null space, respectively.

Worked Example: A Projection onto the Horizontal Axis

Consider

P=\begin{pmatrix}1&2\\0&0\end{pmatrix}.

First check idempotency by multiplication:

P^2= \begin{pmatrix}1&2\\0&0\end{pmatrix} \begin{pmatrix}1&2\\0&0\end{pmatrix} = \begin{pmatrix}1&2\\0&0\end{pmatrix} =P.

For \(x=(x_1,x_2)^T\), \(Px=(x_1+2x_2,0)^T\). The outputs are exactly the vectors on the horizontal axis, so \(\mathcal{R}(P)=\operatorname{span}\{(1,0)^T\}\). To find the null space, solve \(x_1+2x_2=0\). Thus \(\mathcal{N}(P)=\operatorname{span}\{(-2,1)^T\}\). These subspaces have zero intersection and together span \(\mathbb{R}^2\).

For \(x=(3,1)^T\), the projection and residual are

Px=(5,0)^T,\qquad (I-P)x=x-Px=(-2,1)^T.

Their sum is \((3,1)^T\), as required. The range direction \((1,0)^T\) and null direction \((-2,1)^T\) have inner product \(-2\), not zero. This is an oblique projection. The matrix is also nonsymmetric, since \(P^T\ne P\).

When Is a Projection Orthogonal?

In real Euclidean space, symmetry gives an exact test: an idempotent matrix is an orthogonal projection if and only if it is symmetric. For an idempotent matrix, the theorem above already establishes that its range and null space are complementary. Symmetry tells us whether those complementary subspaces are perpendicular.

Theorem: If \(P\) is a real square matrix satisfying \(P^2=P\), then \(P\) is an orthogonal projection if and only if \(P^T=P\). Equivalently, \(\mathcal{N}(P)=\mathcal{R}(P)^\perp\).

Proof. Suppose first that \(P^T=P\). If \(u\in\mathcal{R}(P)\), write \(u=Px\), and let \(z\in\mathcal{N}(P)\). Then

$$ u^Tz=(Px)^Tz=x^TP^Tz=x^TPz=0. $$

Therefore the range and null space are orthogonal. Since they are complementary, the null space is exactly the orthogonal complement of the range. Thus \(P\) is the orthogonal projection onto its range.

Conversely, suppose \(P\) is an orthogonal projection. Write arbitrary vectors as \(x=u+z\) and \(y=w+t\), where \(u,w\in\mathcal{R}(P)\) and \(z,t\in\mathcal{N}(P)=\mathcal{R}(P)^\perp\). Then \(Px=u\) and \(Py=w\). Consequently,

$$ (Px)^Ty=u^T(w+t)=u^Tw=x^Tw=x^TPy. $$

This identity for all \(x,y\) implies \(P^T=P\). \(\square\)

Worked Example: A Rank-One Oblique Projection

Let \(u=(1,1)^T\), \(v=(1,0)^T\), and define \(P=uv^T\). Since \(v^Tu=1\), multiplication gives

P^2=(uv^T)(uv^T)=u(v^Tu)v^T=uv^T=P, \qquad P=\begin{pmatrix}1&0\\1&0\end{pmatrix}.

For any \(x\), \(Px=u(v^Tx)\), so every output is a multiple of \(u\), and \(Pu=u\). Hence \(\mathcal{R}(P)=\operatorname{span}\{(1,1)^T\}\). The equation \(Px=0\) is equivalent to \(v^Tx=x_1=0\), giving \(\mathcal{N}(P)=\operatorname{span}\{(0,1)^T\}\). The range and null directions are not perpendicular: their inner product is \(1\). Also \(P^T\ne P\), confirming that this is oblique.

For \(x=(2,3)^T\),

Px=(2,2)^T,\qquad x-Px=(0,1)^T,\qquad (2,2)^T+(0,1)^T=(2,3)^T.

This construction works more generally whenever \(v^Tu=1\): the matrix \(uv^T\) is idempotent, has range \(\operatorname{span}\{u\}\), and has null space \(\{x:v^Tx=0\}\). The null space is perpendicular to \(v\), not necessarily to \(u\), which explains why the projection need not be orthogonal.

Finding the Range, Null Space, and Complement

For a concrete idempotent matrix, compute the range as the span of its columns and compute the null space by solving \(Px=0\). The theorem guarantees that these two subspaces are complementary. In particular, the null space is a complement of the range selected by this projection. It need not be the only possible complement of the range.

Worked Example: A Projection in Three Dimensions

Consider

P=\begin{pmatrix} 1&0&1\\ 0&1&0\\ 0&0&0 \end{pmatrix}.

Multiplying the matrix by itself gives

P^2= \begin{pmatrix} 1&0&1\\ 0&1&0\\ 0&0&0 \end{pmatrix} \begin{pmatrix} 1&0&1\\ 0&1&0\\ 0&0&0 \end{pmatrix} = \begin{pmatrix} 1&0&1\\ 0&1&0\\ 0&0&0 \end{pmatrix} =P.

For \(x=(x_1,x_2,x_3)^T\), \(Px=(x_1+x_3,x_2,0)^T\). The first two columns are \(e_1\) and \(e_2\), and the third is \(e_1\), so \(\mathcal{R}(P)=\operatorname{span}\{e_1,e_2\}\), the \(xy\)-plane. Solving \(Px=0\) requires \(x_1+x_3=0\) and \(x_2=0\). Thus \(\mathcal{N}(P)=\operatorname{span}\{(-1,0,1)^T\}\). This line is a complement of the \(xy\)-plane: it is not contained in the plane, and the two subspaces have dimensions 1 and 2.

For \(x=(2,3,4)^T\), the decomposition is

Px=(6,3,0)^T,\qquad (I-P)x=(-4,0,4)^T,\qquad (6,3,0)^T+(-4,0,4)^T=(2,3,4)^T.

The projection is oblique because its null direction has inner product \(-1\) with \(e_1\), a direction in its range. Its transpose is not equal to itself, which gives the same conclusion by the symmetry test.

Common Mistake: Equating Idempotent with Symmetric

A frequent error is to say that every projection matrix must be symmetric. This confuses orthogonal projections with projections in general. Idempotency alone guarantees a projection, but not an orthogonal one. For example, the first worked matrix satisfies \(P^2=P\) while \(P^T\ne P\); its range and null space are complementary but not perpendicular.

The correction is to keep the two tests distinct: \(P^2=P\) characterizes projections, while \(P^2=P\) together with \(P^T=P\) characterizes orthogonal projections. Likewise, a vector’s residual \(x-Px\) is always in the null space for an idempotent \(P\), but it is not necessarily perpendicular to the projected vector \(Px\).

Key takeaway: An idempotent matrix projects onto its range along its null space. Those subspaces always form a direct sum; they are orthogonal exactly when the projection matrix is symmetric.

Check Your Understanding

Use the range, null-space, and idempotency criteria to answer the following.

For an idempotent matrix \(P\), why does \(x=Px+(I-P)x\) express every vector as a sum of a range vector and a null-space vector?
What additional condition, beyond idempotency, makes a real projection matrix an orthogonal projection?
For \(P=\begin{pmatrix}1&-1\\0&0\end{pmatrix}\), find the range and null space, and determine whether the projection is orthogonal.
Explain why the null space of an idempotent matrix is a complement of its range, but need not be the only possible complement.
If \(P\) is the projection onto \(U\) along \(V\), what are \(Px\) and \((I-P)x\) when \(x=u+v\), \(u\in U\), \(v\in V\)?