Spectral Projections of a Normal Operator

From Eigenvectors to Projections

A normal operator on a finite-dimensional complex inner-product space has an orthonormal basis of eigenvectors. The spectral theorem gives a particularly useful way to group that basis: collect together the vectors belonging to each distinct eigenvalue. Each group spans an eigenspace, and orthogonal projection onto that eigenspace extracts exactly the part of a vector lying in that group. The resulting projections describe the operator without requiring us to keep track of a particular ordering of all the eigenvectors.

Throughout, let \(V\) be a finite-dimensional inner-product space over \(\mathbb{C}\), and let \(T:V\to V\) be normal, meaning \(T^{*}T=TT^{*}\). We use the convention that the inner product is linear in its first slot. By the Complex Spectral Theorem, \(V\) has an orthonormal basis of eigenvectors of \(T\). In particular, when \(V\ne\{\mathbf{0}\}\), the set \(\sigma(T)\) of distinct eigenvalues is nonempty, and \(V\) is the orthogonal sum of the eigenspaces \(E_{\lambda}\), for \(\lambda\in\sigma(T)\). We use that earlier theorem; the construction and properties of the projections are proved below.

The formula follows from decomposing \(\mathbf{x}\) into its components along the orthonormal basis and a vector orthogonal to that basis. Indeed, the proposed vector \(\sum_j\langle\mathbf{x},\mathbf{u}_j\rangle\mathbf{u}_j\) lies in \(W\). Its difference from \(\mathbf{x}\) is orthogonal to each \(\mathbf{u}_k\), since orthonormality gives \(\langle\mathbf{x},\mathbf{u}_k\rangle-\langle\mathbf{x},\mathbf{u}_k\rangle=0\). Thus the difference lies in \(W^\perp\). Uniqueness follows because \(W\cap W^\perp=\{\mathbf{0}\}\): if a vector lies in both, its inner product with itself is zero, so it is the zero vector.

For each distinct eigenvalue \(\lambda\), define \(P_{\lambda}=P_{E_{\lambda}}\). If \((\mathbf{u}_{\lambda,1},\dots,\mathbf{u}_{\lambda,r_\lambda})\) is an orthonormal basis of \(E_\lambda\), then the construction is explicit:

The definition does not depend on which orthonormal basis of \(E_\lambda\) is chosen: in each case it gives the unique vector in \(E_\lambda\) whose difference from \(\mathbf{x}\) is perpendicular to \(E_\lambda\). In a matrix representation, if the columns of \(\mathbf{U}_\lambda\) are an orthonormal basis of \(E_\lambda\), then \(P_\lambda\) has matrix \(\mathbf{U}_\lambda\mathbf{U}_\lambda^{*}\). Here \(\mathbf{U}_\lambda^{*}\) is the conjugate transpose.

Properties of an Orthogonal Projection

Proof. Write \(P_W\mathbf{x}=\mathbf{w}\), where \(\mathbf{w}\in W\) and \(\mathbf{x}-\mathbf{w}\in W^\perp\). Since \(\mathbf{w}\in W\), its projection onto \(W\) is itself, so \(P_W(P_W\mathbf{x})=P_W\mathbf{w}=\mathbf{w}=P_W\mathbf{x}\). This proves \(P_W^2=P_W\). Every output of \(P_W\) lies in \(W\), and for any \(\mathbf{w}\in W\), \(P_W\mathbf{w}=\mathbf{w}\), so its image is exactly \(W\). Also, \(P_W\mathbf{x}=\mathbf{0}\) holds exactly when the component of \(\mathbf{x}\) in \(W\) is zero, which by the defining decomposition holds exactly when \(\mathbf{x}\in W^\perp\). Thus \(\ker P_W=W^\perp\). To prove self-adjointness, write \(\mathbf{x}=\mathbf{w}+\mathbf{z}\) and \(\mathbf{y}=\mathbf{u}+\mathbf{v}\), with \(\mathbf{w},\mathbf{u}\in W\) and \(\mathbf{z},\mathbf{v}\in W^\perp\). Then \(P_W\mathbf{x}=\mathbf{w}\) and \(P_W\mathbf{y}=\mathbf{u}\). Orthogonality gives \(\langle P_W\mathbf{x},\mathbf{y}\rangle=\langle\mathbf{w},\mathbf{u}\rangle=\langle\mathbf{x},P_W\mathbf{y}\rangle\). This is the defining identity for \(P_W^{*}=P_W\). \(\square\)

For a normal operator, distinct eigenspaces are orthogonal, as established by the earlier result on orthogonality of eigenspaces of normal operators. The next theorem shows how that fact translates into identities among the projections and gives the operator’s spectral decomposition.

Proof. The first claims follow from the proposition applied to \(W=E_\lambda\). Suppose \(\lambda\ne\mu\). For any \(\mathbf{x}\in V\), \(P_\mu\mathbf{x}\in E_\mu\), and \(E_\mu\perp E_\lambda\). A vector in \(E_\lambda^\perp\) projects to zero onto \(E_\lambda\), so \(P_\lambda P_\mu\mathbf{x}=\mathbf{0}\). This holds for every \(\mathbf{x}\), giving \(P_\lambda P_\mu=0\). Interchanging \(\lambda\) and \(\mu\) gives \(P_\mu P_\lambda=0\). By the Complex Spectral Theorem, every \(\mathbf{x}\in V\) can be written as a sum \(\mathbf{x}=\sum_{\lambda\in\sigma(T)}\mathbf{x}_\lambda\) with \(\mathbf{x}_\lambda\in E_\lambda\). Projection \(P_\lambda\) returns \(\mathbf{x}_\lambda\) and annihilates each other component, so \(\sum_\lambda P_\lambda\mathbf{x}=\sum_\lambda\mathbf{x}_\lambda=\mathbf{x}\). This proves the sum identity. Finally, \(T\mathbf{x}_\lambda=\lambda\mathbf{x}_\lambda\) by the definition of eigenspace. Hence \(T\mathbf{x}=\sum_\lambda\lambda\mathbf{x}_\lambda=\sum_\lambda\lambda P_\lambda\mathbf{x}\) for every \(\mathbf{x}\), proving the operator identity. \(\square\)

The theorem says that the projections partition the space into mutually perpendicular eigenspaces. The operator acts on each such component by multiplication by the corresponding eigenvalue. In particular, the coefficients \(\lambda\) in the decomposition are not arbitrary: they are the eigenvalues, and the projections specify where each coefficient acts.

Worked Projection Constructions

Common Pitfall: The Basis Must Be Orthonormal

The sum of outer products \( \mathbf{u}_j\mathbf{u}_j^{*}\) gives the orthogonal projection formula only when the vectors form an orthonormal basis of the subspace. For example, take \(T=\operatorname{diag}(3,3,1)\) on \(\mathbb{C}^3\). Its eigenspace for \(3\) is the first coordinate plane. The vectors \(\mathbf{v}_1=(1,0,0)^\top\) and \(\mathbf{v}_2=(1,1,0)^\top\) are a basis of that plane, but they are not orthonormal.

A tempting but incorrect calculation is to use \(\mathbf{v}_1\mathbf{v}_1^{*}+\mathbf{v}_2\mathbf{v}_2^{*}\) as the projection. It gives \(\begin{bmatrix}2&1&0\\1&1&0\\0&0&0\end{bmatrix}\), which is not a projection: its square is \(\begin{bmatrix}5&3&0\\3&2&0\\0&0&0\end{bmatrix}\), not the original matrix. The correction is to use an orthonormal basis, for example \(\mathbf{e}_1,\mathbf{e}_2\). The actual projection is \(\begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix}\), which is self-adjoint and unchanged by squaring. Thus an orthonormal eigenbasis is not merely convenient notation; it is essential to this simple outer-product formula.

Spectral projections are useful because they separate a normal operator into independent actions on perpendicular subspaces. Once the projections are known, applying \(T\) to any vector means projecting onto each eigenspace and scaling that component by its eigenvalue. The identities \(P_\lambda P_\mu=0\) for distinct eigenvalues and \(\sum_\lambda P_\lambda=\operatorname{id}_V\) also provide direct checks on a proposed collection of spectral projections.