Proof of the Complex Spectral Theorem

The Spectral Theorem as a Two-Step Argument

A normal operator need not look diagonal in an arbitrary orthonormal basis. The complex spectral theorem says, however, that there is an orthonormal basis in which its action is diagonal: every basis vector is an eigenvector. The proof brings together two facts about finite-dimensional complex inner-product spaces. Schur triangularization first gives an upper-triangular matrix in a suitable orthonormal basis. The result that a normal upper-triangular matrix is diagonal then finishes the argument.

Let \(V\) be a finite-dimensional inner-product space over \(\mathbb{C}\). The adjoint \(T^{*}:V\to V\) of a linear operator \(T:V\to V\) is characterized by the inner-product adjoint identity. The operator \(T\) is normal if \(T^{*}T=TT^{*}\). An orthonormal eigenbasis is an ordered basis consisting of eigenvectors whose norms are \(1\) and which are pairwise orthogonal.

Two established results give the proof its structure. Schur's Triangularization Theorem states that every linear operator on a finite-dimensional complex inner-product space has an upper-triangular matrix in some orthonormal basis. The standard result on normal triangular matrices states that a normal upper-triangular complex matrix must be diagonal. Both results are used with their stated hypotheses: finite dimension and complex scalars for Schur triangularization, and both normality and upper-triangular form for the second result.

Why Unitary Changes of Coordinates Preserve Normality

In an orthonormal basis, the matrix of an adjoint is the conjugate transpose of the matrix of the operator. A change between orthonormal bases is represented by a unitary matrix \(\mathbf{Q}\), which satisfies \(\mathbf{Q}^{*}\mathbf{Q}=\mathbf{Q}\mathbf{Q}^{*}=\mathbf{I}\). Before applying the two results above, it is useful to verify that this change of coordinates does not destroy normality.

Proof. Since \(\mathbf{Q}\) is unitary, \(\mathbf{Q}\mathbf{Q}^{*}=\mathbf{I}\). Taking the conjugate transpose of \(\mathbf{B}=\mathbf{Q}^{*}\mathbf{A}\mathbf{Q}\) gives \(\mathbf{B}^{*}=\mathbf{Q}^{*}\mathbf{A}^{*}\mathbf{Q}\). Therefore \(\mathbf{B}^{*}\mathbf{B}=\mathbf{Q}^{*}\mathbf{A}^{*}\mathbf{Q}\mathbf{Q}^{*}\mathbf{A}\mathbf{Q}=\mathbf{Q}^{*}\mathbf{A}^{*}\mathbf{A}\mathbf{Q}\), while \(\mathbf{B}\mathbf{B}^{*}=\mathbf{Q}^{*}\mathbf{A}\mathbf{Q}\mathbf{Q}^{*}\mathbf{A}^{*}\mathbf{Q}=\mathbf{Q}^{*}\mathbf{A}\mathbf{A}^{*}\mathbf{Q}\). Normality of \(\mathbf{A}\) gives \(\mathbf{A}^{*}\mathbf{A}=\mathbf{A}\mathbf{A}^{*}\), so these two products are equal. Thus \(\mathbf{B}\) is normal. \(\square\)

This calculation explains why the choice of an orthonormal basis matters. In such coordinates the adjoint is represented by the conjugate transpose, and unitary changes of orthonormal coordinates preserve the normality equation.

Proof of the Complex Spectral Theorem

Proof. Choose an orthonormal basis of \(V\), and let \(\mathbf{A}\) be the matrix of \(T\) in that basis. Since \(T\) is normal, \(\mathbf{A}^{*}\mathbf{A}=\mathbf{A}\mathbf{A}^{*}\). By Schur's Triangularization Theorem, there is a unitary matrix \(\mathbf{Q}\) such that \(\mathbf{R}=\mathbf{Q}^{*}\mathbf{A}\mathbf{Q}\) is upper triangular. The lemma shows that \(\mathbf{R}\) is normal. The result on normal triangular matrices therefore implies that \(\mathbf{R}\) is diagonal; write \(\mathbf{R}=\operatorname{diag}(\lambda_1,\dots,\lambda_n)\).

Let \(\mathbf{q}_1,\dots,\mathbf{q}_n\) be the columns of \(\mathbf{Q}\). Because \(\mathbf{Q}\) is unitary, these columns are an orthonormal basis of \(\mathbb{C}^n\). From \(\mathbf{R}=\mathbf{Q}^{*}\mathbf{A}\mathbf{Q}\), multiply on the left by \(\mathbf{Q}\) to obtain \(\mathbf{A}\mathbf{Q}=\mathbf{Q}\mathbf{R}\). The \(j\)-th column of this equation is \(\mathbf{A}\mathbf{q}_j=\lambda_j\mathbf{q}_j\). Thus each \(\mathbf{q}_j\) is an eigenvector, and the columns form an orthonormal eigenbasis in coordinates. The corresponding vectors in \(V\) are an orthonormal eigenbasis for \(T\). \(\square\)

The proof does not construct the eigenvectors one at a time from the original matrix. Instead, Schur triangularization organizes the whole operator in orthonormal coordinates, and normality forces the resulting triangular matrix to have no entries above its diagonal. This is why the triangular-matrix result is the decisive second step.

Worked Examples

Common Pitfall: Using the Transpose Instead of the Conjugate Transpose

For complex matrices, normality is defined using \(\mathbf{A}^{*}\), not \(\mathbf{A}^{\top}\). For example, take \(\mathbf{C}=\begin{bmatrix}i&0\\0&2\end{bmatrix}\). A mistaken computation uses \(\mathbf{C}^{\top}\mathbf{C}=\operatorname{diag}(-1,4)\), then may incorrectly conclude that \(\mathbf{C}\) is not normal. The correction is to conjugate the entries when taking the adjoint: \(\mathbf{C}^{*}=\operatorname{diag}(-i,2)\). Therefore \(\mathbf{C}^{*}\mathbf{C}=\operatorname{diag}(1,4)=\mathbf{C}\mathbf{C}^{*}\), so \(\mathbf{C}\) is normal. The complex conjugation is essential in both the normality calculation and the orthogonality check in the preceding example.

The theorem is a sufficient conclusion under a specific hypothesis: finite dimension, complex scalars, and normality. It does not assert that every complex matrix has an orthonormal eigenbasis. For instance, \(\mathbf{J}=\begin{bmatrix}1&1\\0&1\end{bmatrix}\) has only the eigenvalue \(1\), and solving \((\mathbf{J}-\mathbf{I})\mathbf{x}=\mathbf{0}\) gives \(x_2=0\), so its eigenvectors lie in the one-dimensional span of \((1,0)^{\top}\). It cannot have an eigenbasis. This example is not a counterexample to the theorem: \(\mathbf{J}\) is not normal.

What the Proof Gives

In an orthonormal eigenbasis, applying \(T\) to a vector amounts to multiplying each coordinate by the corresponding eigenvalue. The theorem therefore replaces a potentially complicated operator by a diagonal description without changing the inner-product geometry of the space. The proof also clarifies the division of work: Schur triangularization supplies orthonormal coordinates, unitary similarity preserves normality, and the diagonal form of normal triangular matrices turns those coordinates into eigenvectors.

When applying the theorem, keep the logical direction and hypotheses in view. Start with a normal operator on a finite-dimensional complex inner-product space. Triangularize using an orthonormal basis, not an arbitrary basis; then use normality of the triangular matrix. The conclusion is an orthonormal eigenbasis, not merely a list of eigenvalues or a diagonal representation in unspecified coordinates.