A Diagnostic Framework for Complex Normal Matrices
When spectral data are proposed for a complex matrix, several different claims need checking. A number may be a root of the characteristic polynomial, a vector may satisfy the eigenvector equation, and a stated multiplicity may describe either the polynomial or the eigenspace. For a normal matrix, these claims are linked: the complex spectral theorem forces algebraic and geometric multiplicities to agree, and eigenspaces for distinct eigenvalues are orthogonal. These constraints make it possible to detect inconsistent data before accepting a full list of eigenvalues and eigenvectors.
Throughout, \(\mathbf{A}\in\mathbb{C}^{n\times n}\), and \(\mathbf{A}^{*}\) denotes its conjugate transpose. The matrix is normal if \(\mathbf{A}\mathbf{A}^{*}=\mathbf{A}^{*}\mathbf{A}\). The inner product on \(\mathbb{C}^n\) is \(\langle\mathbf{x},\mathbf{y}\rangle=\mathbf{x}^{\top}\overline{\mathbf{y}}\), linear in the first slot. An eigenvalue \(\lambda\) has an eigenvector \(\mathbf{v}\ne\mathbf{0}\) when \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\); its eigenspace is \(E_{\lambda}=\{\mathbf{x}:\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\}\).
For any complex square matrix, a scalar is an eigenvalue exactly when it is a root of \(p_{\mathbf{A}}\): \(\det(\mathbf{A}-\lambda\mathbf{I})=0\) means the matrix is singular, so its homogeneous system has a nonzero solution \(\mathbf{v}\), which satisfies \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). This root test identifies eigenvalues, but by itself it does not verify a proposed eigenvector or impose all the additional restrictions that normality supplies.
What Normality Forces About Multiplicity
Proof. By the complex spectral theorem, there is a unitary matrix \(\mathbf{U}\) and a diagonal matrix \(\mathbf{D}=\operatorname{diag}(\lambda_1,\dots,\lambda_n)\) such that \(\mathbf{A}=\mathbf{U}\mathbf{D}\mathbf{U}^{*}\). Since \(\mathbf{U}^{*}\mathbf{U}=\mathbf{I}\), we have \(\mathbf{A}-\lambda\mathbf{I}=\mathbf{U}(\mathbf{D}-\lambda\mathbf{I})\mathbf{U}^{*}\). The multiplicative property of the determinant and \(\det(\mathbf{U})\det(\mathbf{U}^{*})=1\) give $$ p_{\mathbf{A}}(\lambda)=\det(\mathbf{D}-\lambda\mathbf{I}) =\prod_{j=1}^{n}(\lambda_j-\lambda). $$ Thus the algebraic multiplicity of a value \(\lambda_0\) is the number of indices \(j\) for which \(\lambda_j=\lambda_0\). To find its eigenspace, write \(\mathbf{x}=\mathbf{U}\mathbf{z}\), which is possible uniquely because \(\mathbf{U}\) is invertible. Then \(\mathbf{A}\mathbf{x}=\lambda_0\mathbf{x}\) is equivalent to \(\mathbf{D}\mathbf{z}=\lambda_0\mathbf{z}\). In coordinates, \((\lambda_j-\lambda_0)z_j=0\) for every \(j\). The free coordinates are exactly those with \(\lambda_j=\lambda_0\), so the dimension of this eigenspace is the number of such indices. This equals the algebraic multiplicity. \(\square\)
This theorem supplies a strong consistency check. For a proposed eigenvalue of a normal matrix, the stated eigenspace cannot have a different dimension from the root multiplicity. Also, the distinct eigenvalues’ algebraic multiplicities sum to \(n\), since the diagonalization lists \(n\) diagonal entries. A proposed data set that omits a root, assigns an incorrect multiplicity, or gives the wrong eigenspace dimension cannot describe the matrix as claimed.
Orthogonality as a Second Constraint
Proof. First, for any normal matrix \(\mathbf{M}\) and vector \(\mathbf{x}\), normality gives $$ \|\mathbf{M}\mathbf{x}\|^2 =\langle\mathbf{x},\mathbf{M}^{*}\mathbf{M}\mathbf{x}\rangle =\langle\mathbf{x},\mathbf{M}\mathbf{M}^{*}\mathbf{x}\rangle =\|\mathbf{M}^{*}\mathbf{x}\|^2. $$ Consequently \(\mathbf{M}\mathbf{x}=\mathbf{0}\) implies \(\mathbf{M}^{*}\mathbf{x}=\mathbf{0}\), because the latter vector has norm zero. Now take \(\mathbf{x}\in E_{\lambda}\). The matrix \(\mathbf{M}=\mathbf{A}-\lambda\mathbf{I}\) is normal: expanding both products \(\mathbf{M}\mathbf{M}^{*}\) and \(\mathbf{M}^{*}\mathbf{M}\), the cross terms agree and \(\mathbf{A}\mathbf{A}^{*}=\mathbf{A}^{*}\mathbf{A}\). Since \(\mathbf{M}\mathbf{x}=\mathbf{0}\), the norm identity gives \(\mathbf{M}^{*}\mathbf{x}=\mathbf{0}\), or \(\mathbf{A}^{*}\mathbf{x}=\overline{\lambda}\mathbf{x}\). Likewise, \(\mathbf{A}^{*}\mathbf{y}=\overline{\mu}\mathbf{y}\) for \(\mathbf{y}\in E_{\mu}\). The adjoint identity and the inner product’s conjugate-linearity in the second slot now yield $$ \lambda\langle\mathbf{x},\mathbf{y}\rangle =\langle\mathbf{A}\mathbf{x},\mathbf{y}\rangle =\langle\mathbf{x},\mathbf{A}^{*}\mathbf{y}\rangle =\langle\mathbf{x},\overline{\mu}\mathbf{y}\rangle =\mu\langle\mathbf{x},\mathbf{y}\rangle. $$ As \(\lambda\ne\mu\), subtracting gives \((\lambda-\mu)\langle\mathbf{x},\mathbf{y}\rangle=0\), hence \(\langle\mathbf{x},\mathbf{y}\rangle=0\). \(\square\)
The condition concerns distinct eigenvalues. It says that every vector in one eigenspace is orthogonal to every vector in another; it does not say that every pair of eigenvectors is orthogonal. In particular, two vectors belonging to the same eigenspace need not be orthogonal.
Worked Checks on Proposed Data
Worked Example: Eigenvalues and Eigenvectors of a Complex Normal Matrix
Consider $$ \mathbf{A}=\begin{bmatrix}1&i\\i&1\end{bmatrix}. $$ Check the claims that its eigenvalues are \(1+i\) and \(1-i\), with eigenvectors \(\mathbf{v}_{+}=(1,1)^{\top}\) and \(\mathbf{v}_{-}=(1,-1)^{\top}\).
First verify normality. The conjugate transpose is \(\mathbf{A}^{*}=\begin{bmatrix}1&-i\\-i&1\end{bmatrix}\), and direct multiplication gives $$ \mathbf{A}\mathbf{A}^{*} =\begin{bmatrix}2&0\\0&2\end{bmatrix} =\mathbf{A}^{*}\mathbf{A}. $$ So the normal-matrix constraints apply. The characteristic polynomial is $$ p_{\mathbf{A}}(\lambda) =\det\begin{bmatrix}1-\lambda&i\\i&1-\lambda\end{bmatrix} =(1-\lambda)^2-i^2 =(1-\lambda)^2+1 =(\lambda-(1+i))(\lambda-(1-i)). $$ The two proposed eigenvalues are distinct roots, each with algebraic multiplicity \(1\).
Now check each vector directly: $$ \mathbf{A}\mathbf{v}_{+} =\begin{bmatrix}1+i\\1+i\end{bmatrix} =(1+i)\mathbf{v}_{+}, \qquad \mathbf{A}\mathbf{v}_{-} =\begin{bmatrix}1-i\\i-1\end{bmatrix} =(1-i)\mathbf{v}_{-}. $$ Both vectors are nonzero, so both claims are verified. Their inner product is \(\langle\mathbf{v}_{+},\mathbf{v}_{-}\rangle=1\cdot\overline{1}+1\cdot\overline{-1}=1-1=0\), as required for distinct eigenvalues. Since each eigenspace has a nonzero vector and its algebraic multiplicity is \(1\), the multiplicity theorem also gives geometric multiplicity \(1\) for each.
Worked Example: A Repeated Eigenvalue and a False Eigenvector
Let $$ \mathbf{B}=\begin{bmatrix}3&0&0\\0&3&0\\0&0&-2i\end{bmatrix}. $$ Determine the multiplicities and check whether \(\mathbf{w}=(1,0,1)^{\top}\) is an eigenvector for \(-2i\).
The matrix is diagonal, so \(\mathbf{B}\mathbf{B}^{*}=\mathbf{B}^{*}\mathbf{B}\): both products are \(\operatorname{diag}(9,9,4)\). Its characteristic polynomial is $$ p_{\mathbf{B}}(\lambda)=(3-\lambda)^2(-2i-\lambda). $$ Thus \(3\) has algebraic multiplicity \(2\), and \(-2i\) has algebraic multiplicity \(1\). For \(\mathbf{x}=(x_1,x_2,x_3)^{\top}\), the equation \(\mathbf{B}\mathbf{x}=3\mathbf{x}\) gives \((-2i-3)x_3=0\). Since \(-2i-3\ne0\), \(x_3=0\), so \(E_3=\{(x_1,x_2,0)^{\top}:x_1,x_2\in\mathbb{C}\}\), of dimension \(2\). The equation \(\mathbf{B}\mathbf{x}=-2i\mathbf{x}\) gives \((3+2i)x_1=(3+2i)x_2=0\), so \(E_{-2i}=\{(0,0,x_3)^{\top}:x_3\in\mathbb{C}\}\), of dimension \(1\). Both geometric multiplicities match the algebraic multiplicities.
For the proposed vector, direct substitution gives $$ \mathbf{B}\mathbf{w} =\begin{bmatrix}3\\0\\-2i\end{bmatrix}, \qquad (-2i)\mathbf{w} =\begin{bmatrix}-2i\\0\\-2i\end{bmatrix}. $$ These are not equal, since their first entries differ; \(\mathbf{w}\) is not an eigenvector for \(-2i\). By contrast, \(\mathbf{e}_3=(0,0,1)^{\top}\) is: \(\mathbf{B}\mathbf{e}_3=(0,0,-2i)^{\top}=(-2i)\mathbf{e}_3\). As a further check, \(\mathbf{v}=(1,i,0)^{\top}\in E_3\) is orthogonal to \(\mathbf{e}_3\in E_{-2i}\), because \(\langle\mathbf{v},\mathbf{e}_3\rangle=0\).
Worked Example: Orthogonality Is Not Required Within One Eigenspace
Consider the normal matrix \(\mathbf{C}=\mathbf{I}_2\). Its characteristic polynomial is $$ p_{\mathbf{C}}(\lambda)=\det\begin{bmatrix}1-\lambda&0\\0&1-\lambda\end{bmatrix}=(1-\lambda)^2. $$ The only eigenvalue is \(1\), with algebraic multiplicity \(2\), and \(E_1=\mathbb{C}^2\), with geometric multiplicity \(2\).
Both \(\mathbf{u}=(1,0)^{\top}\) and \(\mathbf{v}=(1,1)^{\top}\) are eigenvectors, since \(\mathbf{C}\mathbf{u}=\mathbf{u}\) and \(\mathbf{C}\mathbf{v}=\mathbf{v}\). But \(\langle\mathbf{u},\mathbf{v}\rangle=1\cdot\overline{1}+0\cdot\overline{1}=1\ne0\). This is valid: both vectors have the same eigenvalue, so the distinct-eigenvalue orthogonality theorem does not apply. For comparison, \((1,0)^{\top}\) and \((0,1)^{\top}\) are also eigenvectors and are orthogonal. Orthogonality can occur within an eigenspace, but it is not forced for every pair of its vectors.
A Practical Order for Checking Spectral Claims
These checks are most reliable when performed in sequence. First establish that the matrix is normal, if normality is being used to justify a claim. For a concrete matrix, this means calculating \(\mathbf{A}\mathbf{A}^{*}\) and \(\mathbf{A}^{*}\mathbf{A}\) and comparing them. Next factor \(p_{\mathbf{A}}(\lambda)\) to identify the roots and their algebraic multiplicities. Then test each proposed eigenvector by calculating \(\mathbf{A}\mathbf{v}\) and comparing it with \(\lambda\mathbf{v}\). Finally, compare the claimed eigenspace dimensions with the algebraic multiplicities and check orthogonality across distinct eigenvalues.
Check that the matrix is square over \(\mathbb{C}\) and, when invoking normal-matrix constraints, that \(\mathbf{A}\mathbf{A}^{*}=\mathbf{A}^{*}\mathbf{A}\).
Factor \(p_{\mathbf{A}}\). A proposed eigenvalue must be a root, and its root multiplicity is its algebraic multiplicity.
Substitute each proposed vector into \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). For a normal matrix, verify that each eigenspace dimension matches the corresponding algebraic multiplicity.
Vectors from eigenspaces for distinct eigenvalues must have inner product zero. Do not impose this condition on every pair from a single eigenspace.
Common Pitfall: Requiring Too Much Orthogonality
A common incorrect conclusion is: “The matrix is normal, so all its eigenvectors are orthogonal.” Applied to \(\mathbf{C}=\mathbf{I}_2\), this would require \(\mathbf{u}=(1,0)^{\top}\) and \(\mathbf{v}=(1,1)^{\top}\) to be orthogonal. The calculation gives \(\langle\mathbf{u},\mathbf{v}\rangle=1\), not \(0\), so that conclusion is false. The correction is to check whether the eigenvalues are distinct: orthogonality is guaranteed between their eigenspaces only when they are distinct. Here both vectors have eigenvalue \(1\), so the theorem imposes no such condition.
Conversely, if proposed vectors belong to distinct eigenspaces of a normal matrix and their inner product is nonzero, the data are inconsistent: at least one of the eigenvector claims, eigenvalue assignments, or normality assumption must be wrong. Multiplicity comparisons give another independent diagnostic. Neither test replaces direct substitution, but together they check different parts of the proposed spectral description.
Check Your Understanding
For each question, distinguish what follows from the characteristic polynomial, what requires direct substitution, and what uses normality.
- A complex normal \(4\times4\) matrix has characteristic polynomial \((2-\lambda)^3(1+i-\lambda)\). What are the algebraic and geometric multiplicities of each eigenvalue?
- For a normal matrix, two proposed eigenvectors belong to eigenvalues \(2\) and \(2+i\). What must their inner product be?
- Can two nonorthogonal eigenvectors of a complex normal matrix have the same eigenvalue? Give an example or explain why not.
- For \(\mathbf{D}=\operatorname{diag}(1,1,3)\), find \(E_1\), \(E_3\), and the algebraic and geometric multiplicities.
- Why does finding that \(\lambda\) is a root of \(p_{\mathbf{A}}\) not verify a particular proposed eigenvector?
