Normality and Orthogonality
For a normal operator, eigenvectors associated with different eigenvalues cannot point in overlapping directions: they are orthogonal. This is a strong geometric consequence of the algebraic condition defining normality. It does not say that every pair of eigenvectors is orthogonal. Vectors in the same eigenspace need not be orthogonal, and the result concerns pairs whose eigenvalues are distinct.
We work over \(\mathbb{C}\) in a finite-dimensional inner-product space \(V\). The inner product is linear in its first slot and conjugate-linear in its second. The adjoint \(T^*\) of a linear operator \(T:V\to V\) is characterized by \(\langle T\mathbf{x},\mathbf{y}\rangle=\langle\mathbf{x},T^*\mathbf{y}\rangle\). The norm is \(\|\mathbf{x}\|=\sqrt{\langle\mathbf{x},\mathbf{x}\rangle}\), and vectors are orthogonal when their inner product is zero.
The key step is to apply normality not just to \(T\), but to a shifted operator. We first establish the norm fact needed for that step. It applies to any normal operator on an inner-product space.
Proof. By the adjoint identity, \(\|S\mathbf{x}\|^2=\langle S\mathbf{x},S\mathbf{x}\rangle=\langle\mathbf{x},S^*S\mathbf{x}\rangle\). Applying the adjoint identity to \(S^*\) gives \(\|S^*\mathbf{x}\|^2=\langle S^*\mathbf{x},S^*\mathbf{x}\rangle=\langle\mathbf{x},SS^*\mathbf{x}\rangle\). Since \(S^*S=SS^*\), these squared norms are equal. Norms are nonnegative, so equality of their squares gives \(\|S\mathbf{x}\|=\|S^*\mathbf{x}\|\). \(\square\)
Shifting a Normal Operator
Fix \(\lambda\in\mathbb{C}\) and set \(S=T-\lambda\operatorname{id}_V\), where \(\operatorname{id}_V\) is the identity operator on \(V\). The adjoint of \(S\) is \(S^*=T^*-\overline{\lambda}\operatorname{id}_V\). The conjugate on the scalar is essential: it follows from conjugate-linearity in the second slot of the inner product.
Proof. Write \(S=T-\lambda\operatorname{id}_V\). Expanding both products, using that scalar multiples of the identity commute with every operator, gives \(S^*S=T^*T-\overline{\lambda}T-\lambda T^*+|\lambda|^2\operatorname{id}_V\) and \(SS^*=TT^*-\lambda T^*-\overline{\lambda}T+|\lambda|^2\operatorname{id}_V\). Normality of \(T\) means \(T^*T=TT^*\), so the two displayed expressions are equal. Thus \(S^*S=SS^*\), which is the definition of normality for \(S\). \(\square\)
Now an eigenvector of \(T\) with eigenvalue \(\lambda\) lies in the kernel of \(T-\lambda\operatorname{id}_V\). For a normal operator, the lemma implies that a vector killed by the operator is also killed by its adjoint: if \(S\mathbf{x}=\mathbf{0}\), then \(\|S^*\mathbf{x}\|=\|S\mathbf{x}\|=0\), and a vector has norm zero only when it is zero. This observation produces an adjoint eigenvector relation without assuming that \(T\) is self-adjoint.
The Orthogonality Theorem
Proof. Take \(\mathbf{x}\in E_\lambda\) and \(\mathbf{y}\in E_\mu\). If either vector is zero, their inner product is zero by the inner-product axioms. Otherwise, the eigenvector equations are \(T\mathbf{x}=\lambda\mathbf{x}\) and \(T\mathbf{y}=\mu\mathbf{y}\). By the proposition, \(S=T-\mu\operatorname{id}_V\) is normal. Since \(S\mathbf{y}=\mathbf{0}\), the norm lemma gives \(\|S^*\mathbf{y}\|=\|S\mathbf{y}\|=0\), and hence \(S^*\mathbf{y}=\mathbf{0}\). As \(S^*=T^*-\overline{\mu}\operatorname{id}_V\), this says \(T^*\mathbf{y}=\overline{\mu}\mathbf{y}\). Now use the adjoint identity and the inner-product conventions: \(\lambda\langle\mathbf{x},\mathbf{y}\rangle=\langle T\mathbf{x},\mathbf{y}\rangle=\langle\mathbf{x},T^*\mathbf{y}\rangle=\langle\mathbf{x},\overline{\mu}\mathbf{y}\rangle=\mu\langle\mathbf{x},\mathbf{y}\rangle\). The last equality uses conjugate-linearity in the second slot. Therefore \((\lambda-\mu)\langle\mathbf{x},\mathbf{y}\rangle=0\). Since \(\lambda\ne\mu\), it follows that \(\langle\mathbf{x},\mathbf{y}\rangle=0\), as required. \(\square\)
In particular, the proof shows that if \(T\mathbf{x}=\lambda\mathbf{x}\) for a normal \(T\), then \(T^*\mathbf{x}=\overline{\lambda}\mathbf{x}\): apply the kernel argument to \(T-\lambda\operatorname{id}_V\). The theorem itself compares two different eigenspaces, and does not require the eigenvalues to be real.
What Orthogonal Eigenspaces Tell Us About Dimension
Distinct eigenspaces are not only pairwise orthogonal; together, any finite collection of them has no linear overlap. This is useful when comparing eigenspace dimensions with the dimension of the whole space. It is a bound, not a claim that the eigenspaces always fill \(V\).
Proof. For each \(j\), choose a basis of \(E_{\lambda_j}\). Take a linear combination of all the chosen basis vectors that equals \(\mathbf{0}\), and group its terms by eigenspace to obtain \(\mathbf{x}_1+\cdots+\mathbf{x}_k=\mathbf{0}\), where \(\mathbf{x}_j\in E_{\lambda_j}\). Fix an index \(r\). Take the inner product of this equation with \(\mathbf{x}_r\) in the second slot. The term \(\langle\mathbf{x}_r,\mathbf{x}_r\rangle\) remains, while every term \(\langle\mathbf{x}_j,\mathbf{x}_r\rangle\) for \(j\ne r\) is zero by the theorem. Thus \(\|\mathbf{x}_r\|^2=0\), so \(\mathbf{x}_r=\mathbf{0}\). This holds for every \(r\), so each group of basis coefficients is zero because the selected vectors form a basis of their eigenspace. The combined list of basis vectors is therefore linearly independent. A linearly independent list in \(V\) has at most \(\dim V\) members, giving the stated inequality. \(\square\)
Worked Example: A Repeated Eigenvalue
Let \(T:\mathbb{C}^3\to\mathbb{C}^3\) have matrix \(\mathbf{A}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&-1\end{bmatrix}\) in the standard orthonormal basis. Find the eigenspaces, their dimensions, and the orthogonality between them.
The matrix is self-adjoint because it is real and diagonal, so \(\mathbf{A}^*=\mathbf{A}\). Consequently \(\mathbf{A}^*\mathbf{A}=\mathbf{A}\mathbf{A}^*\), and \(T\) is normal. For \(\mathbf{x}=(x_1,x_2,x_3)^\top\), the equation \(\mathbf{A}\mathbf{x}=2\mathbf{x}\) becomes \((2x_1,2x_2,-x_3)^\top=(2x_1,2x_2,2x_3)^\top\), so \(3x_3=0\) and \(x_3=0\). Hence \(E_2=\{(x_1,x_2,0)^\top:x_1,x_2\in\mathbb{C}\}=\operatorname{span}\{\mathbf{e}_1,\mathbf{e}_2\}\), with dimension \(2\).
The equation \(\mathbf{A}\mathbf{x}=-\mathbf{x}\) gives \(3x_1=0\), \(3x_2=0\), and no further restriction on \(x_3\). Therefore \(E_{-1}=\operatorname{span}\{\mathbf{e}_3\}\), with dimension \(1\). For arbitrary \(\mathbf{x}=(a,b,0)^\top\in E_2\) and \(\mathbf{y}=(0,0,c)^\top\in E_{-1}\), \(\langle\mathbf{x},\mathbf{y}\rangle=a\overline{0}+b\overline{0}+0\overline{c}=0\). Thus the eigenspaces are orthogonal, and their dimensions add to \(2+1=3=\dim\mathbb{C}^3\).
Check: \(\mathbf{A}(a,b,0)^\top=(2a,2b,0)^\top=2(a,b,0)^\top\), while \(\mathbf{A}(0,0,c)^\top=(0,0,-c)^\top=-(0,0,c)^\top\). Both claimed eigenspaces satisfy their eigenvalue equations.
This example also shows why it is important to distinguish eigenvalues from eigenspace dimensions. The eigenvalue \(2\) has an eigenspace of dimension \(2\), so its eigenvectors do not all have to be pairwise orthogonal. For instance, \(\mathbf{e}_1\) and \(\mathbf{e}_1+\mathbf{e}_2\) both belong to \(E_2\), but their inner product is \(1\), not \(0\). The theorem guarantees orthogonality across distinct eigenvalues, not within one eigenspace.
Checking the Theorem in Matrix Examples
Worked Example: A Normal Matrix with Non-Real Eigenvalues
Consider the operator on \(\mathbb{C}^2\) with matrix \(\mathbf{A}=\begin{bmatrix}0&-1\\1&0\end{bmatrix}\). Verify normality, find eigenvectors for \(i\) and \(-i\), and check their orthogonality.
The conjugate transpose is \(\mathbf{A}^*=\begin{bmatrix}0&1\\-1&0\end{bmatrix}\). Multiplication gives \(\mathbf{A}^*\mathbf{A}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\) and \(\mathbf{A}\mathbf{A}^*=\begin{bmatrix}1&0\\0&1\end{bmatrix}\). Thus \(\mathbf{A}\) is normal. Set \(\mathbf{v}=(1,-i)^\top\). Then \(\mathbf{A}\mathbf{v}=(i,1)^\top=i(1,-i)^\top=i\mathbf{v}\). Set \(\mathbf{w}=(1,i)^\top\). Then \(\mathbf{A}\mathbf{w}=(-i,1)^\top=-i(1,i)^\top=-i\mathbf{w}\). The vectors are nonzero and their eigenvalues are distinct.
Using the inner product linear in the first slot, \(\langle\mathbf{v},\mathbf{w}\rangle=1\overline{1}+(-i)\overline{i}=1+(-i)(-i)=1-1=0\). So these eigenvectors are orthogonal, as the theorem predicts.
Check: The displayed matrix products establish \(\mathbf{A}^*\mathbf{A}=\mathbf{A}\mathbf{A}^*\), and direct multiplication establishes each eigenvector equation. The inner product calculation verifies orthogonality with the specified complex convention.
Worked Example: Why Normality Is Needed
Consider \(\mathbf{B}=\begin{bmatrix}1&1\\0&2\end{bmatrix}\) on \(\mathbb{C}^2\). Its eigenvectors for the distinct eigenvalues \(1\) and \(2\) are not orthogonal.
For \(\mathbf{u}=(1,0)^\top\), \(\mathbf{B}\mathbf{u}=(1,0)^\top=1\mathbf{u}\). For \(\mathbf{v}=(1,1)^\top\), \(\mathbf{B}\mathbf{v}=(2,2)^\top=2\mathbf{v}\). But \(\langle\mathbf{u},\mathbf{v}\rangle=1\overline{1}+0\overline{1}=1\ne0\). The missing hypothesis is normality: \(\mathbf{B}^*=\begin{bmatrix}1&0\\1&2\end{bmatrix}\), and direct products give \(\mathbf{B}^*\mathbf{B}=\begin{bmatrix}1&1\\1&5\end{bmatrix}\) whereas \(\mathbf{B}\mathbf{B}^*=\begin{bmatrix}2&2\\2&4\end{bmatrix}\). These are unequal, so the orthogonality theorem does not apply.
Check: Multiplying \(\mathbf{B}\) by each vector gives the eigenvalue equations above, and multiplying \(\mathbf{u}\) and \(\mathbf{v}\) in the inner product gives \(1\). The example is therefore a genuine failure of orthogonality for a non-normal operator, not a counterexample to the theorem.
Common Pitfall: The Conjugate in the Inner Product
A common error is to use the wrong scalar rule in the second slot. With the convention used here, \(\langle\mathbf{x},c\mathbf{y}\rangle=\overline{c}\langle\mathbf{x},\mathbf{y}\rangle\), not \(c\langle\mathbf{x},\mathbf{y}\rangle\). In the theorem's proof, replacing \(\langle\mathbf{x},\overline{\mu}\mathbf{y}\rangle\) by \(\overline{\mu}\langle\mathbf{x},\mathbf{y}\rangle\) is a wrong calculation; it would yield \((\lambda-\overline{\mu})\langle\mathbf{x},\mathbf{y}\rangle=0\), which does not follow from the hypotheses. The correction is \(\langle\mathbf{x},\overline{\mu}\mathbf{y}\rangle=\overline{\overline{\mu}}\langle\mathbf{x},\mathbf{y}\rangle=\mu\langle\mathbf{x},\mathbf{y}\rangle\). That is why the proof concludes with the factor \(\lambda-\mu\), which is nonzero for distinct eigenvalues.
The dimension bound is useful even before deciding whether a collection of eigenspaces fills the whole space: it rules out incompatible dimension claims. The theorem alone does not prove that every normal operator has enough eigenvectors to span \(V\). It proves that eigenspaces already known to exist for distinct eigenvalues are orthogonal and contribute independent directions. A stronger conclusion about spanning requires additional results.
Check Your Understanding
Use the inner product convention and the hypotheses carefully in each question.
- Why is \(T-\lambda\operatorname{id}_V\) normal whenever \(T\) is normal? Identify where the conjugate of \(\lambda\) occurs.
- If \(T\) is normal and \(T\mathbf{x}=\lambda\mathbf{x}\), what equation does \(\mathbf{x}\) satisfy involving \(T^*\)?
- For distinct eigenvalues of a normal operator, explain why the proof also covers the case in which one chosen vector is \(\mathbf{0}\).
- A normal operator on a \(5\)-dimensional complex inner-product space has eigenspaces of dimensions \(2\) and \(2\) for two distinct eigenvalues. What dimension bound follows, and what does it not establish?
- For the matrix \(\mathbf{B}\) in the final worked example, identify both the nonzero inner product and the failed normality condition.
