Eigenvalues Under Length Preservation
A unitary operator preserves inner products, and therefore preserves lengths. An eigenvector is a nonzero vector whose direction is unchanged by the operator, although the vector may be multiplied by a scalar. Comparing the lengths of an eigenvector and its image places a strong restriction on that scalar: it must have modulus one. This restriction helps identify possible eigenvalues, while solving the eigenvector equation determines which of those possibilities actually occur.
Work over the complex numbers \(\mathbb{C}\). Let \(V\) be a complex inner product space, with inner product linear in its first slot, and let \(U:V\to V\) be unitary. We use the inner-product-preservation property \(\langle U\mathbf{u},U\mathbf{v}\rangle=\langle\mathbf{u},\mathbf{v}\rangle\). The norm is \(\|\mathbf{v}\|=\sqrt{\langle\mathbf{v},\mathbf{v}\rangle}\). For a matrix representation in an orthonormal basis, unitarity is expressed by \(\mathbf{U}^{*}\mathbf{U}=\mathbf{I}\), where \(\mathbf{U}^{*}\) is the conjugate transpose.
The equation defining \(E_\lambda\) is homogeneous and linear, so the eigenspace is a subspace. For example, if \(\mathbf{v},\mathbf{w}\in E_\lambda\) and \(a,b\in\mathbb{C}\), then \(U(a\mathbf{v}+b\mathbf{w})=aU\mathbf{v}+bU\mathbf{w}=\lambda(a\mathbf{v}+b\mathbf{w})\). The distinction between an eigenvalue and an eigenspace matters: the eigenvalue is a scalar, while its eigenspace is a set of vectors.
The Modulus-One Theorem
Proof. Since \(\lambda\) is an eigenvalue, there is an eigenvector \(\mathbf{v}\ne\mathbf{0}\) with \(U\mathbf{v}=\lambda\mathbf{v}\). Inner-product preservation gives \(\|U\mathbf{v}\|^2=\langle U\mathbf{v},U\mathbf{v}\rangle=\langle\mathbf{v},\mathbf{v}\rangle=\|\mathbf{v}\|^2\). On the other hand, the norm of a scalar multiple satisfies \(\|\lambda\mathbf{v}\|=|\lambda|\|\mathbf{v}\|\), so the eigenvector equation gives \(\|U\mathbf{v}\|^2=|\lambda|^2\|\mathbf{v}\|^2\). Hence \((|\lambda|^2-1)\|\mathbf{v}\|^2=0\). Because \(\mathbf{v}\ne\mathbf{0}\), its squared norm is positive, so \(|\lambda|^2=1\) and therefore \(|\lambda|=1\). \(\square\)
The theorem gives a necessary condition, not a guarantee: a scalar of modulus one need not be an eigenvalue of a particular operator. To find the eigenvalues and eigenspaces of a specific matrix, solve \(\mathbf{U}\mathbf{x}=\lambda\mathbf{x}\) and require \(\mathbf{x}\ne\mathbf{0}\). In the examples below the spaces are finite-dimensional, and vectors are written as columns.
Diagonal Unitary Matrices
A diagonal matrix acts coordinate by coordinate, so its eigenvalue equations can be read directly from its diagonal entries. Repeated entries matter: every coordinate position carrying the same diagonal value contributes a direction to that eigenvalue’s eigenspace.
Proof. The conjugate transpose of \(\mathbf{D}\) is \(\mathbf{D}^{*}=\operatorname{diag}(\overline{d_1},\dots,\overline{d_n})\). Thus \(\mathbf{D}^{*}\mathbf{D}=\operatorname{diag}(|d_1|^2,\dots,|d_n|^2)\). This product equals \(\mathbf{I}\) exactly when each \(|d_j|^2=1\), which is equivalent to \(|d_j|=1\). For the eigenspace statement, the equation \(\mathbf{D}\mathbf{x}=\lambda\mathbf{x}\) has coordinate equations \((d_j-\lambda)x_j=0\). If \(d_j\ne\lambda\), this forces \(x_j=0\); if \(d_j=\lambda\), it imposes no further restriction on \(x_j\). A nonzero solution exists exactly when at least one diagonal entry equals \(\lambda\), proving the eigenvalue claim as well. \(\square\)
Worked Example: A Diagonal Matrix with a Repeated Entry
Find the eigenvalues and eigenspaces of \(\mathbf{D}=\begin{bmatrix}1&0&0\\0&i&0\\0&0&i\end{bmatrix}\), where \(i^2=-1\).
The diagonal entries have modulus one: \(|1|=1\) and \(|i|=1\). The theorem therefore shows that \(\mathbf{D}\) is unitary. The coordinate equations for an eigenvalue \(\lambda\) are \((1-\lambda)x_1=0\), \((i-\lambda)x_2=0\), and \((i-\lambda)x_3=0\). For \(\lambda=1\), these equations give \(x_2=x_3=0\), while \(x_1\) is free. Thus \(E_1=\operatorname{span}\{(1,0,0)^{\top}\}\). For \(\lambda=i\), they give \(x_1=0\), while \(x_2,x_3\) are free. Thus \(E_i=\operatorname{span}\{(0,1,0)^{\top},(0,0,1)^{\top}\}\).
Check: \(\mathbf{D}(1,0,0)^{\top}=(1,0,0)^{\top}=1(1,0,0)^{\top}\), and \(\mathbf{D}(0,1,0)^{\top}=(0,i,0)^{\top}=i(0,1,0)^{\top}\). The same equation holds for \((0,0,1)^{\top}\), so every vector in the stated span for \(\lambda=i\) is indeed an eigenvector or the zero vector.
A Cyclic Permutation
A permutation matrix rearranges coordinates. Its eigenvectors are therefore not usually coordinate vectors: the eigenvalue equation instead requires a pattern among the coordinates. A three-coordinate cycle illustrates how the complex scalars make all its eigenvalues visible.
Worked Example: The Three-Cycle
Consider the operator on \(\mathbb{C}^3\) with matrix \(\mathbf{P}=\begin{bmatrix}0&0&1\\1&0&0\\0&1&0\end{bmatrix}\). It sends \((x_1,x_2,x_3)^{\top}\) to \((x_3,x_1,x_2)^{\top}\). Find its eigenvalues and eigenspaces.
The columns of \(\mathbf{P}\) are the standard basis vectors in a different order, so they are orthonormal; hence \(\mathbf{P}^{*}\mathbf{P}=\mathbf{I}\) and \(\mathbf{P}\) is unitary. The eigenvalue equation gives \(x_3=\lambda x_1,\quad x_1=\lambda x_2,\quad x_2=\lambda x_3\). If \(x_3=0\), the third equation gives \(x_2=0\), and then the second gives \(x_1=0\). Such a vector is zero, so any eigenvector must have \(x_3\ne0\). From the second and third equations, \(x_1=\lambda^2x_3\). Substitution into the first gives \(x_3=\lambda^3x_3\), and division by \(x_3\ne0\) yields \(\lambda^3=1\).
The three complex solutions are \(1,\omega,\omega^2\), where \(\omega=(-1+i\sqrt{3})/2\) and \(\omega^2=(-1-i\sqrt{3})/2\). For any one of these values, setting \(x_3=t\) in the equations gives \(x_2=\lambda t\), \(x_1=\lambda^2t\). Consequently \(E_\lambda=\operatorname{span}\{(\lambda^2,\lambda,1)^{\top}\}\) for each \(\lambda\in\{1,\omega,\omega^2\}\).
Check: For each of the three values, \(\lambda^3=1\), and \(\mathbf{P}(\lambda^2,\lambda,1)^{\top}=(1,\lambda^2,\lambda)^{\top}=\lambda(\lambda^2,\lambda,1)^{\top}\). Each displayed vector is nonzero, so each listed scalar really is an eigenvalue. The coordinate equations also showed that no other eigenvalue is possible.
Rotation and Scaling
Over \(\mathbb{C}^2\), a real planar rotation can be studied through complex eigenvectors. Multiplying the rotation matrix by a scalar changes eigenvalue moduli by the scalar’s modulus. This makes it possible to distinguish a unitary rotation from a rotation followed by a non-unit scaling.
Worked Example: A Rotation-Scaling Matrix
For real \(r>0\), let \(\mathbf{A}=r\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}\), acting on \(\mathbb{C}^2\). For \(\theta=\pi/3\), find its eigenvalues and eigenspaces, and determine when it is unitary.
Write \(c=\cos\theta\) and \(s=\sin\theta\). Direct multiplication gives \(\begin{bmatrix}c&-s\\s&c\end{bmatrix}(1,-i)^{\top}=(c+is,-i(c+is))^{\top}=e^{i\theta}(1,-i)^{\top}\). Similarly, \(\begin{bmatrix}c&-s\\s&c\end{bmatrix}(1,i)^{\top}=(c-is,i(c-is))^{\top}=e^{-i\theta}(1,i)^{\top}\). Thus \(\mathbf{A}\) has eigenvalues \(re^{i\theta}\) and \(re^{-i\theta}\), with eigenspaces spanned by the respective displayed vectors. For \(\theta=\pi/3\), these eigenvalues are \(r(1+i\sqrt{3})/2\) and \(r(1-i\sqrt{3})/2\).
The rotation matrix preserves the usual inner product, while multiplication by \(r\) multiplies every vector’s norm by \(r\). Since \(r>0\), \(\mathbf{A}\) preserves lengths exactly when \(r=1\), and it is unitary exactly in that case. For \(r=1\), both eigenvalues have modulus one, as required. If \(r=2\), the eigenvalues have modulus \(2\), and the matrix is not unitary.
Check: With \(\theta=\pi/3\), \(r=1\), and \(\mathbf{v}=(1,-i)^{\top}\), multiplication gives \(\mathbf{A}\mathbf{v}=((1+i\sqrt{3})/2)(1,-i)^{\top}\). The vector is nonzero, verifying the first eigenpair; the calculation above verifies the second.
What the Modulus Condition Does—and Does Not—Say
The modulus-one theorem is a quick way to rule out proposed eigenvalues of a unitary operator: if \(|\lambda|\ne1\), it cannot be an eigenvalue. It does not say that every scalar on the unit circle must occur. For example, the diagonal unitary matrix in the first worked example has eigenvalues \(1\) and \(i\), not every scalar of modulus one. Its eigenspace calculation shows precisely which coordinate positions are available for each eigenvalue.
Repeated eigenvalues also affect the eigenspace, not the modulus restriction. In the diagonal example, \(i\) appears twice and its eigenspace has two independent coordinate directions. The modulus theorem alone does not determine that dimension; it only constrains the possible scalar. Likewise, the cyclic-permutation calculation required solving the coordinate equations to identify both the eigenvalues and their eigenspaces.
Common Pitfall
Wrong conclusion: “The scalar \(e^{i\pi/4}\) has modulus one, so it must be an eigenvalue of every unitary matrix.” The modulus-one theorem proves only that every eigenvalue of a unitary operator has modulus one; it does not establish the converse. For the matrix \(\mathbf{D}=\begin{bmatrix}1&0&0\\0&i&0\\0&0&i\end{bmatrix}\), the eigenvalue equations for \(\lambda=e^{i\pi/4}\) are \((1-\lambda)x_1=0\), \((i-\lambda)x_2=0\), and \((i-\lambda)x_3=0\). Since \(\lambda\ne1\) and \(\lambda\ne i\), all three coordinates must be zero. Therefore \(E_\lambda=\{\mathbf{0}\}\), and \(\lambda\) is not an eigenvalue of this matrix despite having modulus one.
A second check is to verify proposed eigenvectors directly: compute \(U\mathbf{v}\) and compare it with \(\lambda\mathbf{v}\). A vector in an eigenspace must satisfy that equation, and an eigenvector must also be nonzero. These checks catch both incorrect coordinate patterns and the mistake of treating the zero vector as an eigenvector.
Check Your Understanding
Use the eigenvalue equation and the modulus-one theorem where appropriate.
- Prove that if \(U\) is unitary and \(U\mathbf{v}=\lambda\mathbf{v}\) for some nonzero \(\mathbf{v}\), then \(|\lambda|=1\).
- Find all eigenvalues and eigenspaces of \(\operatorname{diag}(-1,1,-1)\) acting on \(\mathbb{C}^3\).
- For the three-cycle matrix, verify directly that \((1,1,1)^{\top}\) is an eigenvector and state its eigenvalue.
- For \(r>0\), what are the eigenvalue moduli of \(r\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}\)? For which \(r\) is the matrix unitary?
- Explain why knowing that \(|\lambda|=1\) does not by itself prove that \(\lambda\) is an eigenvalue of a given unitary operator.
