What a Unitary Operator Preserves
In a real vector space with the usual dot product, an orthogonal transformation preserves lengths, angles, and perpendicularity. In a complex inner-product space, the corresponding transformations are called unitary. The adjoint, rather than the ordinary transpose, is essential: complex conjugation is part of the inner product, so it must also appear in the matrix test.
Throughout this tutorial, \(V\) is a finite-dimensional inner-product space over \(\mathbb{C}\), and \(U:V\to V\) is a linear operator. The inner product is linear in its first slot and conjugate-linear in its second. The adjoint \(U^*\) is characterized by \(\langle U\mathbf{x},\mathbf{y}\rangle=\langle\mathbf{x},U^*\mathbf{y}\rangle\). The identity operator is denoted by \(I\). A list \((\mathbf{b}_1,\dots,\mathbf{b}_n)\) is an orthonormal basis if it is a basis and \(\langle\mathbf{b}_j,\mathbf{b}_k\rangle=1\) when \(j=k\), and \(0\) otherwise.
The two product identities in the definition express that \(U^*\) is a two-sided inverse of \(U\). In finite dimensions, preserving inner products also forces \(U\) to be onto, so one can prove that both identities hold even if one first obtains only \(U^*U=I\). The finite-dimensional hypothesis matters in that step.
The Equivalence Theorem
Proof. First suppose \(U^*U=UU^*=I\). By the adjoint identity, for all \(\mathbf{x},\mathbf{y}\in V\), \(\langle U\mathbf{x},U\mathbf{y}\rangle=\langle\mathbf{x},U^*U\mathbf{y}\rangle=\langle\mathbf{x},\mathbf{y}\rangle\). Thus (1) implies (2).
Now suppose (2) holds. Taking \(\mathbf{x}=\mathbf{y}\) shows that \(\|U\mathbf{x}\|^2=\|\mathbf{x}\|^2\). If \(U\mathbf{x}=\mathbf{0}\), this equality gives \(\|\mathbf{x}\|^2=0\), hence \(\mathbf{x}=\mathbf{0}\). Therefore \(U\) is injective. In a finite-dimensional space, an injective linear map from \(V\) to itself is onto: the images of a basis are linearly independent and, being a list of \(\dim V\) vectors, form a basis. Also, for all \(\mathbf{x},\mathbf{y}\in V\), \(\langle\mathbf{x},U^*U\mathbf{y}\rangle=\langle U\mathbf{x},U\mathbf{y}\rangle=\langle\mathbf{x},\mathbf{y}\rangle\). It follows that \(\langle\mathbf{x},(U^*U-I)\mathbf{y}\rangle=0\) for every \(\mathbf{x}\). Taking \(\mathbf{x}=(U^*U-I)\mathbf{y}\) gives \(\|(U^*U-I)\mathbf{y}\|^2=0\). Hence \(U^*U=I\). For any \(\mathbf{z}\in V\), surjectivity gives \(\mathbf{z}=U\mathbf{y}\) for some \(\mathbf{y}\), and then \(UU^*\mathbf{z}=UU^*U\mathbf{y}=U\mathbf{y}=\mathbf{z}\). Thus \(UU^*=I\), proving (1).
Next suppose (2) holds and \((\mathbf{b}_1,\dots,\mathbf{b}_n)\) is any orthonormal basis. Inner-product preservation gives \(\langle U\mathbf{b}_j,U\mathbf{b}_k\rangle=\langle\mathbf{b}_j,\mathbf{b}_k\rangle\), so the image list is orthonormal. Since \(U\) is onto, that list spans \(V\), and is therefore an orthonormal basis. This proves (2) implies (3), including the stronger assertion about every orthonormal basis.
Finally, suppose \(U\) maps an orthonormal basis \((\mathbf{b}_1,\dots,\mathbf{b}_n)\) to an orthonormal basis. Write \(\mathbf{x}=\sum_{j=1}^n a_j\mathbf{b}_j\) and \(\mathbf{y}=\sum_{j=1}^n c_j\mathbf{b}_j\). Linearity and orthonormality give \(\langle U\mathbf{x},U\mathbf{y}\rangle=\sum_{j=1}^n a_j\overline{c_j}=\langle\mathbf{x},\mathbf{y}\rangle\). Thus (3) implies (2). The implications already proved give (1), completing the equivalence. \(\square\)
The proof distinguishes preservation from surjectivity. Inner-product preservation first gives injectivity; finite dimensionality then supplies surjectivity. In an infinite-dimensional space, that step cannot be assumed without further argument.
Matrix Tests in \(\mathbb{C}^n\)
On \(\mathbb{C}^n\), use \(\langle\mathbf{x},\mathbf{y}\rangle=\sum_{k=1}^n x_k\overline{y_k}\). For a matrix \(\mathbf{A}\), its conjugate transpose \(\mathbf{A}^*\) is obtained by transposing and conjugating each entry. A linear operator \(U:\mathbb{C}^n\to\mathbb{C}^n\) has a square matrix \(\mathbf{A}\), and \(U^*\) has matrix \(\mathbf{A}^*\). The theorem therefore gives the main matrix criterion.
Proof. The operator theorem proves the equivalence of the adjoint identities and inner-product preservation. The columns of \(\mathbf{A}\) are the images of the standard basis vectors \(\mathbf{e}_1,\dots,\mathbf{e}_n\), which form an orthonormal basis. The theorem therefore identifies the column condition with unitarity. Applying the same column condition to \(\mathbf{A}^*\), or reading the entries of \(\mathbf{A}\mathbf{A}^*=\mathbf{I}_n\), gives the row condition: distinct rows have inner product zero and each row has norm one. \(\square\)
For a direct column check, \((\mathbf{A}^*\mathbf{A})_{jk}=\sum_{\ell=1}^n\overline{a_{\ell j}}a_{\ell k}\). This entry is the inner product of column \(k\) with column \(j\), in the stated convention. Thus the product equals \(\mathbf{I}_n\) exactly when the columns are orthonormal. Because the matrix is square, those \(n\) orthonormal columns form a basis. The row test comes from \(\mathbf{A}\mathbf{A}^*=\mathbf{I}_n\).
Worked Example: A Diagonal Unitary Matrix
Test \(\mathbf{A}=\begin{bmatrix} i&0\\0&-1\end{bmatrix}\), where \(i^2=-1\). Conjugating and transposing gives
Multiplying in both orders yields
Both products equal \(\mathbf{I}_2\), so \(\mathbf{A}\) is unitary. As a check on the inner-product meaning, its columns are \((i,0)^{\top}\) and \((0,-1)^{\top}\), each of norm one and orthogonal to the other. For \(\mathbf{x}=(x_1,x_2)^{\top}\) and \(\mathbf{y}=(y_1,y_2)^{\top}\), the transformed inner product is \((ix_1)\overline{iy_1}+(-x_2)\overline{-y_2}=x_1\overline{y_1}+x_2\overline{y_2}=\langle\mathbf{x},\mathbf{y}\rangle\).
Worked Example: A Unitary Matrix with Non-Real Entries
Test the matrix \(\mathbf{B}=\frac{1}{\sqrt{2}}\begin{bmatrix}1&i\\i&1\end{bmatrix}\). Its conjugate transpose is \(\mathbf{B}^*=\frac{1}{\sqrt{2}}\begin{bmatrix}1&-i\\-i&1\end{bmatrix}\). Compute the first product entry by entry:
For the reverse product, the diagonal entries are \(1+i(-i)=2\) and \((-i)i+1=2\), while the off-diagonal entries are \(i-i=0\) and \(-i+i=0\). Therefore \(\mathbf{B}\mathbf{B}^*=\mathbf{I}_2\) as well. The columns \(\frac{1}{\sqrt{2}}(1,i)^{\top}\) and \(\frac{1}{\sqrt{2}}(i,1)^{\top}\) each have squared norm \(\frac12(1+1)=1\); their inner product is \(\frac12(1\cdot\overline{i}+i\cdot\overline{1})=\frac12(-i+i)=0\). The column check agrees with both matrix products, so \(\mathbf{B}\) is unitary.
Worked Example: A Matrix That Does Not Preserve Inner Products
Consider \(\mathbf{C}=\begin{bmatrix}1&1\\0&1\end{bmatrix}\), which is real and hence has \(\mathbf{C}^*=\mathbf{C}^{\top}\). Its product is
The column norms and orthogonality also fail: the columns \((1,0)^{\top}\) and \((1,1)^{\top}\) are not orthogonal. Directly, with \(\mathbf{x}=\mathbf{e}_1\) and \(\mathbf{y}=\mathbf{e}_2\), the original inner product is \(0\), but \(\mathbf{C}\mathbf{x}=(1,0)^{\top}\), \(\mathbf{C}\mathbf{y}=(1,1)^{\top}\), and \(\langle\mathbf{C}\mathbf{x},\mathbf{C}\mathbf{y}\rangle=1\). This confirms the failure by substitution.
Common Mistakes and Why the Criteria Matter
Using transpose instead of conjugate transpose. For the matrix \(\mathbf{A}\) in the first example, the ordinary transpose is \(\mathbf{A}^{\top}=\begin{bmatrix}i&0\\0&-1\end{bmatrix}\). A wrong test computes \(\mathbf{A}^{\top}\mathbf{A}=\begin{bmatrix}-1&0\\0&1\end{bmatrix}\) and concludes that \(\mathbf{A}\) is not unitary. The calculation is correct for that wrong product, but the criterion uses \(\mathbf{A}^*\), not \(\mathbf{A}^{\top}\). The corrected calculation is \(\mathbf{A}^*\mathbf{A}=\mathbf{I}_2\), as verified above. For real matrices, conjugation changes nothing; for complex matrices, it can change the answer.
Confusing a one-sided identity with the square-matrix test. A rectangular matrix can satisfy \(\mathbf{A}^*\mathbf{A}=\mathbf{I}\) without satisfying \(\mathbf{A}\mathbf{A}^*=\mathbf{I}\). For example, \(\mathbf{D}=\begin{bmatrix}1\\0\end{bmatrix}\) has \(\mathbf{D}^*\mathbf{D}=\begin{bmatrix}1\end{bmatrix}\), but \(\mathbf{D}\mathbf{D}^*=\begin{bmatrix}1&0\\0&0\end{bmatrix}\ne\mathbf{I}_2\). It is not a unitary operator from \(\mathbb{C}^2\) to itself: it is not even square. In the theorem, the operator maps a finite-dimensional space to itself; for a square matrix, either product identity implies the other, as the proof of the theorem shows.
The equivalent tests offer different practical routes. The products \(\mathbf{A}^*\mathbf{A}\) and \(\mathbf{A}\mathbf{A}^*\) check the adjoint definition directly; column or row inner products can be quicker when most entries are zero. The inner-product formulation explains the geometric effect: norms are preserved by setting \(\mathbf{x}=\mathbf{y}\), and orthogonality is preserved because a zero inner product remains zero. These facts make unitary transformations useful for changing coordinates in complex inner-product spaces without changing lengths or inner-product relationships.
Check Your Understanding
Use the definitions and the equivalence theorem to justify each conclusion.
- For \(\mathbf{Q}=\frac{1}{\sqrt{2}}\begin{bmatrix}1&1\\1&-1\end{bmatrix}\), compute \(\mathbf{Q}^*\mathbf{Q}\) and decide whether \(\mathbf{Q}\) is unitary.
- Explain why a unitary matrix must send every orthonormal basis to an orthonormal basis.
- Find a nonzero vector \(\mathbf{x}\) that demonstrates \(\mathbf{C}=\begin{bmatrix}1&1\\0&1\end{bmatrix}\) does not preserve lengths.
- For \(\mathbf{D}=\begin{bmatrix}1\\0\end{bmatrix}\), calculate both \(\mathbf{D}^*\mathbf{D}\) and \(\mathbf{D}\mathbf{D}^*\), and explain why the dimensions matter.
- State why replacing \(\mathbf{A}^*\) with \(\mathbf{A}^{\top}\) is not a valid unitary-matrix test over \(\mathbb{C}\).
