Unitary Matrix - Definition, Properties, Examples

What Is A Unitary Matrix?

A unitary matrix is a square matrix $U$ with complex entries whose conjugate transpose is also its inverse. Written as a single condition:

$$U^H U = U U^H = I$$

Here $U^H$ (read "U-Hermitian," also written $U^*$ or $U^\dagger$) is the conjugate transpose: transpose the matrix, then replace every entry by its complex conjugate. $I$ is the identity matrix.

Variable glossary. A complex conjugate of $a + bi$ is $a - bi$ — flip the sign of the imaginary part. The conjugate transpose $U^H$ combines that conjugation with the ordinary transpose. $I$ is the square grid with 1s on the diagonal and 0s elsewhere.

Equivalently, $U$ is unitary exactly when $U^{-1} = U^H$. The defining feature is that its inverse costs nothing to compute: you conjugate and transpose, and you are done.

How Do You Check If A Matrix Is Unitary?

The test is direct: compute $U^H$, multiply it by $U$, and confirm the result is the identity.

The steps, each on its own line:

1. Transpose $U$ to get $U^T$.

2. Conjugate every entry of $U^T$ — flip the sign of each imaginary part — to get $U^H$.

3. Multiply $U^H U$.

4. If the product is $I$, the matrix is unitary.

A faster reading of the same test: a matrix is unitary when its columns are orthonormal — each column has length 1, and any two distinct columns are perpendicular under the complex inner product. The two views agree, because $U^H U = I$ is precisely the statement that column $i$ dotted with column $j$ gives 1 when $i = j$ and 0 otherwise.

What Are The Properties Of A Unitary Matrix?

The properties all flow from the single condition $U^H U = I$.

• It is always square and invertible. A non-square matrix cannot satisfy $U^H U = U U^H = I$, and the determinant is never zero, so the invertible matrix status is automatic.

• The inverse is the conjugate transpose: $U^{-1} = U^H$. No row reduction needed.

• The determinant has absolute value 1: $|\det U| = 1$. The determinant is a complex number on the unit circle, not necessarily $\pm 1$.

• Eigenvalues lie on the unit circle: every eigenvalue $\lambda$ satisfies $|\lambda| = 1$.

• The product of two unitary matrices is unitary. If $U$ and $V$ are unitary, so is $UV$.

• The inverse of a unitary matrix is unitary, and so is its conjugate transpose.

• It preserves length and inner products: $|Ux| = |x|$ for every vector $x$. This is the geometric heart of the definition.

One caution: the sum of two unitary matrices is generally not unitary. The family is closed under multiplication and inversion, not under addition.

Is A Unitary Matrix The Same As An Orthogonal Matrix?

This is the question that comes up most, and the answer is a clean "almost."

A unitary matrix is the complex-number version of an orthogonal matrix. An orthogonal matrix is a real matrix with $A^T A = I$ — its transpose is its inverse. A unitary matrix relaxes "real" to "complex" and replaces the plain transpose with the conjugate transpose.

The relationship is precise:

• If a unitary matrix happens to have only real entries, conjugation does nothing, so $U^H = U^T$, and the unitary condition collapses to the orthogonal condition.

• Therefore every real unitary matrix is orthogonal, and every orthogonal matrix is a real special case of unitary.

• Non-real unitary matrices exist that are not orthogonal — for instance, the diagonal matrix with $i$ on the diagonal.

So orthogonal sits inside unitary, the way the real numbers sit inside the complex numbers.

Examples Of A Unitary Matrix

Example 1

Show that $U = \begin{bmatrix} 0 & i \ i & 0 \end{bmatrix}$ is unitary.

First build $U^H$. Transpose, then conjugate (the conjugate of $i$ is $-i$):

$U^T = \begin{bmatrix} 0 & i \ i & 0 \end{bmatrix}$

$U^H = \begin{bmatrix} 0 & -i \ -i & 0 \end{bmatrix}$

Now multiply $U^H U$:

$U^H U = \begin{bmatrix} 0 & -i \ -i & 0 \end{bmatrix}\begin{bmatrix} 0 & i \ i & 0 \end{bmatrix} = \begin{bmatrix} (-i)(i) & 0 \ 0 & (-i)(i) \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$

Since $(-i)(i) = -i^2 = 1$, the product is $I$.

Final answer: $U^H U = I$, so $U$ is unitary.

Example 2

Is $U = \dfrac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1 \ i & -i \end{bmatrix}$ unitary?

The intuitive move is to skip the conjugation and just test $U^T U = I$, treating it like a real orthogonal check.

Wrong attempt. Using the plain transpose $U^T = \dfrac{1}{\sqrt 2}\begin{bmatrix} 1 & i \ 1 & -i \end{bmatrix}$:

$U^T U = \frac{1}{2}\begin{bmatrix} 1 & i \ 1 & -i \end{bmatrix}\begin{bmatrix} 1 & 1 \ i & -i \end{bmatrix} = \frac{1}{2}\begin{bmatrix} 1 + i^2 & 1 - i^2 \ 1 - i^2 & 1 + i^2 \end{bmatrix} = \frac{1}{2}\begin{bmatrix} 0 & 2 \ 2 & 0 \end{bmatrix}$

That is not the identity, which would suggest $U$ is not unitary.

Why it is wrong. The unitary test uses the conjugate transpose, not the plain transpose. With complex entries, skipping the conjugation tests the wrong condition.

Correct method. Conjugate when transposing: $U^H = \dfrac{1}{\sqrt 2}\begin{bmatrix} 1 & -i \ 1 & i \end{bmatrix}$.

$U^H U = \frac{1}{2}\begin{bmatrix} 1 & -i \ 1 & i \end{bmatrix}\begin{bmatrix} 1 & 1 \ i & -i \end{bmatrix} = \frac{1}{2}\begin{bmatrix} 1 + 1 & 1 - 1 \ 1 - 1 & 1 + 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$

Here $(-i)(i) = 1$ contributes the diagonal 2s, and the off-diagonal terms cancel.

Final answer: With the conjugate transpose, $U^H U = I$, so $U$ is unitary.

Example 3

Confirm $\begin{bmatrix} i & 0 \ 0 & i \end{bmatrix}$ is unitary and find its determinant.

Conjugate transpose: $U^H = \begin{bmatrix} -i & 0 \ 0 & -i \end{bmatrix}$.

$U^H U = \begin{bmatrix} (-i)(i) & 0 \ 0 & (-i)(i) \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$

The determinant of $U$ is $i \cdot i = i^2 = -1$, and $|-1| = 1$.

Final answer: $U$ is unitary, with $\det U = -1$ and $|\det U| = 1$.

Example 4

Verify a real unitary matrix is just orthogonal, using $U = \begin{bmatrix} 0 & 1 \ -1 & 0 \end{bmatrix}$.

All entries are real, so $U^H = U^T = \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix}$.

$U^H U = \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \ -1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$

The test reduces to the orthogonal condition $U^T U = I$, exactly as expected for a real matrix.

Final answer: $U$ is unitary, and because it is real, it is also orthogonal.

Example 5

Show the product of two unitary matrices is unitary, for unitary $U$ and $V$.

Use the product rule for the conjugate transpose, $(UV)^H = V^H U^H$:

$(UV)^H (UV) = V^H U^H U V$

Since $U$ is unitary, $U^H U = I$:

$= V^H I V = V^H V = I$

The last step uses $V$ unitary.

Final answer: $(UV)^H(UV) = I$, so $UV$ is unitary.

Example 6

Check that $U = \dfrac{1}{2}\begin{bmatrix} 1+i & 1-i \ 1-i & 1+i \end{bmatrix}$ is unitary.

Conjugate transpose: conjugate each entry ($\overline{1+i} = 1-i$) and transpose (the matrix is symmetric, so positions are unchanged):

$U^H = \dfrac{1}{2}\begin{bmatrix} 1-i & 1+i \ 1+i & 1-i \end{bmatrix}$

The diagonal of $U^H U$ uses $|1+i|^2 + |1-i|^2 = 2 + 2 = 4$, scaled by $\tfrac14$ to give 1. The off-diagonal entries combine $(1-i)(1-i) + (1+i)(1+i) = -2i + 2i = 0$.

$U^H U = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$

Final answer: $U^H U = I$, so $U$ is unitary.

Why Unitary Matrices Matter: "Transformations that never lose information"

A unitary matrix exists to capture the transformations that preserve length and angle in a complex vector space. Stretching, squashing, or collapsing dimensions all destroy information; a unitary transformation does none of those, so it is perfectly reversible — and its reverse is free to write down.

Where that property carries real weight:

• Quantum mechanics. Every evolution of a quantum state is a unitary transformation, because probabilities must always sum to 1. A non-unitary step would mean probability leaking away, which is physically impossible. The entire formalism rests on this.

• Signal processing. The discrete Fourier transform is unitary (up to scaling). That is why it can move a signal between the time domain and the frequency domain and back with no loss — the transform's inverse is its conjugate transpose.

• Numerical stability. Algorithms that decompose a matrix using unitary factors (the QR and SVD methods) stay numerically stable precisely because unitary steps never amplify rounding errors. Length preservation keeps the arithmetic well-behaved.

Where Students Slip On Unitary Matrices (And How To Fix It)

Mistake 1: Using the transpose instead of the conjugate transpose

Where it slips in: Testing a complex matrix with $U^T U$ as if it were real.

Don't do this: Compute $U^T U$ and conclude the matrix is or isn't unitary from that.

The correct way: Use $U^H U$, the conjugate transpose. The first instinct, especially for someone fresh off orthogonal matrices, is to reach for the plain transpose and skip the conjugation. On real matrices that is fine, because conjugation does nothing — which is exactly why the habit forms and then breaks the moment an $i$ appears.

Mistake 2: Expecting the determinant to be exactly 1

Where it slips in: Reading "$|\det U| = 1$" as "$\det U = 1$."

Don't do this: Reject a valid unitary matrix because its determinant came out as $i$ or $-1$.

The correct way: The determinant of a unitary matrix is a complex number of modulus 1 — it can be $1$, $-1$, $i$, or any point on the unit circle. The memorizer who stored "determinant is 1" misses that it is the absolute value that equals 1, not the determinant itself.

Mistake 3: Assuming the sum of unitary matrices is unitary

Where it slips in: Treating unitary like a property that survives addition.

Don't do this: Add two unitary matrices and assume the result is still unitary.

The correct way: The family is closed under multiplication and inversion, not addition. The second-guesser who notices that $I$ and $-I$ are both unitary but $I + (-I) = 0$ is not, has spotted the counterexample correctly.

Key Takeaways

• A unitary matrix is a complex square matrix with $U^H U = U U^H = I$, so its conjugate transpose is its inverse.

• To test for unitarity, transpose, conjugate every entry, multiply by the original, and check for the identity — or confirm the columns are orthonormal.

• A unitary matrix has $|\det U| = 1$, eigenvalues of modulus 1, and preserves vector length.

• A real unitary matrix is exactly an orthogonal matrix; orthogonal is the real special case of unitary.

• Unitary matrices underpin quantum mechanics, the Fourier transform, and numerically stable matrix algorithms.

A Practical Next Step

Practice these to make the conjugate-transpose test automatic, then return to the test section and write $U^H$ in full if the conjugation step keeps slipping:

1. Check whether $\begin{bmatrix} i & 0 \ 0 & -i \end{bmatrix}$ is unitary.

2. Find the determinant of any unitary matrix you build and confirm its modulus is 1.

3. Take a real rotation matrix and verify it satisfies the unitary condition.

To work through complex matrices with a teacher who builds the geometry first, work with an algebra tutor, a high school math tutor, or through math tutoring. Want a live Bhanzu trainer to walk through more unitary-matrix problems? Book a free demo class.

Read More

• Matrices and determinants — the wider context for the determinant and inverse facts used here.

• Symmetric matrix — the real cousin of the Hermitian matrix the FAQ contrasts with.

• Determinant of a matrix — how the $|\det U| = 1$ property is computed.

• Multiplication of matrices — the row-by-column rule behind the unitary product test.