Hermitian, skew-Hermitian, and unitary matrices

Some matrices are special not because of what they compute but because of what their structure guarantees. A Hermitian matrix, a skew-Hermitian matrix, and a unitary matrix are the complex-number cousins of the familiar real symmetric, skew-symmetric, and orthogonal matrices — and each cousin inherits a strict, predictable eigenvalue behaviour. Hermitian eigenvalues are always real, skew-Hermitian eigenvalues are always purely imaginary, and unitary eigenvalues always sit on the unit circle, $|\lambda|=1$ . These guarantees are not curiosities: they are exactly the facts later diagonalization and quantum-mechanics-flavoured eigenvalue theory will cite without re-proving. This node builds the classification test, proves the eigenvalue guarantee from first principles using the complex inner product, and drills the pattern-recognition skill of reading a matrix's type straight off its conjugate transpose.

The core insight

The conjugate transpose $A^{*T}$ decides everything: $A=A^{*T}$ makes eigenvalues real, $A=-A^{*T}$ makes them purely imaginary, and $A^{*T}A=I$ pins them to the unit circle.

The core idea

For a complex matrix $A$ , the conjugate transpose $A^{*T}=\overline{A}^{T}$ is formed by transposing and conjugating every entry (the order does not matter — the two operations commute). This single object classifies three families at once.

Hermitian: $A=A^{*T}$ , i.e. $a_{ij}=\overline{a_{ji}}$ for every $i,j$ . Diagonal entries must be real, since $a_{ii}=\overline{a_{ii}}$ forces $a_{ii}\in\mathbb R$ .
Skew-Hermitian: $A=-A^{*T}$ , i.e. $a_{ij}=-\overline{a_{ji}}$ . Diagonal entries must be purely imaginary (or zero).
Unitary: $A^{*T}A=I$ , equivalently $A^{-1}=A^{*T}$ ; the columns of $A$ form an orthonormal set under the complex inner product $\langle \mathbf u,\mathbf v\rangle=\sum \overline{u_i}v_i$ .

When $A$ is real, $A^{*T}=A^T$ , so these collapse to the ordinary symmetric ( $A=A^T$ ), skew-symmetric ( $A=-A^T$ ), and orthogonal ( $A^TA=I$ ) matrices — the complex classes are genuine generalizations, not a different idea. The payoff is the eigenvalue guarantee for each family:

A=A^{*T}\ \Rightarrow\ \lambda\in\mathbb R,\qquad A=-A^{*T}\ \Rightarrow\ \lambda\in i\mathbb R,\qquad A^{*T}A=I\ \Rightarrow\ |\lambda|=1.

where: $\lambda$ is any eigenvalue of $A$ , $\mathbb R$ is the real line, and $i\mathbb R$ denotes the purely imaginary axis. A fourth guarantee rides along with the Hermitian/symmetric case: eigenvectors belonging to distinct eigenvalues of a Hermitian matrix are automatically orthogonal under $\langle\cdot,\cdot\rangle$ — a fact used constantly once diagonalization is developed, since it means the eigenvector basis can be chosen orthonormal for free. It helps to see these three tests as one family rather than three unrelated rules. All three compare $A$ (or a product built from $A$ ) against the identity or against $A$ itself, using the same operation — conjugate transpose. Hermitian and skew-Hermitian are mirror images of each other, differing only by a sign; unitary is a step up in strength, comparing a product $A^{*T}A$ to the identity rather than comparing $A$ to its own transpose. This is also why the classification step should always be done in a fixed order: transpose first, conjugate second (or the reverse — the result is identical), then compare the result to $A$ , to $-A$ , and finally check whether $A^{*T}A$ collapses to $I$ . Skipping the conjugate step is the single most common way this classification goes wrong on a complex matrix, since a matrix can easily satisfy $A^T=A$ (real-style symmetry) while failing $A^{*T}=A$ once its complex entries are properly conjugated.

Figure

The complex plane showing eigenvalue locations for the three matrix types, real axis points for Hermitian, imaginary axis points for skew-Hermitian, and unit circle points for unitary matrices

Why the eigenvalues are pinned down this way

Derivation AReal eigenvalues of a Hermitian matrix

The goal is to prove, using nothing beyond the definition $A=A^{*T}$ and the positivity of the complex inner product $\langle\mathbf v,\mathbf v\rangle=\mathbf v^{*T}\mathbf v=\sum|v_i|^2$ , that every eigenvalue of a Hermitian matrix must equal its own complex conjugate and is therefore a real number, with the skew-Hermitian and unitary cases following the identical pattern once the defining relation is swapped in.

Step 1. Start from the eigenvalue equation. Let $A\mathbf v=\lambda\mathbf v$ with $\mathbf v\neq\mathbf 0$ , and form the scalar $s=\mathbf v^{*T}A\mathbf v$ . Substituting the eigenvalue equation directly gives $s=\mathbf v^{*T}(\lambda\mathbf v)=\lambda\,\mathbf v^{*T}\mathbf v$ .

Step 2. Take the conjugate transpose of the same scalar. A $1\times1$ matrix equals its own transpose, so $s=\overline{s}$ is automatic once we compute $\overline s$ directly: $\overline{s}=\overline{\mathbf v^{*T}A\mathbf v}=\mathbf v^{*T}A^{*T}\mathbf v$ (conjugate-transposing a product reverses order and conjugate-transposes each factor). Since $A$ is Hermitian, $A^{*T}=A$ , so $\overline{s}=\mathbf v^{*T}A\mathbf v=s$ — consistent, and it also equals $\overline\lambda\,\mathbf v^{*T}\mathbf v$ by conjugating Step 1's result.

Step 3. Equate the two expressions for $s$ . From Step 1, $s=\lambda\,\mathbf v^{*T}\mathbf v$ ; from Step 2, $s=\overline\lambda\,\mathbf v^{*T}\mathbf v$ . Therefore $\lambda\,\mathbf v^{*T}\mathbf v=\overline\lambda\,\mathbf v^{*T}\mathbf v$ .

Step 4. Divide out the positive scalar. Since $\mathbf v\neq\mathbf 0$ , $\mathbf v^{*T}\mathbf v=\sum|v_i|^2>0$ , so it is safe to divide both sides by it, leaving $\lambda=\overline\lambda$ . A complex number equal to its own conjugate has zero imaginary part, so $\lambda\in\mathbb R$ .

Step 5. Read off the sibling results. Replacing $A^{*T}=A$ with $A^{*T}=-A$ throughout the same argument gives $\lambda=-\overline\lambda$ , i.e. $\lambda$ is purely imaginary — the skew-Hermitian case. Starting instead from $\langle A\mathbf v,A\mathbf v\rangle=\langle\mathbf v,\mathbf v\rangle$ (unitary matrices preserve length) and substituting $A\mathbf v=\lambda\mathbf v$ gives $\overline\lambda\lambda\,\mathbf v^{*T}\mathbf v=\mathbf v^{*T}\mathbf v$ , so $|\lambda|^2=1$ and $|\lambda|=1$ — the unitary case.

Worked example

Problem. Classify $A=\begin{pmatrix}2 & 3-i \\ 3+i & 5\end{pmatrix}$ and $U=\dfrac{1}{\sqrt2}\begin{pmatrix}1 & i\\ i & 1\end{pmatrix}$ , then verify each matrix's eigenvalues obey the guarantee proved above.

Step 1. Classify $A$ . Transpose: $A^T=\begin{pmatrix}2 & 3+i\\3-i & 5\end{pmatrix}$ ; conjugate each entry: $\overline{A^T}=\begin{pmatrix}2&3-i\\3+i&5\end{pmatrix}=A$ . Since $A^{*T}=A$ , $A$ is Hermitian (diagonal entries $2,5$ are real, as required).

Step 2. Find $A$ 's eigenvalues. $\det(A-\lambda I)=(2-\lambda)(5-\lambda)-(3-i)(3+i)=0$ . Since $(3-i)(3+i)=9-i^2=10$ , this expands to $\lambda^2-7\lambda+10-10=0$ , i.e. $\lambda(\lambda-7)=0$ , so $\lambda=0$ or $\lambda=7$ — both real, confirming the Hermitian guarantee.

Step 3. Classify $U$ . Transpose leaves it unchanged (it is symmetric); conjugating gives $U^{*T}=\dfrac{1}{\sqrt2}\begin{pmatrix}1&-i\\-i&1\end{pmatrix}$ . Multiplying, $U^{*T}U=\dfrac12\begin{pmatrix}2&0\\0&2\end{pmatrix}=I$ , so $U$ is unitary.

Step 4. Find $U$ 's eigenvalues. $\det(U-\lambda I)=\left(\tfrac1{\sqrt2}-\lambda\right)^2-\left(\tfrac{i}{\sqrt2}\right)^2=0$ . Since $\left(\tfrac{i}{\sqrt2}\right)^2=-\tfrac12$ , this gives $\left(\tfrac1{\sqrt2}-\lambda\right)^2=-\tfrac12$ , so $\lambda=\dfrac{1\mp i}{\sqrt2}$ . Checking modulus: $|\lambda|=\sqrt{\left(\tfrac1{\sqrt2}\right)^2+\left(\tfrac1{\sqrt2}\right)^2}=\sqrt1=1$ .

Both guarantees confirmed: $A$ (Hermitian) has real eigenvalues $\{0,7\}$ , and $U$ (unitary) has eigenvalues $\dfrac{1\mp i}{\sqrt2}$ , each of modulus exactly $1$ .

Competitive stretch

Problem. Without finding eigenvalues, decide whether $B=\begin{pmatrix}0 & 2+i\\ -2+i & 0\end{pmatrix}$ can have a real, nonzero eigenvalue.

Step 1. Classify $B$ first, since the eigenvalue guarantee follows immediately from the class. Transpose: $B^T=\begin{pmatrix}0&-2+i\\2+i&0\end{pmatrix}$ ; conjugate: $\overline{B^T}=\begin{pmatrix}0&-2-i\\2-i&0\end{pmatrix}$ . Compare to $-B=\begin{pmatrix}0&-2-i\\2-i&0\end{pmatrix}$ — they match, so $B^{*T}=-B$ and $B$ is skew-Hermitian.

Step 2. Apply the guarantee directly, no determinant needed. A skew-Hermitian matrix's eigenvalues are purely imaginary or zero, so $B$ cannot have a real, nonzero eigenvalue — the only real eigenvalue it could possibly own is $0$ . This is the practical payoff of the classification: once a matrix's type is identified, its eigenvalue behaviour is known instantly, without solving a single characteristic equation.

Common traps

Checking only $A^T=A$ (real symmetry) on a complex matrix and forgetting to conjugate — this misclassifies almost every genuinely complex Hermitian or non-Hermitian matrix.
Assuming a Hermitian matrix must have real entries — off-diagonal entries can be complex; only the diagonal entries and the eigenvalues are guaranteed real.
Mixing up the three guarantees: real eigenvalues go with Hermitian/symmetric, purely imaginary with skew-Hermitian/skew-symmetric, modulus-1 with unitary/orthogonal — they are three distinct results, not interchangeable.
Treating "orthogonal eigenvectors for distinct eigenvalues" as automatic for any matrix — it specifically requires Hermitian/symmetric structure; a generic matrix's eigenvectors need not be orthogonal even when the eigenvalues are distinct.

Ask the tutor: want to classify a matrix of your own, or walk through the skew-Hermitian and unitary eigenvalue proofs step by step? Ask the AI tutor.

Why does a real symmetric matrix count as Hermitian?

Because for a real matrix the conjugate transpose is just the ordinary transpose — conjugating a real number does nothing. So the Hermitian condition $A=A^{*T}$ reduces exactly to the symmetric condition $A=A^T$ when every entry is real. Real symmetric, real skew-symmetric, and real orthogonal matrices are not a separate topic; they are the special case of the complex definitions where the imaginary parts happen to vanish, and all the same eigenvalue guarantees apply.

Do I need to memorize a separate proof for skew-Hermitian and unitary eigenvalues?

No — the same three-step pattern from the derivation works for all three, only the defining relation you substitute changes. For skew-Hermitian, swap $A^{*T}=A$ for $A^{*T}=-A$ in Step 2 and you get $\lambda=-\overline\lambda$ instead of $\lambda=\overline\lambda$ . For unitary, start from length preservation, $\langle A\mathbf v,A\mathbf v\rangle=\langle\mathbf v,\mathbf v\rangle$ , and substitute the eigenvalue equation to get $|\lambda|^2=1$ . Learn the one argument, then track which defining relation you are plugging in.

If a matrix is neither Hermitian, skew-Hermitian, nor unitary, does it still have some eigenvalue guarantee?

No general guarantee applies — a matrix outside these three families can have any complex eigenvalues at all, and its eigenvectors for distinct eigenvalues need not be orthogonal. That is exactly why these three structural conditions matter: they are the specific algebraic tests that unlock the real, purely-imaginary, or modulus-1 conclusions. Always check $A^{*T}$ against $A$ and $-A$ , and $A^{*T}A$ against $I$ , before assuming any eigenvalue behaviour.