Hermitian and skew-Hermitian matrices

At a Glance

• Frequency: 3 sub-parts across 2 of 14 years (2013, 2016)
• Priority tier: T3
• Marks (count): 10 (1), 8 (2)
• Average solve time: ~6 min
• Difficulty mix: medium 2, easy 1
• Section: A | Dominant type: proof

Why This Chapter Matters

Hermitian matrix questions are short proof items (8–10 marks) that test a small number of standard arguments. Three proofs cover every past question: the “sandwich” argument that Hermitian eigenvalues are real, the cyclic trace identity that gives $\text{tr}(AA^*)=\text{tr}(A^*A)$, and the orthogonality-implies-independence chain for distinct-eigenvalue eigenvectors. All three can be written in under 8 minutes once the structure is internalised. These also appear as stepping stones inside larger eigenvalue and diagonalisation problems.

Minimum Theory

Definitions. The conjugate transpose of a matrix $A$ is $A^*=\bar{A}^\top$ (complex conjugate of each entry, then transpose). A square matrix is Hermitian if $A^*=A$; skew-Hermitian if $A^*=-A$.

Key properties. (i) $AA^*$ and $A^*A$ are both Hermitian for any $A$, since $(AA^*)^*=A^{**}A^*=AA^*$. (ii) Hermitian matrices have real eigenvalues; skew-Hermitian matrices have purely imaginary (or zero) eigenvalues. (iii) The inner product $\langle u,v\rangle=u^*v$ satisfies $\langle Au,v\rangle=\langle u,A^*v\rangle$ — this identity drives the eigenvalue-reality proof.

Trace identity. $\text{tr}(AA^*)=\sum_{i,j}|A_{ij}|^2=\|A\|_F^2=\text{tr}(A^*A)$. The quick proof: expand $(AA^*)_{ii}=\sum_j|A_{ij}|^2$ and sum over $i$; the result is $\sum_{i,j}|A_{ij}|^2$, symmetric in swapping $AA^*\leftrightarrow A^*A$.

Orthogonality of eigenvectors. For a Hermitian matrix with distinct eigenvalues $\lambda_i\ne\lambda_j$, the corresponding eigenvectors satisfy $\langle X_i,X_j\rangle=0$. Proof: $\langle AX_i,X_j\rangle=\lambda_i\langle X_i,X_j\rangle=\lambda_j\langle X_i,X_j\rangle$ (using $A^*=A$ and real eigenvalues), so $(\lambda_i-\lambda_j)\langle X_i,X_j\rangle=0$.

Question Archetypes

hermitian-real-eigenvalues (2 question(s); 2013, 2016)

Recognition Cues

• “Show the eigenvalues of $AA^*$ (or $A^*A$, or a Hermitian matrix $H$) are real.”
• “Show $\text{tr}(AA^*)=\text{tr}(A^*A)$.”
• The word “adjoint” in UPSC Linear Algebra means conjugate transpose $A^*$, not the adjugate matrix.

Solution Template

1. Show the matrix is Hermitian. For $AA^*$: compute $(AA^*)^*=A^{**}A^*=AA^*$. For a given $H$: use $H^*=H$.
2. Sandwich argument. Let $\lambda$ be an eigenvalue, $Hv=\lambda v$, $v\ne 0$. Form $v^*Hv$:
   • From the left: $v^*Hv=v^*(\lambda v)=\lambda\|v\|^2$.
   • From the Hermitian property: $(v^*Hv)^*=v^*H^*v=v^*Hv$, so $v^*Hv$ is real.
3. Cancel $\|v\|^2>0$. Conclude $\lambda\in\mathbb R$.
4. For the trace identity. Expand $(AA^*)_{ii}=\sum_j|A_{ij}|^2$; sum over $i$. Then expand $(A^*A)_{jj}=\sum_i|A_{ij}|^2$; sum over $j$. Both equal $\sum_{i,j}|A_{ij}|^2$.

Worked Example

2013 Paper 1, 2013-P1-Q1b (10 marks)

Let $A$ be a square matrix with conjugate transpose $A^*$. Show the eigenvalues of $AA^*$ and $A^*A$ are real, and show $\text{tr}(AA^*)=\text{tr}(A^*A)$.

Step 1 — Both matrices are Hermitian.

$(AA^*)^*=(A^*)^{*}A^*=AA^*$, so $AA^*$ is Hermitian. Similarly $(A^*A)^*=A^*(A^*)^*=A^*A$, so $A^*A$ is Hermitian.

Step 2 — Hermitian matrices have real eigenvalues. Let $H$ be Hermitian ($H^*=H$) with eigenvalue $\lambda$ and eigenvector $v\ne 0$. Pre-multiply $Hv=\lambda v$ by $v^*$:

$v^*Hv=\lambda\|v\|^2.\qquad(1)$

Take the conjugate transpose of the scalar $v^*Hv$:

$(v^*Hv)^*=v^*H^*(v^{**})=v^*Hv.$

So $v^*Hv$ equals its own conjugate, hence is real. Since $\|v\|^2>0$, (1) gives $\lambda=\dfrac{v^*Hv}{\|v\|^2}\in\mathbb R$.

Applying this to $H=AA^*$ and $H=A^*A$ completes the first part.

Step 3 — Trace identity.

$(AA^*)_{ii}=\sum_j A_{ij}(A^*)_{ji}=\sum_j A_{ij}\overline{A_{ij}}=\sum_j|A_{ij}|^2.$

$\text{tr}(AA^*)=\sum_i\sum_j|A_{ij}|^2.$

$(A^*A)_{jj}=\sum_i(A^*)_{ji}A_{ij}=\sum_i|A_{ij}|^2.$

$\text{tr}(A^*A)=\sum_j\sum_i|A_{ij}|^2=\sum_i\sum_j|A_{ij}|^2=\text{tr}(AA^*).$

$\boxed{\;\text{tr}(AA^*)=\text{tr}(A^*A)=\|A\|_F^2.\;}\qquad\blacksquare$

2016 Paper 1, 2016-P1-Q2b-ii (8 marks)

Prove that eigenvalues of a Hermitian matrix are all real.

Step 1 — Setup. Let $A^*=A$. Let $Av=\lambda v$, $v\ne 0$.

Step 2 — Sandwich. Pre-multiply by $v^*$:

$v^*Av=\lambda\|v\|^2.\qquad(\star)$

Step 3 — Hermitian forces $v^*Av$ to be real. Since $v^*Av$ is $1\times 1$, it equals its own conjugate transpose:

$(v^*Av)^*=v^*A^*v=v^*Av.$

So $v^*Av$ is real; its conjugate equals itself.

Step 4 — Conclude $\lambda\in\mathbb R$. Taking the conjugate of $(\star)$: $\overline{v^*Av}=\overline\lambda\|v\|^2$. Since $v^*Av$ is real, $v^*Av=\overline{v^*Av}$, giving $\lambda\|v\|^2=\overline\lambda\|v\|^2$. As $\|v\|^2>0$:

$\lambda=\overline\lambda,\quad\text{i.e.}\quad\lambda\in\mathbb R.\qquad\blacksquare$

Common Traps

• “Adjoint” in UPSC means conjugate transpose $A^*=\bar A^\top$, not the adjugate (matrix of cofactors). If you use the wrong definition, the entire argument collapses.
• In the sandwich: write $v^*Av$, not $v^\top Av$. Over $\mathbb C$, the plain transpose gives a bilinear form, not the Hermitian inner product, and $v^\top Av$ need not be real.
• Must explicitly state $\|v\|^2=v^*v>0$ (eigenvector is nonzero) before cancelling.
• Skew-Hermitian ($A^*=-A$) gives purely imaginary eigenvalues (the same argument gives $\lambda=-\overline\lambda$, i.e.\ $\text{Re}(\lambda)=0$). Do not confuse the two cases.

hermitian-eigenvector-independence (1 question(s); 2013)

Recognition Cues

• “Hermitian matrix with distinct eigenvalues $\lambda_1,\ldots,\lambda_n$; show the eigenvector matrix $C$ is non-singular.”
• The proof uses: orthogonality $\Rightarrow$ independence $\Rightarrow$ full rank $\Rightarrow$ non-singular.

Solution Template

1. Prove pairwise orthogonality. For $i\ne j$: compute $\langle AX_i,X_j\rangle$ two ways (eigenvalue of $X_i$; Hermitian move to $A^*=A$ giving eigenvalue of $X_j$). Get $(\lambda_i-\lambda_j)\langle X_i,X_j\rangle=0$; since $\lambda_i\ne\lambda_j$, conclude $\langle X_i,X_j\rangle=0$.
2. Prove linear independence. Suppose $\sum c_k X_k=0$. Inner-product with $X_k$: all cross terms vanish (Step 1), leaving $c_k\|X_k\|^2=0$, so $c_k=0$.
3. Conclude non-singularity. Columns of $C$ are linearly independent, so $\text{rank}(C)=n$, so $\det C\ne 0$.

Worked Example

2013 Paper 1, 2013-P1-Q2c-i (8 marks)

Let $A$ be Hermitian with distinct eigenvalues $\lambda_1,\ldots,\lambda_n$ and corresponding eigenvectors $X_1,\ldots,X_n$. Let $C$ be the matrix with $k$-th column $X_k$. Show $C$ is non-singular.

Step 1 — Orthogonality. Take $i\ne j$. Compute $\langle AX_i,X_j\rangle$ two ways:

Way 1: $AX_i=\lambda_i X_i$, so $\langle AX_i,X_j\rangle=\lambda_i\langle X_i,X_j\rangle$.

Way 2: using $A^*=A$: $\langle AX_i,X_j\rangle=\langle X_i,A^*X_j\rangle=\langle X_i,AX_j\rangle=\langle X_i,\lambda_j X_j\rangle=\overline{\lambda_j}\langle X_i,X_j\rangle$.

Since eigenvalues of $A$ are real (Hermitian), $\overline{\lambda_j}=\lambda_j$. Equating:

$(\lambda_i-\lambda_j)\langle X_i,X_j\rangle=0.$

As $\lambda_i\ne\lambda_j$, we get $\langle X_i,X_j\rangle=0$.

Step 2 — Linear independence. Suppose $\sum_{k=1}^n c_k X_k=0$. Take inner product with $X_m$:

$0=\sum_{k=1}^n c_k\langle X_k,X_m\rangle=c_m\|X_m\|^2$

since $\langle X_k,X_m\rangle=0$ for $k\ne m$ (Step 1). As $\|X_m\|^2>0$, we get $c_m=0$ for each $m$.

Step 3 — Non-singularity. The $n$ columns of $C$ are linearly independent, so $C$ has rank $n$, so $\det C\ne 0$, i.e.\ $C$ is non-singular.

$\boxed{\;C\text{ is non-singular.}\;}\qquad\blacksquare$

Common Traps

• The hypothesis distinct eigenvalues is essential. Without it, eigenvectors for the same eigenvalue need not be orthogonal (though they can be made so by Gram-Schmidt). Always invoke distinctness explicitly.
• The inner product in Way 2 is $\langle X_i,AX_j\rangle=\langle X_i,\lambda_j X_j\rangle=\overline{\lambda_j}\langle X_i,X_j\rangle$, not $\lambda_j\langle X_i,X_j\rangle$. The conjugate $\overline{\lambda_j}$ arises from the second slot. Then use real eigenvalues to drop the bar.
• In Step 2, after inner-producting with $X_m$, all cross terms vanish by Step 1, leaving only the $k=m$ term. Write this out explicitly.

Marks-Aware Writing

8-mark questions (2013-Q2c-i, 2016-Q2b-ii): Each step (orthogonality, or sandwich argument) should be one clear paragraph with the inner-product computation written out in full. An answer that states “Hermitian matrices have real eigenvalues” without proof earns 0 marks for that part. For the eigenvector independence question: Steps 1, 2, and 3 carry roughly 4 + 2 + 2 marks. Missing Step 2 (the linear independence argument) and jumping directly to “non-singular” loses 2 marks.

10-mark question (2013-Q1b): Part 1 (real eigenvalues) and Part 2 (trace identity) each carry 5 marks. For the trace identity: the computation $\text{tr}(AA^*)=\sum_{i,j}|A_{ij}|^2=\text{tr}(A^*A)$ is 4 marks; citing the Frobenius norm interpretation (optional) adds context but is not required for full marks.

Practice Set

• 2013-P1-Q2b-i (8 m) — — Hint: show every square matrix can be written as $H+S$ where $H=(A+A^*)/2$ is Hermitian and $S=(A-A^*)/2$ is skew-Hermitian; verify directly.
• 2021-P1-Q4a-ii (10 m) — — Hint: the same sandwich argument applies; check whether the given matrix satisfies $A^*=A$.