Orthogonal and unitary matrices

At a Glance

• Frequency: 3 sub-parts across 3 of 13 years (2014, 2019, 2020)
• Priority tier: T3
• Marks (count): 15 (1), 20 (1), 7 (1)
• Average solve time: ~9 min
• Difficulty mix: medium 2, easy 1
• Section: A | Dominant type: proof

Why This Chapter Matters

Orthogonal and unitary matrices appear roughly once every four years in Paper 1, always as a proof-type question. The 2020 Householder reflector question is the most marks-dense item in this topic (20 marks, four sub-parts) and rewards anyone who knows the two key identities cold: $u^Tu=1$ (inner product) and $uu^Tuu^T=uu^T$ (rank-1 collapse). The 2019 singularity question is a one-trick proof using the determinant factorisation $A+B=A(A+B)^TB$; once you see it, it takes under five minutes. The 2014 unitary eigenvalue proof is a three-line argument that examiners accept at 7 marks — easiest possible full-marks return.

Minimum Theory

Orthogonal matrices. A real matrix $A$ is orthogonal if $AA^T = A^TA = I$. Equivalently, its columns (and rows) form an orthonormal set. For any orthogonal $A$: $\det A = \pm 1$, $A^{-1}=A^T$, and $A$ preserves Euclidean norms: $\|Av\|=\|v\|$.

Unitary matrices. A complex matrix $U$ is unitary if $UU^* = U^*U = I$, where $U^*=\overline{U}^T$ is the conjugate transpose. Unitary matrices preserve the standard inner product: $\langle Uu, Uv\rangle = \langle u,v\rangle$. Every unitary eigenvalue satisfies $|\lambda|=1$. Real unitary matrices are exactly orthogonal matrices.

Householder reflector. For a real unit column vector $u$ (so $u^Tu=1$), the matrix $H=I-2uu^T$ is symmetric ($H^T=H$), orthogonal ($H^TH=I$), and an involution ($H^2=I$). Its trace is $n-2$, because $\operatorname{tr}(uu^T)=u^Tu=1$. The key simplification throughout: $uu^T\cdot uu^T = u(u^Tu)u^T = uu^T$.

Question Archetypes

householder-properties (1 question(s); 2020)

Recognition Cues

• Question defines $A=I-2uu^T$ (or $H=I-2uu^T$) where $u$ is a unit column vector.
• Asks you to check symmetry and/or orthogonality.
• May ask for trace, or to compute the explicit matrix for a given $u$.
• The phrase “reflector” or “Householder” may or may not appear.

Solution Template

1. State the unit-vector condition $u^Tu=1$ (inner product, scalar) versus $uu^T$ (outer product, rank-1 matrix).
2. Symmetry: Transpose: $(I-2uu^T)^T = I - 2(uu^T)^T = I-2uu^T = A$. Done.
3. Orthogonality: Use $A^T=A$, so check $A^2=I$. Expand $(I-2uu^T)^2$; the crucial step: $uu^T\cdot uu^T = u(u^Tu)u^T = uu^T$. So $A^2 = I - 4uu^T + 4uu^T = I$.
4. Trace: $\operatorname{tr}(A)=\operatorname{tr}(I)-2\operatorname{tr}(uu^T)=n-2\cdot1=n-2$ (use cyclic trace: $\operatorname{tr}(uu^T)=\operatorname{tr}(u^Tu)=u^Tu=1$).
5. Explicit matrix: Compute $uu^T$ as outer product, multiply by 2, subtract from $I$.

Worked Example

2020 Paper 1, 2020-P1-Q2b (20 marks)

Define an $n\times n$ matrix as $A=I-2u\cdot u^T$, where $u$ is a unit column vector. (i) Examine if $A$ is symmetric. (ii) Examine if $A$ is orthogonal. (iii) Show that $\operatorname{trace}(A)=n-2$. (iv) Find $A_{3\times3}$, when $u=(1/3,\,2/3,\,2/3)^T$.

Step 1 — Setup. $u$ is a real unit column vector, so $u^Tu = \|u\|^2 = 1$. Note $uu^T$ is an $n\times n$ rank-1 matrix; $u^Tu=1$ is a scalar.

Step 2 — Symmetry (part i).

$A^T = (I-2uu^T)^T = I^T - 2(uu^T)^T = I - 2(u^T)^Tu^T = I - 2uu^T = A.$

$\boxed{A \text{ is symmetric.}}$

Step 3 — Orthogonality (part ii). Since $A^T=A$, we need $A^2=I$.

$(I-2uu^T)^2 = I - 2uu^T - 2uu^T + 4\,uu^T\,uu^T.$

The last term: $uu^T\cdot uu^T = u(u^Tu)u^T = u(1)u^T = uu^T$. Therefore:

$A^2 = I - 4uu^T + 4uu^T = I.$

$\boxed{A^TA = A^2 = I,\text{ so }A\text{ is orthogonal (and }A^{-1}=A).}$

Step 4 — Trace (part iii). By linearity of trace and the cyclic property $\operatorname{tr}(uu^T)=\operatorname{tr}(u^Tu)=u^Tu=1$:

$\operatorname{tr}(A) = \operatorname{tr}(I) - 2\operatorname{tr}(uu^T) = n - 2\cdot 1 = n-2.\qquad\blacksquare$

Step 5 — Explicit matrix (part iv). Verify $u=(1/3,2/3,2/3)^T$ is unit: $1/9+4/9+4/9=1$ ✓.

Outer product:

$uu^T = \frac{1}{9}\begin{pmatrix}1\\2\\2\end{pmatrix}\begin{pmatrix}1&2&2\end{pmatrix} = \frac{1}{9}\begin{pmatrix}1&2&2\\2&4&4\\2&4&4\end{pmatrix}.$

$A = I - 2uu^T = \frac{1}{9}\begin{pmatrix}9&0&0\\0&9&0\\0&0&9\end{pmatrix} - \frac{1}{9}\begin{pmatrix}2&4&4\\4&8&8\\4&8&8\end{pmatrix} = \boxed{\frac{1}{9}\begin{pmatrix}7&-4&-4\\-4&1&-8\\-4&-8&1\end{pmatrix}.}$

Verify: $\operatorname{tr}=\tfrac{1}{9}(7+1+1)=1=n-2=3-2$ ✓. Row 1 norm: $\tfrac{1}{81}(49+16+16)=1$ ✓. Row 1 $\cdot$ Row 2: $\tfrac{1}{81}(-28-4+32)=0$ ✓.

Common Traps

• The inner product $u^Tu$ (a scalar, equals 1) is not the same as the outer product $uu^T$ (an $n\times n$ matrix). Confusing the two destroys the simplification $uu^Tuu^T = uu^T$.
• In part (ii), the cleanest path is to show $A^2=I$ (using $A^T=A$ from part i), not to compute $A^TA$ from scratch.
• For the trace, the cyclic property gives $\operatorname{tr}(AB)=\operatorname{tr}(BA)$, so $\operatorname{tr}(uu^T)=\operatorname{tr}(u^Tu)=1$. Alternatively sum the diagonal entries of $uu^T$ directly.
• In part (iv), the $(2,3)$ and $(3,2)$ entries are $-8/9$, not $-4/9$. The factor of 2 from $2uu^T$ doubles the $4/9$ from the outer product.

orthogonal-singularity-proof (1 question(s); 2019)

Recognition Cues

• Two orthogonal matrices $A,B$ of the same order with $\det A + \det B = 0$.
• Asked to prove $A+B$ is singular.
• The key insight: determinants of orthogonal matrices are $\pm1$, so $\det A+\det B=0$ forces $\det A \cdot \det B = -1$.

Solution Template

1. Argue about determinants. Since $A,B$ are orthogonal, $\det A,\det B\in\{+1,-1\}$. The condition $\det A+\det B=0$ forces them to be opposite: $\det A=-\det B$, so $\det A\cdot\det B=-1$.
2. Establish the factorisation. Show $A+B = A(A+B)^T B$ using $AA^T=I$ and $B^TB=I$.
3. Take determinants of both sides. Use $\det(M^T)=\det M$ to get $\det(A+B)=\det A\cdot\det(A+B)\cdot\det B$.
4. Conclude. Substituting $\det A\cdot\det B=-1$: $\det(A+B)=-\det(A+B)$, so $\det(A+B)=0$.

Worked Example

2019 Paper 1, 2019-P1-Q2b (15 marks)

Let $A$ and $B$ be two orthogonal matrices of same order and $\det A + \det B = 0$. Show that $A+B$ is a singular matrix.

Step 1 — Determinants of orthogonal matrices. Since $AA^T=BB^T=I$, taking determinants: $(\det A)^2=(\det B)^2=1$, so $\det A,\det B\in\{+1,-1\}$. The hypothesis $\det A+\det B=0$ forces:

$\det A = -\det B, \qquad\text{hence}\qquad \det A\cdot\det B = -1.$

Step 2 — The key factorisation. Claim: $A+B = A(A+B)^T B$.

Verify: expand the right side using $B^TB=I$ and $AA^T=I$:

$A(A+B)^T B = A(A^T+B^T)B = AA^TB + AB^TB = IB + AI = B + A = A+B.\quad\checkmark$

Step 3 — Take determinants.

$\det(A+B) = \det A \cdot \det\!\big((A+B)^T\big)\cdot\det B = \det A\cdot\det(A+B)\cdot\det B,$

since $\det(M^T)=\det M$. Substituting $\det A\cdot\det B=-1$:

$\det(A+B) = -\det(A+B).$

$\therefore\quad 2\det(A+B)=0 \quad\Longrightarrow\quad \det(A+B)=0.$

$\boxed{A+B \text{ is singular.}}\qquad\blacksquare$

Common Traps

• Do not assume $A$ and $B$ are $2\times2$; the proof must work for any $n\times n$ orthogonal matrices. A specific example confirms the result but does not constitute a proof.
• The condition $\det A+\det B=0$ with both values in $\{+1,-1\}$ forces one to be $+1$ and the other $-1$. They cannot both be $0$ (orthogonal determinants are never $0$).
• The factorisation $A+B=A(A+B)^TB$ is the non-obvious step. Verify it by expanding; students who skip this verification and just write “taking determinants” lose the key marks.
• Do not confuse $A^T(A+B)B^T$ with $A(A+B)^TB$; the latter is what gives $\det(A+B)$ on the left with coefficient $\det A\cdot\det B$.

unitary-eigenvalue-modulus (1 question(s); 2014)

Recognition Cues

• Unitary matrix $U$ (satisfying $U^*U=I$, $U^*$ = conjugate transpose).
• Asked to prove every eigenvalue has modulus 1.
• Sometimes phrased as: “eigenvalues lie on the unit circle in $\mathbb C$.”

Solution Template

1. Setup. Let $\lambda$ be an eigenvalue with eigenvector $v\neq 0$: $Uv=\lambda v$.
2. Compute $\|Uv\|^2$ two ways. Via unitary property: $\|Uv\|^2=\langle Uv,Uv\rangle=\langle v,U^*Uv\rangle=\langle v,v\rangle=\|v\|^2$. Via eigenvalue: $\|Uv\|^2=\|\lambda v\|^2=|\lambda|^2\|v\|^2$.
3. Equate and divide. $|\lambda|^2\|v\|^2=\|v\|^2$, and $\|v\|^2>0$, so $|\lambda|^2=1$, giving $|\lambda|=1$.

Worked Example

2014 Paper 1, 2014-P1-Q3c-ii (7 marks)

Prove that the eigenvalues of a unitary matrix have absolute value 1.

Let $U$ be an $n\times n$ unitary matrix, so $U^*U=I$. Let $\lambda\in\mathbb C$ be an eigenvalue with eigenvector $v\neq 0$, so $Uv=\lambda v$.

Step 1 — Compute $\|Uv\|^2$ via the unitary property.

$\|Uv\|^2 = \langle Uv,Uv\rangle = \langle v, U^*Uv\rangle = \langle v,Iv\rangle = \langle v,v\rangle = \|v\|^2.$

Step 2 — Compute $\|Uv\|^2$ via the eigenvalue equation.

$\|Uv\|^2 = \|\lambda v\|^2 = \langle\lambda v,\lambda v\rangle = \lambda\overline\lambda\,\langle v,v\rangle = |\lambda|^2\|v\|^2.$

Step 3 — Equate and conclude.

$|\lambda|^2\|v\|^2 = \|v\|^2.$

Since $v\neq0$, we have $\|v\|^2>0$. Dividing:

$|\lambda|^2 = 1 \quad\Longrightarrow\quad |\lambda|=1.$

$\boxed{\text{Every eigenvalue of a unitary matrix has absolute value 1.}}\qquad\blacksquare$

Common Traps

• The inner product step $\langle Uv,Uv\rangle=\langle v,U^*Uv\rangle$ uses the sesquilinearity of the inner product (conjugate-linear in the first argument in the physics convention, or the second in the math convention — be consistent). The conclusion is the same.
• The key identity is $|\lambda|^2=\lambda\overline\lambda$. Students who only write $|\lambda\|$ without connecting it to the eigenvalue equation may not get full credit.
• The result extends to real orthogonal matrices: since the only real numbers with $|\lambda|=1$ are $\pm1$, eigenvalues of a real orthogonal matrix are in $\{+1,-1,e^{i\theta},e^{-i\theta}\}$, with complex ones appearing in conjugate pairs.

Marks-Aware Writing

7-mark question (unitary eigenvalue, 2014): The three-step argument (unitary property → $\|Uv\|^2=\|v\|^2$; eigenvalue equation → $\|Uv\|^2=|\lambda|^2\|v\|^2$; equate + divide) is exactly what the marking scheme expects. State $U^*U=I$ explicitly and write both computations of $\|Uv\|^2$ before equating.

15-mark question (singularity, 2019): Two marks-bearing sub-arguments — (a) deducing $\det A\cdot\det B=-1$ from the hypothesis, and (b) establishing the factorisation $A+B=A(A+B)^TB$ and taking determinants. Both must be present. The factorisation step must be verified by expanding; just stating it without proof loses points.

20-mark question (Householder, 2020): Four sub-parts with roughly equal weight. Sub-parts (i)–(iii) are each a few lines; sub-part (iv) is the calculation. Common deductions: missing the $u^Tu=1$ step in the orthogonality proof (2–3 marks), arithmetic error in the outer product (1–2 marks per entry).

Practice Set

(No unseen questions in the corpus for this atom; all three are worked above.)