Symmetric and Skew-Symmetric Matrices

At a Glance

• Frequency: 1 sub-part across 1 of 13 years (2021)
• Priority tier: T4
• Marks (count): 8 (1)
• Average solve time: ~12 min
• Difficulty mix: medium 1
• Section: A | Dominant type: proof

Why This Chapter Matters

Symmetric and skew-symmetric matrices are the two canonical self-adjoint structures in real linear algebra: every square matrix decomposes uniquely into these two parts. UPSC 2021 asked an 8-mark Section A question — most likely a proof of a property (e.g., the unique decomposition, reality of eigenvalues, or the orthogonality of eigenvectors) or a decomposition exercise. Mastery here also directly supports the spectral theorem and quadratic forms, which recur across the syllabus.

Minimum Theory

Definitions.

• $A$ is symmetric if $A^T = A$ (equivalently, $a_{ij} = a_{ji}$ for all $i, j$).
• $A$ is skew-symmetric if $A^T = -A$ (equivalently, $a_{ij} = -a_{ji}$; in particular $a_{ii} = 0$ for all $i$ over $\mathbb{R}$).

Unique decomposition. Every $n\times n$ real matrix $A$ decomposes as $A = S + K, \quad S = \frac{A + A^T}{2}, \quad K = \frac{A - A^T}{2},$ where $S$ is symmetric and $K$ is skew-symmetric. The decomposition is unique.

Proof of uniqueness: If $A = S' + K'$ is another such decomposition, then $S - S' = K' - K$. The left side is symmetric and the right side is skew-symmetric; the only matrix that is both is $0$. Hence $S = S'$ and $K = K'$. $\square$

Properties of real symmetric matrices.

• All eigenvalues are real.
• Eigenvectors corresponding to distinct eigenvalues are orthogonal.
• $A$ is orthogonally diagonalizable (Spectral Theorem): $A = Q\Lambda Q^T$ with $Q$ orthogonal and $\Lambda$ diagonal real.

Properties of real skew-symmetric matrices.

• Diagonal entries are all zero.
• Eigenvalues are either $0$ or purely imaginary (i.e., of the form $i\mu$, $\mu \in \mathbb{R}$).
• $x^T A x = 0$ for all $x \in \mathbb{R}^n$.

Proof that eigenvalues of a real symmetric matrix are real. Let $Av = \lambda v$ with $v \neq 0$, $v \in \mathbb{C}^n$. Then $\lambda \bar{v}^T v = \bar{v}^T(\lambda v) = \bar{v}^T A v = (A\bar{v})^T v = \overline{(Av)}^T v = \overline{(\lambda v)}^T v = \bar\lambda \bar{v}^T v.$ Since $\bar{v}^T v = \|v\|^2 > 0$, we get $\lambda = \bar\lambda$, so $\lambda \in \mathbb{R}$. $\square$

Proof that eigenvectors of a real symmetric matrix for distinct eigenvalues are orthogonal. Suppose $Av_1 = \lambda_1 v_1$ and $Av_2 = \lambda_2 v_2$ with $\lambda_1 \neq \lambda_2$. Then $\lambda_1 v_1^T v_2 = (Av_1)^T v_2 = v_1^T A^T v_2 = v_1^T A v_2 = \lambda_2 v_1^T v_2.$ Since $\lambda_1 \neq \lambda_2$, we conclude $v_1^T v_2 = 0$. $\square$

Question Archetypes

unique-decomposition-proof (1 question(s); 2021)

Recognition Cues

• The question says “prove” or “show” something about symmetric/skew-symmetric structure.
• May provide a specific matrix and ask to decompose, then verify.
• 8 marks: expects a clean proof (half the marks) plus an illustrative example or explicit computation (the other half).
• Key phrase: “uniquely expressible” or “every square matrix.”

Solution Template

1. Define $S = (A + A^T)/2$ and $K = (A - A^T)/2$; verify $A = S + K$.
2. Check $S^T = S$: $(A + A^T)^T/2 = (A^T + A)/2 = S$. Check $K^T = -K$: $(A - A^T)^T/2 = (A^T - A)/2 = -K$.
3. Prove uniqueness: assume another decomposition $A = S' + K'$; derive $S - S' = K' - K$ is both symmetric and skew-symmetric, hence zero.
4. (If asked for a concrete decomposition) Apply the formulas to the given $A$.

Worked Example

2021 Paper 1, 2021-P1-Q1e (8 marks)

Prove that every square matrix can be uniquely expressed as the sum of a symmetric matrix and a skew-symmetric matrix. Hence decompose $A = \begin{pmatrix} 3 & 5 & -2 \\ 1 & 4 & 6 \\ 0 & -3 & 7 \end{pmatrix}$ into its symmetric and skew-symmetric parts.

Part 1: Existence.

Define $S = \frac{A + A^T}{2}, \qquad K = \frac{A - A^T}{2}.$

Then $S + K = \dfrac{(A+A^T) + (A-A^T)}{2} = A$. We verify:

$S^T = \left(\frac{A+A^T}{2}\right)^T = \frac{A^T + A}{2} = S, \quad \text{so } S \text{ is symmetric.}$

$K^T = \left(\frac{A-A^T}{2}\right)^T = \frac{A^T - A}{2} = -K, \quad \text{so } K \text{ is skew-symmetric.}$

Part 2: Uniqueness.

Suppose $A = S_1 + K_1$ where $S_1^T = S_1$ and $K_1^T = -K_1$. Then $A^T = S_1 - K_1$, giving

$S_1 = \frac{A + A^T}{2} = S, \qquad K_1 = \frac{A - A^T}{2} = K.$

The decomposition is unique. $\square$

Part 3: Explicit decomposition.

$A^T = \begin{pmatrix}3 & 1 & 0\\ 5 & 4 & -3\\ -2 & 6 & 7\end{pmatrix}.$

$S = \frac{1}{2}\begin{pmatrix}6 & 6 & -2\\ 6 & 8 & 3\\ -2 & 3 & 14\end{pmatrix} = \begin{pmatrix}3 & 3 & -1\\ 3 & 4 & \frac{3}{2}\\ -1 & \frac{3}{2} & 7\end{pmatrix}.$

$K = \frac{1}{2}\begin{pmatrix}0 & 4 & -2\\ -4 & 0 & 9\\ 2 & -9 & 0\end{pmatrix} = \begin{pmatrix}0 & 2 & -1\\ -2 & 0 & \frac{9}{2}\\ 1 & -\frac{9}{2} & 0\end{pmatrix}.$

Verification: $S + K = A$. $\checkmark$

$\boxed{A = S + K \text{ as above, uniquely.}}$

Common Traps

• Forgetting to prove uniqueness — existence alone earns at most half the marks when the question says “uniquely.”
• Errors in computing $A^T$: off-diagonal entries swap ($a_{12}$ becomes $(A^T)_{21}$), not the diagonal.
• Claiming the decomposition is “obvious by definition” without writing the algebraic verification of $S^T = S$ and $K^T = -K$ — examiners require this.
• For the eigenvalue-reality proof: not using the Hermitian inner product $\bar{v}^T v$ and instead assuming $v$ is real (which is circular).

Marks-Aware Writing

At 8 marks (Section A, ~12 min), split roughly: 2 marks for defining $S$ and $K$ and showing $A = S+K$; 2 marks for verifying $S^T = S$ and $K^T = -K$; 2 marks for the uniqueness argument; 2 marks for the concrete decomposition (if asked). Write the uniqueness proof as a clean 3-line derivation — do not merely assert it. For eigenvalue-property questions, the Hermitian inner product argument above is the canonical 4-line proof worth 4–5 marks.

Practice Set

Only one historical question on this atom (shown above).