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UPSC Maths 2013 Paper 1 Q2b-i — Solution

8 marks · Section A

Question

Let $A=\begin{bmatrix}1 & 1 & 1\\ 1 & \omega^{2} & \omega\\ 1 & \omega & \omega^{2}\end{bmatrix}$ where $\omega(\ne 1)$ is a cube root of unity. If $\lambda_1,\lambda_2,\lambda_3$ denote the eigenvalues of $A^{2}$, show that $|\lambda_1|+|\lambda_2|+|\lambda_3|\le 9$.

Technique

Compute $A^{*}A$; recognise the scalar-matrix outcome; use normality to convert singular values $\to$ eigenvalue moduli.

Solution

Strategy. Compute $A^{*}A$ and exploit that it turns out to be a scalar matrix. This makes $A$ (and hence $A^{2}$) a scalar multiple of a unitary matrix, so all eigenvalues lie on a circle.

Step 1 — Key facts about $\omega$

$\omega\ne 1$ and $\omega^{3}=1$, so $1+\omega+\omega^{2}=0$. Conjugation: $\overline{\omega}=\omega^{2}$ and $\overline{\omega^{2}}=\omega$ (the two non-real cube roots of unity are complex conjugates).

Also, $A$ is symmetric ($A^{T}=A$), so $A^{*}=\overline{A^{T}}=\overline A$.

Step 2 — Compute $A^{*}A$

$\overline A=\begin{bmatrix}1&1&1\\1&\omega&\omega^{2}\\1&\omega^{2}&\omega\end{bmatrix}$. Compute $(A^{*}A)_{ij}=(\overline A A)_{ij}=\sum_{k}\overline{A_{ki}}A_{kj}=\sum_{k}\overline{A_{ik}}A_{kj}$ (using symmetry $A_{ki}=A_{ik}$). Equivalently, dot the $i$th row of $\overline A$ with the $j$th column of $A$:

Therefore

So $\tfrac{1}{\sqrt 3}A$ is unitary; equivalently, the singular values of $A$ are all $\sqrt 3$.

Step 3 — $A^{2}$ is also a scalar-times-unitary

$(A^{2})^{*}(A^{2})=A^{*}A^{*}AA=A^{*}(A^{*}A)A=A^{*}(3I)A=3\,A^{*}A=9\,I$.

So $\tfrac{1}{3}A^{2}$ is unitary; equivalently, singular values of $A^{2}$ are all $3$.

Step 4 — $A^{2}$ is normal; eigenvalues have modulus equal to singular values

$A^{2}$ commutes with its adjoint: $(A^{2})^{*}(A^{2})=9I=(A^{2})(A^{2})^{*}$ (the second equality is the same calculation applied to $AA^{*}$, which also equals $3I$ by an identical computation using symmetry of $A$).

For a normal matrix $B$, the absolute values of the eigenvalues equal the singular values (this is the spectral theorem — $B=U^{*}DU$ unitarily, so $|\lambda_i(B)|=\sigma_i(B)$).

Therefore $|\lambda_i(A^{2})|=\sigma_i(A^{2})=3$ for $i=1,2,3$.

Step 5 — Sum up

In particular,

Answer