← 2013 Paper 1

UPSC Maths 2013 Paper 1 Q1b — Solution

10 marks · Section A

Question

Let $A$ be a square matrix and $A^*$ be its adjoint, show that the eigenvalues of matrices $AA^*$ and $A^*A$ are real. Further show that trace $(AA^*) =$ trace $(A^*A)$.

Technique

Self-adjointness of $AA^*$ and $A^*A$ via $(MN)^*=N^*M^*$ and $A^{**}=A$; standard Hermitian-implies-real-eigenvalue inner-product argument.

Solution

(Here $A^*$ denotes the conjugate transpose — the standard “adjoint” in linear algebra over $\mathbb{C}$.)

Part 1 — Eigenvalues of $AA^*$ and $A^*A$ are real

Strategy. Both matrices are Hermitian; Hermitian matrices have real eigenvalues.

Step 1 — Hermitian-ness.

For any square matrix $A$,

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

Step 2 — Hermitian $\Rightarrow$ real eigenvalues.

Let $H$ be a Hermitian matrix ($H^*=H$) and $\lambda$ an eigenvalue with eigenvector $v\ne 0$. Then $Hv=\lambda v$, and the inner product

But also $\langle Hv,v\rangle=\langle v,H^*v\rangle=\langle v,Hv\rangle=\overline{\langle Hv,v\rangle}$, so $\langle Hv,v\rangle$ is real. Since $\|v\|^{2}>0$ is real, $\lambda$ must be real.

Applying this to $H=AA^*$ and $H=A^*A$ gives the desired conclusion.

Part 2 — $\operatorname{tr}(AA^*)=\operatorname{tr}(A^*A)$

Strategy. Compute both traces directly as sums of squared moduli of $A$‘s entries.

Step 3 — Trace of $AA^*$.

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

Step 4 — Trace of $A^*A$.

$(A^*A)_{jj}=\sum_{i}(A^*)_{ji}A_{ij}=\sum_{i}\overline{A_{ij}}A_{ij}=\sum_{i}|A_{ij}|^{2}$. Summing:

The two double sums differ only in the order of summation, so they are equal:

Answer