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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 $\operatorname {trace}(AA^*)=\operatorname {trace}(A^*A)$.

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Let $\lambda \in \mathbb C$ be an eigenvalue of $A^*A$ and $x \in \mathbb C^n$ a corresponding eigenvector. Then $$ \lambda(x,x) = (\lambda x,x) = (A^*Ax,x) = (x,A^*Ax) = (x,\lambda x) = \bar\lambda(x,x)$$ Dividing by $(x,x) \ne 0$ gives $\lambda = \bar \lambda$, hence $\lambda \in \mathbb R$. The same proof works for $AA^*$.

The fact about the trace follows from a more general fact, namely: If $A \in {\rm Mat}(n,k)$ and $B\in {\rm Mat}(k,n)$ are matrices over some field $K$ then $\def\tr{\mathop{\rm trace}}\tr(AB) = \tr(BA)$. This can be seen as follows: \begin{align*} \tr(AB) &= \sum_{i=1}^n (AB)_{ii}\\ &= \sum_{i=1}^n \sum_{j=1}^k A_{ij}B_{ji}\\ &= \sum_{j=1}^k \sum_{i=1}^n B_{ji}A_{ij}\\ &= \sum_{j=1}^k (BA)_{jj}\\ &= \tr(BA). \end{align*}

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  • $\begingroup$ forgive my ignorance. What does λ(x,x) stand for ? I am not able follow the first part is there any other method? $\endgroup$ – Swami Jul 2 '14 at 9:07
  • $\begingroup$ $(\cdot, \cdot)$ denotes the inner product, so $\lambda(x,x)$ is $\lambda$ multiplied with the norm-square of $x$. $\endgroup$ – martini Jul 2 '14 at 9:09
  • $\begingroup$ (λx,x)=(A∗Ax,x) How is this possible λ is a scalar while A∗A is matrix. Please explain the rationale behind this step. $\endgroup$ – Swami Jul 2 '14 at 9:38
  • $\begingroup$ @Swami $x$ is an eigenvector for $A$ with eigenvalue $\lambda$, hence $A^*Ax = \lambda x$. $\endgroup$ – martini Jul 2 '14 at 9:40
  • $\begingroup$ If x is an eigenvector for A with eigenvalue λ, then we should have Ax = λx then how did we get A∗Ax=λx ? $\endgroup$ – Swami Jul 2 '14 at 9:47
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Let $(\lambda,x)$ ne an eigenpair of $A^*A$, wlog we may suppose $\|x\|_2 = 1$, so that then $$\lambda = \lambda \|x\|^2_2 = <\lambda x,x> = <A^*Ax,x> = <Ax,Ax> = \|Ax\|_2^2 \in \mathbb{R},$$ so $\lambda$ must be real. For the same reason every eigenvalue of $AA^*$ are reals. For the trace, see its properties here.

Shorter proof: $(AA^*)=A^{**}A^* = AA^*$ and $(A^*A)^* = A^*A$, thus these are self adjoint operators and have therefore only real eigenvalues.

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  • $\begingroup$ "thus these are self adjoint operators and have therefore only positive eigenvalues." - Surely you mean real? $\endgroup$ – Sten Jul 2 '14 at 9:43
  • $\begingroup$ @Sten Yes obviously, this is a typo... thanks! $\endgroup$ – Surb Jul 2 '14 at 9:44

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