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I want to show that all the non-zero s-numbers, i.e. singular values $s_j(A):=(\lambda_j(A^*A))^{1/2}$, of A (a bounded linear operator of finite rank acting on a separable Hilbert space $H$) are equal to 1 if and only if $A^*=A^*AA^*$ and $A=AA^*A$.

$A$ is an operator of finite rank, hence it is compact. So, I can use the following proposition:

If $A:H\rightarrow H$ is a compact operator and $B,C:H\rightarrow H$ are bounded and linear, then $$ s_j(BAC)\le \|B\|\|C\|s_j(A). $$

Thus,

$$ s_j(A^*)=s_j(A^*AA^*)\le \|A^*\|^2s_j(A). $$

$A$ is compact, therefore $A^*$ is compact. So,

$$ s_j(A)=s_j(AA^*A)\le \|A\|^2s_j(A^*). $$

Here I get stuck. I have no idea how to prove this. Any hints are appreciated.

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Note that the two equalities $A=AA^*A$ and $A^*=A^*AA^*$ are the same, since you can obtain one from the other by taking adjoints.

Assume first that $A=AA^*A$. By multiplying by $A^*$ on the left, we get $$ A^*A=(A^*A)^2.$$ It follows that the eigenvalues of $A^*A$ all satisfy the equation $\lambda=\lambda^2$, so only $0$ and $1$ are possible.

Conversely, assume that all nonzero eigenvalues of $A^*A$ are equal to $1$. Because $A^*A$ is selfadjoint, it is diagonalizable, $A^*A=UDU^*$ with $D$ diagonal with zeros and ones in the diagonal. Then $D^2=D$, and it follows that $(A^*A)^2=A^*A$. We can write this as $$\tag{1}A^*(A-AA^*A)=0.$$ Now, since $\text{ran}\,(A)^\perp=\ker A^*$, the equality $(1)$ implies that $A-AA^*A=0$ (because $A^*$ is injective on $\text{ran}\,A$).

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