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Let $A$ be an $m\times n$ complex matrix of rank $r$ and let $$\lambda_1 (AA^*)\geq \lambda _2(AA^*)\geq \ldots \lambda_r(AA^*)$$ denote the non-zero eigen values of $AA^*$. If the real scalar $\alpha$ satisfies $$0<\alpha<\frac 2{\lambda_1(AA^*)}$$ then prove that the eigen values of $\alpha AA^*-AA^\dagger$ are $$\begin{cases} \alpha \lambda_i(AA^*)-1 & i=1,2,3,\ldots ,r\\ 0 & i =r+1,\ldots,m \end{cases}$$

Here the notation $A^*$ is complex conjugate transpose of $A$ and $A^\dagger$ is moore penrose inverse of $A$. That is, $A^\dagger$ satisfies, $$AA^\dagger A=A\quad A^\dagger AA^\dagger=A^\dagger\quad (AA^\dagger)^*=AA^\dagger\quad (A^\dagger A)^*=A^\dagger A$$

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Let $A=USV^*$ be the reduced SVD of $A$ ($U\in\mathbb{C}^{m\times r}$, $S:=\mathrm{diag}(\sigma_1,\ldots,\sigma_r)$ is $r\times r$ a diagonal positive definite matrix, $V\in\mathbb{C}^{n\times r}$). In your notation, you can choose $\sigma_i$ such that $\sigma_i^2=\lambda_i(AA^*)$, $i=1,\ldots,r$. Then $A^\dagger=VS^{-1}U^*$ and $$ \alpha AA^*-AA^\dagger=\alpha USV^*VSU^*-USV^*VS^{-1}U^* =U\left(\alpha S^2-I_r\right)U^*. $$ Now let $U^\perp\in\mathbb{C}^{m\times (m-r)}$ be such that $\tilde{U}:=[U,U^\perp]$ is a square unitary matrix. Then $$ \alpha AA^*-AA^\dagger=U\left(\alpha S^2-I_r\right)U^*=\tilde{U}\pmatrix{\alpha S^2-I_r\\&0_{m-r}}\tilde{U}^* $$ is the spectral decomposition of the Hermitian matrix $\alpha AA^*-AA^\dagger$.

Note that the assumption on the bounds for $\alpha$ is irrelevant.


Another way without SVD by using the definition of the pseudoinverse

It is sufficient to show, that:

  • if $AA^*v=\lambda v$, $\lambda\neq 0$, $v\neq 0$, then $AA^\dagger v=v$,
  • if $AA^*w=0$, then $AA^\dagger w=0$.

Take this as a fact for now. The first item implies that $$ (\alpha AA^*v-AA^\dagger) v=(\alpha\lambda-1)v $$ and the second item gives $$ (\alpha AA^*-AA^\dagger)w=0-0=0. $$ This shows what you already wanted.

Now to show the first item is true, assume $AA^*v=\lambda v$ and $\lambda\neq 0$. Then $$ AA^\dagger v=AA^\dagger(\lambda^{-1}AA^*v)=\lambda^{-1}AA^\dagger AA^*v =\lambda^{-1}AA^*v=\lambda^{-1}\lambda v=v. $$ For the second item, note that $AA^*w=0$ if and only if $A^*w=0$. We show that the kernels of $A^\dagger$ and $A^*$ are identical. Since $A^\dagger=A^\dagger AA^\dagger=A^\dagger (AA^\dagger)^*=A^\dagger (A^\dagger)^* A^*$, we have $\mathrm{ker}(A^*)\subseteq\ker(A^\dagger)$. On the other hand, $A^*=(AA^\dagger A)^*=A^*(AA^\dagger)^*=A^*AA^\dagger$ and hence $\mathrm{ker}(A^\dagger)\subseteq\mathrm{ker}(A^*)$.

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  • $\begingroup$ thank you for the answer.. its awesome.. I am thinking no need to take $U_{m\times r}$ rather we can take $U_{m\times m}$ itself. That is also possible!!(similarly for $V$). and $S=diag(\sigma_1,\ldots,\sigma_r,0,\ldots,0)$. So that computations are becoming simple. $\endgroup$ – David Nov 28 '15 at 15:56

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