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Let $A$ be a $n\times n$ complex matrix. Assume that $A$ is self-adjoint and $B$ denote the inverse of $A+iI$, where $I$ is identity matrix of order $n\times n$. Then all eigenvalues of $(A-iI)B$ are

  1. purely imaginary

  2. real

  3. of modulus one

  4. of modulus less than one

My attempt: I discarded options $1,4$ by taking $A=O$ $($null matrix$)$. Let '$a$' be an eigenvalue of $A$ then the eigenvalues of $(A-iI)B$ are $(a-i)(a+i)^{-1}$. But I'm not sure about it. Please help me.

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With $A$ self-adjoint and

$Av = \lambda v, \tag 1$

we have

$\lambda \in \Bbb R \tag 2$

and

$(A + iI)v = (\lambda + i)v; \tag 3$

since every eigenvalue of $A$ is real, no eigenvalue of $A + iI$ vanishes, hence it is invertible; we use this fact to write (3) as

$(A + iI)^{-1}v = \dfrac{1}{\lambda + i} v = \dfrac{\lambda - i}{\lambda^2 + 1}v; \tag 4$

then

$(A - iI)(A + iI)^{-1} = (A - iI) \dfrac{\lambda - i}{\lambda^2 + 1}v = \dfrac{\lambda - i}{\lambda^2 + 1}(A - iI)v$ $= \dfrac{\lambda - i}{\lambda^2 + 1}(\lambda - i)v = \dfrac{(\lambda - i)^2}{\lambda^2 + 1}v; \tag 5$

we have

$\left \vert \dfrac{(\lambda - i)^2}{\lambda^2 + 1} \right \vert = \dfrac{1}{\lambda^2 + 1} \vert(\lambda - i)^2 \vert$ $= \dfrac{1}{\lambda^2 + 1} \vert(\lambda - i) \vert^2 = \dfrac{1}{\lambda^2 + 1} (\sqrt{\lambda^2 + 1})^2 = \dfrac{\lambda^2 + 1}{\lambda^2 + 1} = 1; \tag 6$

thus option (3) is correct.

The transformation

$A \to (A - iI)(A + iI)^{-1} \tag 7$

is a useful one, since in maps self-adjoint $A$ to the unitary $(A - iI)(A + iI)^{-1}$, yielding additional insights into such operators. This is especially true since it may be extended to unbounded $A$ on a Hilbert space, and allows $A$ to be addressed in terms of a bounded (unitary) operator.

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  • 1
    $\begingroup$ If A and B be two complex matrices. If a and b are eigen values of A and B respectively. Then eigen values of AB is ab. Is this true? Please clear my doubt. $\endgroup$ – Mathforjob Feb 7 at 7:34
  • $\begingroup$ @Mathforjob: If $A$ and $B$ have a common eigenvector $v$, so that $Av = av$ and $Bv = bv$, then $(AB)v = A(Bv) = A(bv) = b(Av) = (ba)v = (ab)v$, so $ab$ is an eigenvalue of $AB$. I don't think this holds in the absense of a common eigenvector in general. You need a common eigenvector for each pair of such eigenvalues of $A$ and $B$. If $AB = BA$, $A$ and $B$ share eigespaces so something like this still works. $\endgroup$ – Robert Lewis Feb 7 at 7:43
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Since $A$ is self-adjoint, the matrix $B=(A-iI)(A+iI)^{-1}$ is unitary. Now let $ \mu \in \mathbb C$, $x \ne 0$ and $Bx= \mu x$. Then

$||x||^2=(x,x)=(B^*Bx,x)=(Bx,Bx)=(\mu x,\mu x)=|\mu|^2 ||x||^2.$

This gives $|\mu|=1.$

( $ (\cdot, \cdot)$ denotes the usual inner oroduct on $\mathbb C^n$.)

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