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.


With $A$ self-adjoint and

$Av = \lambda v, \tag 1$

we have

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


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

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

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


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

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.

  • 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

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 $\mathbb C^n$.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.