Eigenvalues of product of two complex matrices

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

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

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.

• 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. – Mathforjob Feb 7 at 7:34
• @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. – 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 $$\mathbb C^n$$.)