Eigenvalues of A-adjoint and A

[Shackleford]

Eigenvalues of A* and A

Show that the eigenvalues of A* are conjugates of the eigenvalues of A.

I know this is an easy problem, but I've just been spinning my wheels manipulating the equations with the transpose, conjugate, and adjoint properties.

[itex]
\begin{align}

A^* = \bar{A}^T\\

A\vec{x} = \lambda\vec{x}\\

A^*\vec{x} = \bar\lambda\vec{x}\\
\end{align}

[/itex]

 

[Dick]

[itex]\lambda[/itex] is an eigenvalue of [itex]A[/itex] if [itex] det(A-\lambda I)=0[/itex]. Try taking conjugate and transpose on that.

 

[Shackleford]

[itex]\lambda[/itex] is an eigenvalue of [itex]A[/itex] if [itex] det(A-\lambda I)=0[/itex]. Try taking conjugate and transpose on that.

You know, I thought about doing it like that but didn't because I figured the other way would be easier.

[itex]
\begin{align}
det(A-\lambda I) = 0\\\\
\overline{det(A-\lambda I)} = \bar0\\
det(\overline{A-\lambda I)} = 0\\
= det(\bar A-\bar\lambda I) = 0\\\\
= det(\bar A-\bar\lambda I)^T = 0^T\\
= det(\bar A^T-\bar\lambda I) = 0\\
= det(\bar A^T-\bar\lambda I) = 0\\
= det(A^*-\bar\lambda I)=0\\
\\

\end{align}
[/itex]

 

[Dick]

Just to make sure you are clear on this. [itex]A-\lambda I[/itex] is NOT EQUAL to [itex]A^* - \bar{\lambda} I[/itex]. What is equal is [itex]det(A-\lambda I)=det(A^* - \bar{\lambda} I)[/itex].

 

[Shackleford]

Just to make sure you are clear on this. [itex]A-\lambda I[/itex] is NOT EQUAL to [itex]A^* - \bar{\lambda} I[/itex]. What is equal is [itex]det(A-\lambda I)=det(A^* - \bar{\lambda} I)[/itex].

Yes, I know that.

The characteristic polynomials for A and A* are equal?

 

[Dick]

Yes.

The characteristic polynomials are equal?

Well, no. That's not what they are asking you to show. If A=[[1,0],[0,i]], then how does the characteristic polynomial of A compare with the characteristic polynomial of A*?

 

[Shackleford]

Well, no. That's not what they are asking you to show. If A=[[1,0],[0,i]], then how does the characteristic polynomial of A compare with the characteristic polynomial of A*?

I know what they're asking me. You're right. The characteristic polynomials aren't equal. I was overlooking a small detail.

The two characteristic polynomials are complex conjugates, right?

I'm not quite sure why this is true: [itex]det(A-\lambda I)=det(A^* - \bar{\lambda} I)[/itex].

 

[A. Bahat]

Those two determinants are equal because they are both zero.

 

[Dick]

Those two determinants are equal because they are both zero.

Right. Since they are complex conjugates of each other, if one equals 0 then the other equals zero.

 

[Shackleford]

Right. Since they are complex conjugates of each other, if one equals 0 then the other equals zero.

Oh, okay. I didn't know the characteristic polynomials were complex conjugates of each other.

I think I have it written out more correctly now in post #3.

 

[Dick]

Oh, okay. I didn't know the characteristic polynomials were complex conjugates of each other.

I think I have it written out more correctly now in post #3.

Post 3 still looks a little funny. The properties of determinant that you need are [itex]det(\bar M)=\overline{det(M)}[/itex] and [itex]det(M^T)=det(M)[/itex].

 

[Shackleford]

Post 3 still looks a little funny. The properties of determinant that you need are [itex]det(\bar M)=\overline{det(M)}[/itex] and [itex]det(M^T)=det(M)[/itex].

Is it okay now?

 

[Dick]

Is it okay now?

Okay. I wouldn't bother with transposing a scalar. Notice if you start with [itex]det(A-\lambda I)=c[/itex] then you wind up with [itex]det(A^*-\bar{\lambda} I)=\bar c[/itex]

 

[Shackleford]

Okay. I wouldn't bother with transposing a scalar. Notice if you start with [itex]det(A-\lambda I)=c[/itex] then you wind up with [itex]det(A^*-\bar{\lambda} I)=\bar c[/itex]

Oh, you're right. I mistakenly treated it as a vector.

Thanks for the help.