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.

$\begin{align}<br /> <br /> A^* = \bar{A}^T\\<br /> <br /> A\vec{x} = \lambda\vec{x}\\<br /> <br /> A^*\vec{x} = \bar\lambda\vec{x}\\<br /> \end{align}<br />$

 

[Dick]

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

 

[Shackleford]

$\lambda$ is an eigenvalue of $A$ if $det(A-\lambda I)=0$. 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.

$\begin{align}<br /> det(A-\lambda I) = 0\\\\<br /> \overline{det(A-\lambda I)} = \bar0\\<br /> det(\overline{A-\lambda I)} = 0\\<br /> = det(\bar A-\bar\lambda I) = 0\\\\<br /> = det(\bar A-\bar\lambda I)^T = 0^T\\<br /> = det(\bar A^T-\bar\lambda I) = 0\\<br /> = det(\bar A^T-\bar\lambda I) = 0\\<br /> = det(A^*-\bar\lambda I)=0\\<br /> \\<br /> <br /> \end{align}$

 

[Dick]

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

 

[Shackleford]

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

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: $det(A-\lambda I)=det(A^* - \bar{\lambda} I)$.

 

[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 $det(\bar M)=\overline{det(M)}$ and $det(M^T)=det(M)$.

 

[Shackleford]

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

Is it okay now?

 

[Dick]

Is it okay now?

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

 

[Shackleford]

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

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

Thanks for the help.