Finding a matrix, given one eigenvalue

[vellum93]

Homework Statement

Suppose B is a real 2x2 matrix with the following eigenvalue:

$\frac{√3}{2} + \frac{3i}{2}.$

Find B$^3$.

Homework Equations

One of the hints is to consider diagonalization over C together with the fact
that $(\frac{1}{2} + \frac{√3}{2}i)^3 = -1$.

The Attempt at a Solution

I found B by going backwards through the quadratic formula, and found B =$\begin{pmatrix} 0 & 3\\-1 & √3\end{pmatrix}$

How can I do this without the method I used? How can I apply the hint given above?

 

[Mark44]

Homework Statement

Suppose B is a real 2x2 matrix with the following eigenvalue:

$\frac{√3}{2} + \frac{3i}{2}.$

Find B$^3$.

Homework Equations

One of the hints is to consider diagonalization over C together with the fact
that $(\frac{1}{2} + \frac{√3}{2}i)^3 = -1$.

The Attempt at a Solution

I found B by going backwards through the quadratic formula, and found B =$\begin{pmatrix} 0 & 3\\-1 & √3\end{pmatrix}$

How can I do this without the method I used? How can I apply the hint given above?

What your hint is telling you is that 1/2 + √3i/2 is a cube root of -1. Obviously, one cube root is -1. The third cube root is the complex conjugate of the one you have. Use it as another eigenvalue to get a second eigenvector.

 

[HallsofIvy]

I am confused by fact that the given eigenvalue is $\sqrt{3}/2+ 3i/2$, NOT $1/2+ \sqrt{3}i/2$.

 

[Mark44]

My guess is that there's a typo, and the the given eigenvalue should be 1/2 + √3i/2, which actually is a cube root of -1.

 

[Dick]

My guess is that there's a typo, and the the given eigenvalue should be 1/2 + √3i/2, which actually is a cube root of -1.

I would guess it means what it says, so one eigenvalue is sqrt(3) times a cube root of -1. They have a cube root of -1 correct in the hint. Since you know what the other eigenvalue must be it's pretty easy to find B^3 when you consider the diagonalization.

 

[vellum93]

Ok, thanks for the help everyone. Does this mean then that B^3 has eigenvalues that are equal to -3√3? I just cubed the eigenvalues for B after factoring out a √3.

 

[Dick]

Ok, thanks for the help everyone. Does this mean then that B^3 has eigenvalues that are equal to -3√3? I just cubed the eigenvalues for B after factoring out a √3.

B^3 for sure has one eigenvalue that -3√3. But that's not the question. The question is what is the matrix of B^3. There's a little bit more to do.

 

[vellum93]

The other eigenvalue is the same isn't it? So there's only one distinct eigenvalue? For the matrix B^3, I put the eigenvalues on the diagonal and then 0's in the other two spots.

 

[Dick]

The other eigenvalue is the same isn't it? So there's only one distinct eigenvalue? For the matrix B^3, I put the eigenvalues on the diagonal and then 0's in the other two spots.

Sure. But a matrix can have only one eigenvalue and not be diagonalizable. That's not the case here. You just have to say why. That's all.