Matrix equation

**sakodo** · January 7th 2011, 05:06 PM

Let M = { [a -b] , a,b E R }
..............[b a]

Solve X^4+X^2+I=0, where X E M

I attempted this question a while ago. I remember finding the eigenvalues and eigenvectors (X^4+X^2=-I and treat X^4+X^2 as a matrix itself) but it led to nowhere. I then tried to use a linear map but no luck either.

I would love to hear ideas on how to solve this one.

Thanks.

**HallsofIvy** · January 7th 2011, 05:17 PM

You would solve for the eigenvalues by solving the numeric equation $x^4+ x^2- 1= 0$ . If we let $y= x^2$ , then we have $y^2+ y- 1= 0$ which, by the quadratic formula, has roots $\frac{-1\pm i\sqrt{3}}{2}$ . The eigenvalues of the original equation are the roots of those and so will be very messy!

**snowtea** · January 7th 2011, 05:20 PM

$M = \begin{bmatrix}a & -b \\ b & a\end{bmatrix} = \gamma\begin{bmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{bmatrix}$

Where $\gamma = \sqrt{a^2 + b^2}$ and $\theta = \tan^{-1}(b/a)$ .

So M is a rotation by $\theta$ with a scaling by $\gamma$ .

Immediately using a geometric interpretation, we can see that a rotation by 60 degrees is a solution for X. Also a rotation by 120 degrees is another solution.
There are more solutions, but hopefully this helps you get started finding roots.

**FernandoRevilla** · January 8th 2011, 01:54 AM

An alternative: we can use the natural field isomorphism:

$\varphi: \mathcal{M}\rightarrow{\mathbb{C}},\quad \varphi\begin{bmatrix}{a}&{-b}\\{b}&{\;\;a}\end{bmatrix}=a+bi$

so, solving

$X^4+X^2+I=0$ in $\mathcal{M}$

it is equivalent to solve

$z^4+z^2+1=0$ in $\mathbb{C}$ .

But,

$z^4+z^2+1=(z^2+1)^2-z^2=(z^2+z+1)(z^2-z+1)=0$

$z=\dfrac{-1\pm\sqrt{3}i}{2},\;z=\dfrac{1\pm\sqrt{3}i}{2}$

As a consequence the roots of the given equation are the elements of the set:

$\mathcal{S}=\left\{{\dfrac{1}{2}\begin{bmatrix}{-1}&{-\sqrt{3}}\\{\sqrt{3}}&{-1}\end{bmatrix},\;\dfrac{1}{2}\begin{bmatrix}{-1}&{\sqrt{3}}\\{-\sqrt{3}}&{-1}\end{bmatrix}, \;\dfrac{1}{2}\begin{bmatrix}{1}&{-\sqrt{3}}\\{\sqrt{3}}&{1}\end{bmatrix},\;\dfrac{1} {2}\begin{bmatrix}{1}&{-\sqrt{3}}\\{\sqrt{3}}&{1}\end{bmatrix}}\right\}$

Fernando Revilla

**sakodo** · January 8th 2011, 12:54 PM

Thanks for your help guys.

FernandoRevilla, I haven't learn anything about natural field isomorphism so I don't understand what you did there...Is that some kind of a linear map? Is there any link about it so I can have a read? I tried google but nothing really came up.

EDIT: you can actually prove that its a linear map. =) Thanks that's a very pretty solution!

snowtea, could you please explain how you got 60 as a solution using geometric interpretation? I didn't quite see that =S

I get everything up to "So M is a rotation by with a scaling by gamma." X^4 and X^2 mean rotations by 4*theta and 2*theta respectively(with scalings too). But what about X^4+X^2+I? How did you see that 60 is a solution so quickly?

**snowtea** · January 8th 2011, 01:00 PM

Let X be a rotation by 60 degrees (no scaling).

Draw a vector v.
Multiplying by I keeps it the same.
Multiplying it by X^2 rotates it by 120 degrees.
Multiplying it by X^4 rotates it by 240 degrees.

(X^4+X^2+I) v is the sum of all the above vectors.
You can show this is always 0 using some simple trigonometry.

**sakodo** · January 8th 2011, 01:53 PM

Originally Posted by snowtea

Let X be a rotation by 60 degrees (no scaling).

Draw a vector v.
Multiplying by I keeps it the same.
Multiplying it by X^2 rotates it by 120 degrees.
Multiplying it by X^4 rotates it by 240 degrees.

(X^4+X^2+I) v is the sum of all the above vectors.
You can show this is always 0 using some simple trigonometry.

Thanks that makes sense now.

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