Find all matrices which satisfy $M^2-3M+3I = 0$

Question

I am trying to find all matrices which solve the matrix equation

$$M^2 -3M +3I=0$$

Since this doesn't factor I tried expanding this in terms of the coordinates of the matrix. It also occurs to me to put it into "vertex" form:

$$M^2 - 3M + \frac{9}{4}I+\frac{3}{4}I=0$$

$$(M-\frac{3}{2}I)^2 = -\frac{3}{4}I$$

but this doesn't look much better.

What I found from expanding by coordinates was, if $M=\pmatrix{a & b \\ c & d}$ then

$$\pmatrix{a^2+bc -3a + 3& ab + bd - 3b \\ ac+cd-3c & bc+d^2-3d+3} = \pmatrix{0&0\\0&0}$$

From the off-diagonal entries I get that either

$$a+d-3=0$$

or

$$b=c=0$$

If $a+d-3\not=0$ then $a^2-3a+3=0$ and likewise for $d$. Then we get more cases for $a$ and $d$.

If $a+d-3=0$ the upper-left is unchanged and the lower-right is

$$bc + (3-a)^2-3(3-a)+3 = 0$$

which simplifies to the same thing from the upper-left and so is redundant. In the off-diagonals

$$ac+c(a-3)-3c = 0 \Rightarrow $$ $$2ac-6c = 0$$

We again get cases, and I suppose after chasing cases enough you get the solution set.

However, it just feels like this can't be the intended solution given how tedious and uninformative all of this case-chasing is. Is there some bigger idea I'm missing?

Answer 1

You already found the "completion of the square" $$ \left( {M - {3 \over 2}I} \right)^{\,2} = - {3 \over 4}I $$

Then you can write $$ \left( {i{2 \over {\sqrt 3 }}\left( {M - {3 \over 2}I} \right)} \right)^{\,2} = X^{\,2} = I $$

So we are essentially looking for the square roots of the unit matrix, also complex, or for the square roots of $- \, I$.

You can find various papers dealing with this subject, for example this related post or this thesis.

-- p.s. --
I thought you were interested in the general case of $n \times n$ matrices.
If your question is limited to $2 \times 2$ then the $\sqrt{\pm I}$ is easily found on the net (e.g.,see the hint on Pauli matrices).

Answer 2

hint

By Cayley-Hamilton,

if the caracteristic polynom is $$x^2-3x+3$$

then

$$M^2-3M+3I=0$$

then

$$(a-x)(d-x)-bc=x^2-3x+3$$

Answer 3

$m^2 - 3m + 3 = 0\\ \lambda = \frac {3}{2} \pm i\frac {\sqrt {3}}{2}$

You could say that it is all matrices with eigenvalues equal to $\frac {3}{2} + i\frac {\sqrt {3}}{2},\frac {3}{2} - i\frac {\sqrt {3}}{2}$

If we restrict our universe to real $2\times 2$ matrices.

Then it would be all matrices with characteristic equations equal to:

$\lambda^2 - 3\lambda + 3 = 0$

We are looking for matrices with trace equal to 3, and determinant 3.

$\begin{bmatrix} a & b\\ -\frac {a^2 -3a + 3}{b} & 3-a \end {bmatrix}$

Answer 4

Minimal polynomial of $M, m_M(x),$ is a factor of $x^2-3x+3=[x-(\frac{3+i\sqrt3}2)][x-(\frac{3-i\sqrt3}2)]$

Either $m_M(x)=x-(\frac{3+i\sqrt3}2)\implies M=[\frac{3+i\sqrt3}2]$

or $m_M(x)=x-(\frac{3-i\sqrt3}2)\implies M=[\frac{3-i\sqrt3}2]$

or $m_M(x)=x^2-3x+3\implies$ the eigenvalues of $M$ are $\frac{3\pm i\sqrt3}2$

In case of $2\times2$ matrices, product of eigenvalues $=\det(M)=3$, sum of eigenvalues $=\text{Tr}(M)=3$

We have $M=\begin{bmatrix}a&b\\c&3-a\end{bmatrix}$ and $3a-a^2-bc=3; a,b,c\in\Bbb C$.

You could go for $3\times3,4\times4,...$ matrices by defining the same eigenvalues and conditions. In case you are looking for real matrices, you just have to take the real subset of these matrices.