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?

Have you thought about what the Cayley-Hamilton theorem tells you here? — , Dec 14 '18 at 21:45

G Cab · Accepted Answer · 2018-12-17T23:35:05.100

2

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).

edited Dec 17 '18 at 23:35

answered Dec 14 '18 at 23:36

G Cab

35,964

The links don't work. – YiFan Tey Dec 17 '18 at 22:26
1

@YiFan I think the Pauli matrices satisfy this condition. – Mustafa Said Dec 17 '18 at 22:41
@MustafaSaid sorry if I wasn't clear. The two links at the bottom of the post don't refer to anything (as viewed on my android phone). I wasn't referring to the math content. – YiFan Tey Dec 17 '18 at 22:45
1

@YiFan you can google "Pauli matrices" and I am pretty sure they satisfy the condition $X^2 = -I$. – Mustafa Said Dec 17 '18 at 22:48
@YiFan: sorry ! I forgot to add the links: added now – G Cab Dec 17 '18 at 23:14
@Diger: oh yes, sorry, thanks for signalling! – G Cab Dec 17 '18 at 23:20
I think you mixed up the denominator and nominator of $\sqrt{3}/2$. (accidentally deleted) – Diger Dec 17 '18 at 23:20
@MustafaSaid: good hint, thanks. However I was trying to find a solution in the more general case ($n \times n$) – G Cab Dec 17 '18 at 23:24

Doug M · Answer 2 · 2018-12-14T23:02:43.480

1

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

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.

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

edited Dec 14 '18 at 23:02

answered Dec 14 '18 at 21:55

Doug M

58,694

1

How is $0$ an eigenvalue? – Shubham Johri Dec 14 '18 at 22:00
$0$ is an eigenvalue for $n\times n$ matrices $n>2$. Rather if $m^2 - 3m + 3 = 0$ then so does $m^3 - 3m^2 + 3m = 0$. – Yadati Kiran Dec 14 '18 at 22:19
@YadatiKiran ? this is obvisouly false. – Thinking Dec 14 '18 at 22:21
@Thinking: And why would that be? – Yadati Kiran Dec 14 '18 at 22:31
@YadatiKiran The question is rather. Why $0$ is an eigenvalue of every matrix ? Just take an invertible matrix and $0$ is never an eigenvalue... – Thinking Dec 14 '18 at 22:32
Consider a diagonal matrix that has $0$ somewhere on its diagonal. Filling it in does not result in the zero matrix. – Klaas van Aarsen Dec 14 '18 at 22:33
@Thinking: But the matrix must still satisfy the polynomial $m^2 - 3m + 3 = 0$ be it of any order. So what can the other eigenvalues be? – Yadati Kiran Dec 14 '18 at 22:34
Why is it being assumed that $M$ is a $2 \times 2$ matrix? – Kavi Rama Murthy Dec 14 '18 at 23:54
@KaviRamaMurthy It can be any matrices with those eigenvalues. But the $2\times2$ with real eigenvalues is an nice example. – Doug M Dec 15 '18 at 02:24
If $0$ is an eigenvalue of any matrix, $m^2-2m+3\ne0$, because any annihilating polynomial must be divisible by the minimal polynomial. – Shubham Johri Dec 15 '18 at 07:47

Shubham Johri · Answer 3 · 2018-12-17T22:06:54.033

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.

score 0 · Answer 4 · answered Dec 14 '18 at 21:49

0

hint

By Cayley-Hamilton,

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

then

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

then

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

answered Dec 14 '18 at 21:49

hamam_Abdallah

63,719

Cayley-Hamilton only works in one direction doesn't it? So if the characteristic polynomial is $x-\frac{3+i\sqrt 3}{2}$, the result will also hold. And it can for instance be $(x-\frac{3+i\sqrt 3}{2})^2$ as well with certain restrictions. – Klaas van Aarsen Dec 14 '18 at 22:30

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

4 Answers4