8
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ Have you thought about what the Cayley-Hamilton theorem tells you here? $\endgroup$ – user296602 Dec 14 '18 at 21:45
  • 4
    $\begingroup$ All matrices means all $2\times2$ matrices, I guess. $\endgroup$ – egreg Dec 14 '18 at 22:34
0
$\begingroup$

hint

By Cayley-Hamilton,

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

then

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

then

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

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ – Klaas van Aarsen Dec 14 '18 at 22:30
1
$\begingroup$

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

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.

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

$\endgroup$
10
  • 1
    $\begingroup$ How is $0$ an eigenvalue? $\endgroup$ – Shubham Johri Dec 14 '18 at 22:00
  • $\begingroup$ $0$ is an eigenvalue for $n\times n$ matrices $n>2$. Rather if $m^2 - 3m + 3 = 0$ then so does $m^3 - 3m^2 + 3m = 0$. $\endgroup$ – Yadati Kiran Dec 14 '18 at 22:19
  • $\begingroup$ @YadatiKiran ? this is obvisouly false. $\endgroup$ – Thinking Dec 14 '18 at 22:21
  • $\begingroup$ @Thinking: And why would that be? $\endgroup$ – Yadati Kiran Dec 14 '18 at 22:31
  • $\begingroup$ @YadatiKiran The question is rather. Why $0$ is an eigenvalue of every matrix ? Just take an invertible matrix and $0$ is never an eigenvalue... $\endgroup$ – Thinking Dec 14 '18 at 22:32
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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).

$\endgroup$
8
  • $\begingroup$ The links don't work. $\endgroup$ – YiFan Dec 17 '18 at 22:26
  • 1
    $\begingroup$ @YiFan I think the Pauli matrices satisfy this condition. $\endgroup$ – Mustafa Said Dec 17 '18 at 22:41
  • $\begingroup$ @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. $\endgroup$ – YiFan Dec 17 '18 at 22:45
  • 1
    $\begingroup$ @YiFan you can google "Pauli matrices" and I am pretty sure they satisfy the condition $X^2 = -I$. $\endgroup$ – Mustafa Said Dec 17 '18 at 22:48
  • $\begingroup$ @YiFan: sorry ! I forgot to add the links: added now $\endgroup$ – G Cab Dec 17 '18 at 23:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.