![logo%20ENS[1].jpg](https://i.imgur.com/38TERUD.jpg "logo%20ENS[1].jpg")
forget putnam exams, let's get some worthy oral exams in here
I will try to post regularily some funny challenges.
If no one has found in less than say 24h, I'll post a solution. Will try to add hints before if necessary.
If you already know the solution, please let the others try before posting.
Show that every matrix [math] M \in \mathcal{M}_2(\mathbb{R}) [/math] can be written as [math] A^2 + B^2[/math], with [math]A,B \in \mathcal{M}_2(\mathbb{R}) [/math]
What is [math] \mathcal{M}_2(\mathbb{R}) [/math]?
Is it the ring of [math]2 \times 2[/math] matrices with real entries or what?
>>7801917
yes
sorry I thought it was like that everywhere.
>>7801905
> pesage de couilles façon normale sup'
FGN ftw
>>7801905
Every matrix with real entries is similiar to a matrix in real Jordan normal form. So we only have to consider three cases:
Case 1:
[eqn] M = P^{-1} \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} P [/eqn]
Case 2:
[eqn] M = P^{-1} \begin{pmatrix} a & 1 \\ 0 & a \end{pmatrix} P [/eqn]
Case 3:
[eqn] M = P^{-1} \begin{pmatrix} a & b \\ -b & a \end{pmatrix} P [/eqn]
In case 1 find a number c>0 such that a + c > 0 and b + c > 0 then
[eqn] M = P^{-1} \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} P = \left( P^{-1} \begin{pmatrix} \sqrt{a + c} & 0 \\ 0 & \sqrt{b + c} \end{pmatrix} P \right)^2 + \left( P^{-1} \begin{pmatrix} 0 & \sqrt{c} \\ -\sqrt{c} & 0 \end{pmatrix} P \right)^2[/eqn]
In case 3 calculate [math]c + i d := \sqrt{a + i b} [/math] then
[eqn] M = P^{-1} \begin{pmatrix} a & b \\ -b & a \end{pmatrix} P = \left( P^{-1} \begin{pmatrix} c & d \\ -d & c \end{pmatrix} P \right)^2 [/eqn]
q.e.d.
>>7801960
In case 2 again find a number c>0 such that a+c > 0 then
[eqn] M = P^{-1} \begin{pmatrix} a & 1 \\ 0 & a \end{pmatrix} P = \left( P^{-1} \begin{pmatrix} \sqrt{a + c} & \frac{1}{2 \sqrt{a + c}} \\ 0 & \sqrt{a + c} \end{pmatrix} P \right)^2 + \left( P^{-1} \begin{pmatrix} 0 & \sqrt{c} \\ -\sqrt{c} & 0 \end{pmatrix} P \right)^2 [/eqn]
>>7801960
I don't understand why your cases cover all possible cases
>>7802049
The characteristic polynomial of M is
[math] t^2 - \text{tr}(M) t + \det(M) [/math].
>If it has two distinct real roots we're in case 1 with [math]a \neq b [/math].
>If it has only a single real root and the minimal polynomial is different from the characteristic polynomial we're in case 1 with [math]a = b [/math].
>If it has only a single real root and the minimal polynomial is equal with the characteristic polynomial we're in case 2.
>If it has two complex conjugate roots we're in case 3.
There is no other case.
>>7802081
thanks anon
haven't seen a lot about jordan forms yet, that's why I couldn't see the possible splitting
OK
let's try another one since everyone is convinced.
I have another method for the first question if anyone is interested.
Let [math](P,Q) \in \mathbb{C}[X]^2[/math] be two non constant polynomials that both have the same set of roots.
Assume [math]P-1[/math] and [math]Q-1[/math] also both have the same set of roots.
Show that [math]P=Q[/math]
Will start posting tips if anyone needs them.
bump for glory
come on faggits
>>7803575
this should work, i think. maybe i should detail the part where i mention that C is algebraically closed.
>>7803575
use Extended Euclid's algorithm. pretty easy desu
