ENS CHALLENGE

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

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>>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

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come on faggits

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>>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