ENS CHALLENGE

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Anonymous

ENS CHALLENGE 2016-01-22 13:04:29 Post No. 7801905

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ENS CHALLENGE 2016-01-22 13:04:29 Post No. 7801905

[Report] Image search: [iqdb] [SauceNao] [Google]

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?

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>>7801917

yes

sorry I thought it was like that everywhere.

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>>7801905

> pesage de couilles façon normale sup'

FGN ftw

>>

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.

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>>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]

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>>7801960

I don't understand why your cases cover all possible cases

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>>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.

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>>7802081

thanks anon

haven't seen a lot about jordan forms yet, that's why I couldn't see the possible splitting

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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.

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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.

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>>7803575

use Extended Euclid's algorithm. pretty easy desu

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