/sci/ - Science & Math » Thread #13081203

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Anonymous Wed 05 May 18:35:49 2021 No.13081203 View Reply Original Report

Quoted By: >>13081341 >>13081942

Hi, math bros! I'm pretty brainlet at linear algebra and I have to prove this.

"Suppose that u and v are unit vectors of Rn (vector space) such that u and v are orthogonal.
Q = u*ut + v*vt.
Prove that Q is an idempotent matrix".

(t comes from transpose)

Could you please help me, or at least give me a hint?
I am a boomer studying data science online and I don't know anyone to ask for help in real life. Thank you very much.

Anonymous

Anonymous Wed 05 May 2021 18:41:07 No.13081240 Report

Quoted By:

(pic not related btw)

Anonymous

Anonymous Wed 05 May 2021 18:42:09 No.13081248 Report

Quoted By: >>13081268

https://yutsumura.com/unit-vectors-and-idempotent-matrices/#b_Prove_that_Qmathbfumathbfutransmathbfvmathbfvtrans_is_an_idempotent_matrix

Anonymous

Anonymous Wed 05 May 2021 18:44:42 No.13081268 Report

Quoted By:

>>13081248
WTF!!!! How can you be so BASED?
Thank you bro :3

Anonymous

Anonymous Wed 05 May 2021 18:46:55 No.13081282 Report

Quoted By: >>13081295

idempotent means that you get the same result if you apply the matrix to some vector any number of times. in other words, you want to show that multiplying by the matrix twice is the same as multiplying by it once.

you already know $\mathbf{Q} = \mathbf{u} \mathbf{u}^T + \mathbf{v} \mathbf{v}^T$ , so squaring that is $\mathbf{Q}^2 = [\mathbf{u} \mathbf{u}^T + \mathbf{v} \mathbf{v}^T][\mathbf{u} \mathbf{u}^T + \mathbf{v} \mathbf{v}^T]$ . simply distribute
$\mathbf{Q}^2 = \mathbf{u} \mathbf{u}^T \mathbf{u} \mathbf{u}^T + \mathbf{u} \mathbf{u}^T \mathbf{v} \mathbf{v}^T + \mathbf{v} \mathbf{v}^T \mathbf{u} \mathbf{u}^T + \mathbf{v} \mathbf{v}^T \mathbf{v} \mathbf{v}^T$
and take advantage of your orthonormal vectors and simplify the inner products
$\mathbf{Q}^2 = \mathbf{u} 1 \mathbf{u}^T + \mathbf{u} 0 \mathbf{v}^T + \mathbf{v} 0 \mathbf{u}^T + \mathbf{v} 1 \mathbf{v}^T = \mathbf{u} \mathbf{u}^T + \mathbf{v} \mathbf{v}^T] = \mathbf{Q}$
done!

Anonymous

Anonymous Wed 05 May 2021 18:49:42 No.13081295 Report

Quoted By:

>>13081282
Thank you too, anon! That's so clear!

Anonymous

Anonymous Wed 05 May 2021 18:54:52 No.13081341 Report

Quoted By:

>>13081203
the thing is that Q is the orthogonal projection onto the 2-dimensional subspace spanned by u and v. an orthogonal projection is clearly idempotent (actually that's the definition of a projection)

Anonymous

Anonymous Wed 05 May 2021 20:19:08 No.13081942 Report

Quoted By:

>>13081203
Can anyone explain polynomic diffeomorphisms to me please? Am also brainlet

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