#permutation matrices
leaden acornBOT
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spark tendon
if $(e_i){1 \le i \le p}$ is the canonical basis of $\bR^{p}$ what is $A{\sigma} e_i$ ? $\forall \sigma \in S_p, \forall i \in {1,…,p}$
tawdry thornBOT
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spark tendon
you can also try expressing the coefficients of $A_{\sigma}$ using the kronecker delta notation
tawdry thornBOT
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spark tendon
For any permutation $\sigma$
tawdry thornBOT
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dry sundial
We didn't have the kronecker delta notation in the lecture, which is why we are not allowed to use it. Regarding your other suggestion. A_σ is the unit matrix, but with swapped rows, so that A_σ has a 1 in the 𝜎(i)-th position and a zero otherwise. However, I don't know exactly how I can deduce Aσ - Aτ = A_σ◦τ and, above all, how this can be written down formally / mathematically.
spark tendon
We didn't have the kronecker delta notation in the lecture, which is why we are ...
Prove that $\forall i \in {1,…,p}$ A_{\sigma} e_i=e_{\sigma(i)}$
tawdry thornBOT
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Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
spark tendon
$A_{\sigma} A_{\tau}=A_{\sigma \circ \tau}$ if and only if $\forall i \in {1,…,p}, A_{\sigma} A_{\tau} e_i=A_{\sigma \circ \tau} e_i$
tawdry thornBOT
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spark tendon
a linear map is uniquely determined by its image for each element of the basis
@dry sundial
spark tendon
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Try proving the right proposition
spark tendon
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Compile Error! Click the <:errors:50426655605496219...
After proving this
dry sundial
Does e_i mean the unit vector, in other words a vector that has a 1 in the i-th position and a 0 in the remaining positions?
spark tendon
Does e_i mean the unit vector, in other words a vector that has a 1 in the i-th ...
Yes
spark tendon
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.
spark tendon
I proposed to do it with the canonical basis but you can also show that if $x=(x_1,….,x_p)$ then $A_{\sigma} x=(x_{\sigma(1)},….,x_{\sigma(p)})$
tawdry thornBOT
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dry sundial
Can I send my proof and you tell me if I've done something wrong?
spark tendon
Yeah go right ahead
dry sundial
The light blue one says that e_i only has a 1 at the i-th position and a 0 otherwise, from which it follows that only if j=i, the multiplication is not 0.
The green one says that E_σ has a 1 at σ(i), i has a 0 otherwise and e_i has a 1 only at i, from which it follows that the multiplication results in a single-column matrix that is not equal to 0 only at σ(i), consequently e_σ(i).
Behauptung, Beweis = Assertion, proof
@spark tendon
Are you still online?
spark tendon
Yeah sorry I was eating
dry sundial
No problem. I was just wondering why there was no answer.
spark tendon
Looks good
spark tendon
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Although I’d explain a bit why this is
At the beginning of your proof
dry sundial
What exactly would you explain? Why I calculate A_σ*e_i or do you mean something else?
spark tendon
What exactly would you explain? Why I calculate A_σ*e_i or do you mean something...
I mean, why to prove that both matrices are equal, does it suffice do prove that their image is the same on the canonical basis, if it’s a property you’ve seen in class, then you just need to mention that is what you are using, if you haven’t seen it, it requires more justification
dry sundial
Something like that, and if so, where would you best insert that into the proof.
spark tendon
Right before $A_{\sigma} e_i=e_{\sigma(i)}$ me thinks
tawdry thornBOT
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spark tendon
You are very welcome
grand compassBOT
hoary radish
@dry sundial here