How is idempotent related to the rank of Matrix?
#Basic Linear Algebra
steady drum
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How is idempotent related to the rank of Matrix?
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lament vale
How is idempotent related to the rank of Matrix?
If we have an n⨯n matrix A such that A^k = 0, then rk(A) ≤ ⌊n(k - 1)/k⌋.
Sources:
https://math.stackexchange.com/a/863077
https://math.stackexchange.com/a/863087
steady drum
If we have an n⨯n matrix A such that A^k = 0, then rk(A) ≤ ⌊n(k - 1)/k⌋.
Sources...
Aah I see
indigo bone
@steady drum you can also use the classification of nilpotent matrices in terms of integer partitions
(that's probably where this result come from)
A² ≠ 0 and A⁴=0, thus you know that the sequence λi = dim (Ker(A^(i)) / Ker(A^(i-1))) satisfies λ1≥λ2≥ … ≥ λ6 with Σ λi = 6, λ5 = λ6 = 0. Then doing all the partitions (there are 5 of them) gives that the dimension of the kernel is either 2 or 3, hence that the rank is either 3 or 4.
steady drum
A² ≠ 0 and A⁴=0, thus you know that the sequence λi = dim (Ker(A^(i)) / Ker(A^(i...
Aah what exactly kernel mean here?
indigo bone
ker A = {x st Ax = 0}
jade summit
A² ≠ 0 and A⁴=0, thus you know that the sequence λi = dim (Ker(A^(i)) / Ker(A^(i...
how do you prove that for the sequence? #math-discussion
indigo bone
how do you prove that for the sequence? <#1154179069806641192>
you can do this in an elementary way, it is called the iterated kernels lemma. Denoting Ki=Ker A^i, then
- Ki c Ki+1
- there is an injective map Ki+1/Ki-> Ki/Ki-1
Thus, the sequence lambdai satisfies the announced relations because since A is nilpotent, Kn=V for some n <= 6.
Thus,you car just en umerate integer partitions.
Conversely, given any integer partition, it is not that hard to construct an explicit matrix having these lambdai.
jade summit
thank you for the explanation
steady drum
ker A = {x st Ax = 0}
Aah basically map to null vectors 😭
indigo bone
thank you for the explanation
(you start from the map A : Ki+1->Ki, then A^(-1) (Ki-1) = Ki, thus you get the injection using the first isomorphism theorem)
steady drum
Man how rusty I got in LA
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@steady drum