on matrices

[analog fox]

what would be adj ( adj ( adj A))) and adj(adj(adj (adj (A )))) and n times adj(adj(adj(adj...n (A))

[astral gust]

We know:
A^(-1) = |A| adj(A)
So:
adj(A) = |A| A^(-1)
adj(A^(-1)) = |A^(-1)| A = |A|^(-1) A
So:
adj(adj(A)) = |adj(A)| adj(A^(-1)) = ||A| A^(-1)| |A|^(-1) A
Now, suppose that A is an n⨯n matrix. In that case |aA| = a^n |A|. So:
adj(adj(A)) = |A|^(n - 1 - 1) A = |A|^(n - 2) A
And now you can just continue from here.

[analog fox]

can you just explain for 4 times k

I got it for 2 times

adj

[astral gust]

Just apply the result twice.

[analog fox]

hm

[analog fox]

like there is a formula for k times

k times adj

is there any pattern?

[astral gust]

Hm... Well, we can probably look at several cases, then maybe try induction.

[analog fox]

hm

I got one for even power

adj adj adj... k times A = [|A|^((n-1)^k - 1 / n) * A

like this

but k must be even

for odd power there must be another one

[analog fox]

not sure if its correct

[astral gust]

We already have adj(A) = |A| A^(-1) and adj(adj(A)) = |A|^(n - 2) A. So, it's obvious that an odd amount of adjugated irls proportional to A^(-1) and even to A.

[astral gust]

Let's see...

[analog fox]

hmm

[astral gust]

Hold on, let me move to my PC.

[analog fox]

okayy

[astral gust]

Alright, let's try to see a pattern.
For brevity, let d = |A| and let k adjoints of A be adj(k, A).
adj(3, A) = adj(2, adj(A)) = |adj(A)|^(n - 2) adj(A) = |dA^(-1)|^(n - 2) dA^(-1) = d^(n(n - 2) - (n - 2) + 1) A^(-1) = d^(n^2 - 3n + 3) A^(-1)
adj(4, A) = adj(2, adj(2, A)) = |adj(2, A)|^(n - 2) adj(2, A) = |d^(n - 2) A|^(n - 2) d^(n - 2) A = d^(n(n - 2)^2 + (n - 2) + (n - 2)) A = d^(n^3 - 4n^2 + 6n - 4) A
Clearly, those are just entries from Pascal's triangle. Looking at the coefficients, I hypothesize the following:
adj(k, A) = |A|^(((n - 1)^k - (-1)^k)/n) A^((-1)^k)
To remind you: k is the amount of adjugates taken, n is the size of the matrix.
If true, I think we can prove it by induction, but I'm too lazy 😄

[analog fox]

alr

lemme read

[astral gust]

Here you go - a bit easier to read.

[analog fox]

alright

[astral gust]

This works for k = 0, 1, 2, 3, 4.

[analog fox]

👍

[astral gust]

Interesting formula!

Haven't really thought about doing such an operation. Turned out to be interesting.

[analog fox]

so

the formula I got was right, right?

for even power

Yes got it

thanks @astral gust

[granite urchin]

@analog fox has given 1 rep to @astral gust

[analog fox]

thanks @astral gust

[astral gust]

You're welcome! As I said, this was quite interesting for me, too. I've saved this result.

[analog fox]

pro

[analog fox]

@astral gust

is A inverse

in last part

@astral gustping me when u r here

[astral gust]

@kind latch , what do you mean?

[analog fox]

the

A^(-1)^k

here

is it A inverse

@astral gust

[astral gust]

Well, (-1)^k = -1 for odd k and 1 for even k.
So, A^(-1)^k = A^(-1) for odd k and A for even k.

[analog fox]

I get that

for example if A is 1

then is it

A inverse

or just

1/A

?

[astral gust]

1/A doesn't make sense for matrices.

You can't divide matrices.

[analog fox]

yeah

thats what I asked

[astral gust]

You can, however, raise them to an integer power. So, for example A^(-1).

[analog fox]

hm

so

not inverse right

[astral gust]

As I just said, inverse for odd k and just the matrix for even k.

[analog fox]

A inverse is denoted by A^-1

[analog fox]

yeah got it

[analog fox]

can we get those without pascal's triangle?

just using properties

ping me when here

[analog fox]

how did u get (-1)^k in (n-1)^k - (-1)^k

@astral gust lmk when u r here

[astral gust]

Hm... Well, maybe.
But at this point it's just easier to guess a formula and prove it by induction.

[astral gust]

That's because each time I recognize the expression for (n - 1)^k, but without the final term, which is (-1)^k, and all the powers of n are 1 less than in (n - 1)^k.

[analog fox]

yeah

[analog fox]

yeah I understood

I tried again before some time

[analog fox]

yeah its easier but I was trying to get the same using properties

[merry thorn]

This doesn't work for non-invertible matrices though

In the case that n>2 (where n is the size of the matrix) you get that adj(adj(A))=0 if A is non-invertible

[astral gust]

Oh, yeah, true. I assumed that A is invertible, of course.

[merry thorn]

And adj(adj(A))=A if n=2

[astral gust]

Hm.
I think that if A is singular, then if rk(A) = n - 1, then adj(A) ≠ Z, but rk(adj(A)) = 1, and if rk(A) < n - 1, then adj(A) = Z.

[merry thorn]

All matrices worth dealing with are

Z meaning the zero matrix?

[astral gust]

Yes.

[merry thorn]

If rk(A)=n-1 then rk(adj(A))=1

And the rest is true yeah

I just stumbled across this a while back when playing around with cross products

[analog fox]

ooo

[analog fox]

is there something general for all cases

[astral gust]

Well, in case of A not being invertible the formulas might become a bit easier, but I don't care about those cases, honestly.

[analog fox]

hm yeah true

lets leave it hear then ig

[analog fox]

using properties
since A^-1 is adjA )/ |A|
I wrote like
adj(adj...k times(A)) is

[|A|^((n-1)^k - 1 / n) * A if k is even
[|A|^((n-1)^k - (n-1) / n) * adj(A) if k is odd

so yeah we can prove using

properties as well

[merry thorn]

You need to split it up into cases as we did above

[analog fox]

yeah

[analog fox]

thanks @astral gust

[granite urchin]

@analog fox has given 1 rep to @astral gust

[analog fox]

thanks @astral gust

[granite urchin]

@analog fox has given 1 rep to @astral gust

[livid summit]

LordDarpinger: 0 Rep (##ω)

[astral gust]

Well, you've already thanked me 😄