Inverse of a matrix

[gleaming elm]

Help with this one.

[lethal whale]

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[flat jewel]

what have you tried?

[low narwhal]

Try to rearrange the equality such that you have A * something = some constant * identity

there is an obvious way to do it, you'll see it

it suffices to note that for square matrices AB = I implies BA= I

@gleaming elm

[terse gale]

Note that a matrix satisfies its characteristic polynomial. And a matrix is invertible if it doesn't have any zero eigenvalues.

[low narwhal]

this is an annihilating polynomial, which need not be a multiple of characteristic

Cayley Hamilton implies only that characteristic is annihilating, not the other way round

As an example, (x-2)(x-1) annihilates the 2x2 identity, but it does not divide [nor is divided by] (x-1)(x-1) , which is the characteristic polyn. of the identity

it is true that determinant is zero if and only if 0 is an eigenvalue

[terse gale]

Oh, interesting! I didn't know about that.

[candid silo]

Yes and for any polynomial P in Ann(A) Sp(A) C root(P) so here the annulator polynomial obviously doesn’t have 0 as a root thus 0 is not an eigenvalue thus it’s indeed invertible

I find it really cool that you can deduce so many properties of the matrice just by looking at one of its annulator polynomials

[low narwhal]

there is a much more elementary solution to this particular problem, though

involves a bit of algebraic manipulation and it's done

[terse gale]

How do we ensure the RHS isn't degenerate, though?
Also, you forgot I near the 2.

[sacred lagoon]

Ah yea

True 🙁

[flat jewel]

you lot need to learn to let the user respond before discussing the ideas lol

[sacred lagoon]

I thought it was allowed

Sorry!

[flat jewel]

It's fine, but like having an entire discussion in a help channel isn't useful

it drowns out the actual helper's message, since we still dont know what the OP has tried/thought

[low narwhal]

this ensures det A is nonzero

so there's no problems with existence of A-1

[terse gale]

Oh, true!

[low narwhal]

to be fair I also implicitly assume here that det AB = det A * det B, which is true for all square matrices

[gleaming elm]

I like this one the best. Because this year we only have elementary linear algebra. We will study eigenvalues the next semester

[gleaming elm]

I also understand this and it's fair enough

[terse gale]

Well, that information isn't relevant here, after all, due to what aL said.

[gleaming elm]

I didn't understand that annihilating thing. Reading about it rn.

[gleaming elm]

.

[low narwhal]

the annihilating polynomial talk is redundant here, I commented on it because there was a comment regarding characteristic polynomials

but we are only given an arbitrary annihilating polynomial, we can't deduce much about the characteristic polynomial from that alone

[gleaming elm]

Ok

[flat jewel]

yeah, aL has a tendency to just talk, forgetting the point of helping

[gleaming elm]

😅

[flat jewel]

But yeah, as aL just handed out to you, you can write I as a product of A(something else), and that something else will be in terms of A, to which that must be the inverse