To be clear, it got the first question right for the wrong reasons, and the second question wrong for the wrong reasons. Its math is nonsense and doesn't resemble linear algebra at all. I would guess it's just saying "yes" without understanding and then regurgitates some stuff that resembles math, but is not recognizable linear algebra.
I've relied on GPT4 a lot while learning math, it has been very precise and has not lead me astray. I do use a text book as well.
I shared my story about it being wrong, but it was wrong in a very open minded manner. It knew C(A) = C(AA^T) might be true for special cases, but it couldn't see that it was true for all cases until I explained the proof. Even when wrong it still made compelling arguments that made sense. (Of course, arguments that are wrong but make sense are the most dangerous kind.)
GPT4 is still state-of-the-art. Claude has a ways to go.
OP is using the notation C(A) to mean "the column space of matrix A", whereas here that notation is being interpreted as "multiplication of matrix C by matrix A, which happens to be inside of parentheses". These are totally different questions. It's a reasonable misunderstanding as OP's notation is nonstandard.
But also the following math/logic is nonsense.
and it answered:
Yes, that is correct for appropriate matrix sizes. If A is an m x n matrix and C is an n x m matrix, then (CA) = C(AA^T) where A^T is the transpose of A.
This comes from the property that for any matrices A, B and C of appropriate sizes:
Applying that to your example, with B = A^T, we have: So in short, yes your statement (A) = C(AA^T) is true for matrices A and C of the appropriate dimensions.Then I asked it: "in linear algebra, is the following true? C(A) = C((A^T)A)"
and it answered:
Yes, that is also true for appropriate matrix sizes. This follows from a similar argument as before.
If A is an n x m matrix and C is an m x n matrix, then:
So the statement C(A) = C((A^T)A) is true for matrices A and C of the appropriate dimensions.This works because post-multiplying a matrix A by its transpose A^T results in a square matrix, (A^T)A, of the same dimension as C. So C can be multiplied with that resulting square matrix.
In summary, both of the statements you provided are true linear algebra identities for appropriately sized matrices:
Hope this explanation helps! Let me know if you have any other questions.[1] - https://poe.com/Claude-instant