3 ms·
It's a good initial thought, GPT had the same thought and said similar things. (I had no initial thought, since the text book spoon fed me the proof.) The proo
by Buttons840 3y ago
It's a good initial thought, GPT had the same thought and said similar things. (I had no initial thought, since the text book spoon fed me the proof.)
The proof is (briefly): A and AA^T have the same rank. AA^T is a linear transformation of the columns in A and spans the same number of dimensions and is thus the same subspace. It applies to non-square matrices as well.
- picatepica 3y agoYou are right, A can be non-square because A and AA^t both have the same number of rows. To prove that A and AA^t have the same range you apply that if AA^Tv = 0 then v^TAA^Tv = 0, this implies that (A^Tv)^T (A^tv) = 0, so that A^Tv = 0 hence dim(nullspace AA^T) = dim(nullspace A^t) and so range(AA^T) = range(A^T).