Given ATAx = 0 A T A x → = 0, we have that Ax A x → is in the image of A A as well as in the kernel of AT A T. Since the kernel of AT A T is the orthogonal complement of the image of A …

Sep 3, 2019 · Suppose some matrices A ∈Rn×n A ∈ R n × n with rank(A) = r r a n k (A) = r and B ∈Rn×(n−r) B ∈ R n × (n − r) satisfies Ker(A) = Im(B) K e r (A) = I m (B). In this situation, I …

Today we were discussing how for an nxn orthogonal projection matrix from Rn R n onto a subspace W, Ker (A A)= (Im (I m A)⊥) ⊥ = W⊥ W ⊥ and that Ker (AT A T) is also W⊥ W ⊥. …

You have figured out that in order for a vector ⎡⎣⎢x y z⎤⎦⎥ [x y z] to be in the kernel, you must have x = y x = y. From here, it is personal preference how you explicitly write the kernel, but a …

Mar 25, 2015 · If A A is a n × m n × m matrix, is the formula (ker A)⊥ = im AT (ker A) ⊥ = im A T necessarily true? I'm thinking that rank-nullity would be the simplest and easiest way to prove …

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I'm trying to prove the following: Let A A and B B be two commutative square matrices (AB = BA A B = B A) over a commutative field such that Im(A) = ker(A) Im (A) = ker (A) and Im(B) = ker(B) …

Let A A be a square matrix. Show that im(A⊤) = ker(A)⊥ im (A ⊤) = ker (A) ⊥.

How do I show that ker(A) = Im(A∗)⊥ ker (A) = Im (A ∗) ⊥ for any square matrix A A. I have done this problem before with the linear operator T T on a hermitian space but I can't seem to apply …

Feb 10, 2016 · Consider the following true/ false qustion: There exists a 2 × 2 2 × 2 matrix A A such that im(A) = ker(A) im (A) = ker (A). I know that this is true, but I am not sure how to show …