I have the following theorem plus proof, could you please explain this more, I dont understand the proof, I have no idea or grasp on what they're doing and why, so please explain the proof in more detail Theorem : For every linear subspace V of Rn there exists an m ×n matrix A such that V is the null space of A. Moreover, we can choose m = n − dim(V ).
I have the following theorem plus proof, could you please explain this more, I dont understand the proof, I have no idea or grasp on what they're doing and why, so please explain the proof in more detail
Theorem : For every linear subspace V of Rn there exists an m ×n matrix A such that V is the null space of A. Moreover, we can choose m = n − dim(V ).
Proof: Choose a basis of V and write the basis vectors in row form as a dim(V ) × n matrix that we call B. Choose a basis of the null space of B. This consists of n − dim(V ) elements. Write these basis vectors as columns in a matrix C. So BC = 0. After transposing, CtBt = 0. Now let A = Ct. We have the following facts, rank(A) = rank(C) = n − dim(V ) and V is contained in the null space of A due to ABt = CtBt = 0. Furthermore, dim(Null(A)) = n − (n − dim(V)) = dim(V ).
It follows that: Null(A) = V.
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