One-to-one ProofDate: 06/25/98 at 05:49:32 From: will Subject: One-to-one How can I do the following proof? Let T(x) = Ax, where A is a 2 x 2 matrix. Show that T is one-to-one if and only if the determinant of A is not zero. Date: 06/27/98 at 14:03:13 From: Doctor Joe Subject: Re: one-to-one Dear Will, I shall assume that you know the standard theorems: (A) n x n matrix has determinant not zero if and only if the matrix has row rank = n. (B) For an n x n matrix A, row rank(A) + nullity (A) = n. Along the way, I shall leave some work to you. This means I shall start the proof and leave enough hints for you to go on. Now let's start to prove the statement. (=>) If the map T is one-one. Suppose that the det A = 0. Take any v from Ker A (which is defined as the subspace of R^2 which is mapped to the 0-vector of R^2). Then, T(v) = 0. But T(0) = 0. This implies that T(v) = T(0). Exercise: Make use of the fact that T is one-to-one and invoke Theorem (A) to deduce the nullity(A). By using Theorem (B), complete the proof. (<=) Assume that det A is not zero. Then we know that the nullity (A) is zero, by Theorem (A) and (B) by the same reasoning. Therefore Ker(A) = {0}. Take any two vectors u and v such that T(u) = T(v). Therefore, T(u)-T(v) = 0. Exercise: Use the linear property of T to show that u-v is an element of the Ker A. Show that T is one-one. I hope this helps. - Doctor Joe, The Math Forum http://mathforum.org/dr.math/ |
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