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### One-to-one Proof

```
Date: 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/
```
Associated Topics:
College Linear Algebra

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