Geoffrey Klok

Math 232A

Lord Byron is reported to have written jokingly in a letter to his future wife, "I know that two and two make four--and I should be glad to prove it too if I could--though I must say if by any sort of process I could convert two and two into five it would give me much greater pleasure"(Mathematics; Byron, Lord). The idea of proving that such a prospect is impossible would seem irrational, even pointless to most people. But for linear algebra, even the most fundamental of concepts must be proven prior to their use. Let us consider a formula:

Given that T:Rⁿ®R^m is a linear transformation, prove that T(u) = T(v)

if and only if T(u-v) = 0, where u and v are vectors in Rⁿ.

What does this statement mean? Simply put, our objective is to prove both that if T(x) minus T(y) equals zero, then T(x) must equal T(y), and that the converse of this statement (if T(x) equals T(y), then T(x) minus T(y) equals zero) are true. Any difficulty in the proof is a result of its application to linear algebra and vectors. However, the benefits of proving a hypothesis fully compensate for the work involved, since once a statement is proven, it may be used without question in any future application.

Motivated by these facts, let us begin by considering what is involved in this proof. In reality, there are only two possible outcomes: either the statement is true, or the statement is false. We shall be using two methods of proof in our argument; the first half, involving the "if" of "if and only if" in the statement above, will be a direct proof (A implies B); the second half of our proof refers to the phrase "and only if" in the centered statement above and warrants a brief discussion. This second point could be approached once again from the standard direct line of proof (B implies A, in this case) so that "if and only if" means "not only does A imply B, but B implies A". However, to make things more interesting, let us prove the second half of the statement by contrapositive, i.e. NOT B implies NOT A. Typically this method of proof is more difficult, but for our purposes, it is very effective, simple, and produces the same results. "If and only if" explained, we shall begin under the hypothesis that the statement is true, which implies that

1.If T(u-v) equals 0, then T(u) must equal T(v).

2.For any transformation T, if T(u-v) does not equal 0, then T(u) is not equal to T(v).

Let us now consider the proof of part one:

• We are given the formula T(u-v) = 0 and that T is linear.
• According to Otto Bretscher, we know:

"A transformation T from Rⁿ to R^m is linear if (and only if)

a) T(v + w) = T(v) + T(w), for all v, w, in Rⁿ, and

b) T(kv) = kT(v), for all v in Rⁿ and all scalars k." (Bretscher 66).

Given that the above is a requirement for linearity, and given that T is a linear transformation, we can infer that

T(u-v) = 0 Þ T(u) - T(v) = 0.

• Since we know that T(u) - T(v) = 0, we can add T(v) to both sides of the equality so that

T(u) - T(v) + T(v) = 0 + T(v).

• Logically T(v) - T(v) = 0, so we can replace the appropriate values in our formula, so that

T(u) - T(v) + T(v) = 0 + T(v) Þ T(u) - 0 = 0 + T(v),

• and it is obvious then that

T(u) - 0 = 0 + T(v) Þ T(u) = T(v).

We have just proven part one by showing that

if T(u-v) = 0 then T(u) = T(v).

Now we must show that if T(u-v) does not equal 0, T(u) is not equal to T(v). The proof of this point is very similar to the proof of part one. Let us consider the following:

• We are given T(u-v)¹ 0, and that T is once again a linear transformation.
• Since T is linear, the requirements for linearity hold true again (see Part one, step two)

and therefore T(u-v)¹ 0 can be written as T(u) - T(v) ¹ 0.

• Just as in part one, step three, we add the value of T(v) to both sides of the inequality, which implies

T(u) - T(v) + T(v) ¹ 0 + T(v).

• Once again, we can see that T(v) - T(v) = 0, so let us change the corresponding values in our equation so that T(u) - T(v) + T(v) ¹ 0 + T(v) Þ T(u) - 0 ¹ 0 + T(v),

and it becomes obvious that T(u)¹ T(v).

We have just shown that if T(u-v) does not equal 0 then T(u) does not equal T(v). By proving this second part, we have also succesfully proven that both part one and part two hold true, and therefore, the entire statement is correct. In other words,

Given that T:Rⁿ®R^m is a linear transformation, T(u) = T(v)

if and only if T(u-v) = 0, where u and v are vectors in Rⁿ.

What are the applications of the statement we have just proven? Consider the following:

The kernel of a linear transformation is defined as the set of all zeroes of the transformation. In other words, ker(N) = {x ÎRⁿ : N(x) = 0}. Suppose that the kernel of our linear transformation T is the set which contains only the zero vector, meaning ker(T) = {0}. In the course of this proof we have considered the case where T(u - v) = 0. If this is the case, then (u - v) is in the kernel of T. This leads us to the following:

T(u - v) = 0 Þ ker(T) = {(u - v)}.

But since ker(T) = {0}, (u - v) = 0.

(u - v) = 0 Þ u = v.

So for any linear transformation T, if T(u - v) = 0 and ker(T) = {0}, u = v.

But let us take this reasoning even further.

Otto Bretscher provides an interesting proof of an important quality found in many linear transformations. If we have T(u - v) = 0 and u = v, then ker(T) = {0}. If T(x) = Ax, where A is an nxn matrix, then we can deduce the following.

ker(A) = {0} Þ im(A) = Rⁿ,

im(A) = Rⁿ Þ rank(A) = n,

rank(A) = n Þ reduced row echelon form (rref) of A = I_n,

rref(A) = I_n Þ for any b ÎRⁿ of Ax = b, x = A^-1b,

x = A^-1b Þ for every output b, there is only one input x Þ T is defined as one-to-one

over all of Rⁿ.

So, we can state the following fact:

Given T:Rⁿ®Rⁿ is a linear transformation and vectors u and v in Rⁿ,

if T(u - v) = 0 and ker(T) = {0}, then u = v and T is one-to-one over all of Rⁿ.

Note: This extensive proof is explained well on page 139 of Otto Bretscher’s Linear Algebra with Applications.

These steps open up a wide variety of options to consider, too numerous to mention here. But in essence, the properties we have considered today are fundamental to a great deal of concepts. This is truly the rationale behind considering such elementary proofs: they provide the basis for moving on to far more advanced ideas, and sometimes, they contain a hidden wealth of information themselves.

SUMMARY OF ALGEBRA

1) T: Rⁿ ® R^m, {u, v ÎRⁿ}, T(u-v) = 0.

2) T(u-v) = 0 Þ T(u) - T(v) = 0.

3) T(u) - T(v) = 0 Þ T(u) - T(v) + T(v) = 0 + T(v).

4) T(u) - T(v) + T(v) = 0 + T(v) Þ T(u) = 0 + T(v).

5) T(u) = 0 + T(v) Þ T(u) = T(v).

1) T: Rⁿ ® R^m, {u, v ÎRⁿ}, T(u-v) ¹ 0.

2) T(u-v) ¹ 0 Þ T(u) - T(v) ¹ 0.

3) T(u) - T(v) ¹ 0 Þ T(u) - T(v) + T(v) ¹ 0 + T(v).

4) T(u) - T(v) + T(v) ¹ 0 + T(v) Þ T(u) ¹ 0 + T(v).

5) T(u) ¹ 0 + T(v) Þ T(u) ¹ T(v).

Bretscher, Otto. Linear Algebra with Applications. Upper Saddle, NJ: Prentice Hall, 1997.

"Mathematics; Byron, Lord." Microsoft Bookshelf ‘98. CD-ROM. Microsoft, Inc., 1997.