Example 1.

Prove that if a square matrix A is a zero divisor (that is AB=0 for some non-zero matrix B) then det(A)=0.

Proof. By contradiction, assume that the assumption that A is a zero divizor is true and in addition assume that the conclusion is false, that is det(A) is not 0. Since A is a zero divisor, there exists a non-zero matrix B such that AB=0. Since det(A) is not zero, by the third theorem about determinants A is an invertible matrix. Multiply the equality AB=0 by A^-1 on the left: A^-1*(A*B)=0. Since A^-1A=I and I*B=B, we get B=0, a contradiction (B is not zero by our assumption). Thus we assumed that our statement is false and deduced an obvious contradiction. This means that our statement is in fact true.

The main feature of a proof by contradiction is that we are allowed to operate with all the assumptions of the original statement A plus the new assumption that the conclusion of A is false.

Example 2.

Prove that if A is a square matrix and det(A)=0 then A is a zero divisor.

Proof. Indeed, assume that det(A)=0 (the assumption of the original statement) and A is not a zero divisor (that is the conclusion of our statement is wrong). By the third theorem about determinants A is not invertible matrix (since det(A)=0). Therefore by the second theorem about inverses the reduced row echelon form C of our matrix A is not the identity matrix. Therefore C contains a zero row. Since C is in the reduced row echelon form, the n-th row of C is zero where n is the size of C.

We want to show that there exists a non-zero matrix B such that CB=0. Let B be a "generic" n by n matrix with unknown entries B(i,j), the number of entries in B is, of course, n², so we have n² unknowns. Represent C as a column of rows r₁,...,r_n and represent B as a row of columns c₁,...,c_n. Then the equation CB=0 is equivalent to a system of n² linear equations r_ic_j=0. If i=n then such an equation has the form 0=0 since r_n=0, so we can delete these equations from our system. Thus we have a system of (n²-n) homogeneous equations with n² unknowns. Since n²-n < n², we can apply the theorem about homogeneous systems of equations and conclude that this system has a non-zero solution. Thus indeed, there exists a non-zero matrix B such that CB=0.

We know that C is obtained from A by a series of row operations. Then A also can be obtained from C also by a series of row operations (see the lemma about inverses of elementary matrices and the paragraph before that lemma). By the theorem about elementary matrices A is equal to E₁E₂...E_kC for some elementary matrices E₁,...,E_k. Therefore

AB=(E₁E₂...E_kC)B=(E₁...E_k)CB=0
This contradicts our assumption that A is not a zero divizor (indeed, B is not 0 and AB=0). This contradiction completes the proof.

Proofs by case-analyzis

This method consists of dividing a statement that we want to prove into several sub-statements and proving them separately. We have used this method in the proof of the theorem about elementary matrices.

The following is an easier example.

Example. Prove that |-x|=|x| for every real number x. Here |.| denotes the absolute value of a number.

Proof. The definition of absolute value suggests that we need to consider two cases: in the first case x is a non-negative number and in the second case x is a negative number.

Case 1. x is non-negative. Then by the definition of absolute value |-x|=x and |x|=x, so |-x|=|x| as required.

Case 2. x is negative. Then again by the definition of absolute value |x|=-x and |-x|=-x, so again |-x|=|x|.

The statement is proved.

To be continued

Proof 2

2. Prove that if all entries of a square matrix A are integers and det(A) =1 then all entries of A-1 are also integers.

Since det(A) =1, we know that A is invertible, and from Theorem 2.4.2 about invertible matricies, we know:

A-1 = (1/det(A)) * ( adj(A))

Since it was given that det(A) = 1, we can rewrite the above equation as:

A-1 = adj(A)

In order to prove that the entries of A-1 are integers, we must show that the entries of adj(A) are integers. Since A has only integer entries, each of the cofactors will be computed as the sum of integer products, and thus, will be integers themselves. By definition, the adjoint of A is the transpose of cofactors from A. Since taking the transpose of the matrix of cofactors changes only the location of the entries within the matrix, and does not affect the entries themselves, we know that the entries of the adjoint of matrix A are also integers. From above, we know that A-1 = adj(A), so since the entries of adj(A) are integers, so also are the entries of A-1.

Thus, our proof is complete!

Problem 1

1. Let A=(1,2,3,4,5), B=(2,1,3,4,4), C=(1,2,1,3,1), D=(2,1,3,1,1) be vectors from R⁵. Find all numbers x, y, z such that:

xA+yB+zC=D

By simply using the multiplication by a scalar operation of vectors (multiplying vector A by x, vector B by y, and vector C by z) we can obtain the following:
xA = (x , 2x , 3x , 4x , 5x) yB = (2y , y , 3y , 4y , 4y) zC = (z , 2z , z , 3z , z) D = (2,1,3,1,1)

Then, by using the addition operation of vectors, we can obtain the following two vectors:

xA+yB+zC = (x+2y+z, 2x+y+2z, 3x+3y+z, 4x+4y+3z, 5x+4y+z)

D = (2,1,3,1,1)

Given the above two vectors (xA+yB+zC and D), we can set up a system of 5 linear equations in 3 variables by setting the components of xA+yB+zC equal to the corresponding components of D:
x + 2y + z = 2 2x + y + 2z = 1 3x + 3y + z = 3 4x + 4y + 3z = 1 5x + 4y + z = 1

Now, we will proceed to set up these 5 linear equations in matrix form.

Step 1: We will begin by opening up the "linalg" package in Maple.

> with(linalg):

Step 2: Now we will define the augmented matrix of the system to Maple:

> A:=matrix([[1,2,1,2],[2,1,2,1],[3,3,1,3],[4,4,3,1],[5,4,1,1]]);

A :=	[ 1	2	1	2 ]
	[ 2	1	2	1 ]
	[ 3	3	1	3 ]
	[ 4	4	3	1 ]
	[ 5	4	1	1 ]

Step 3: Now we will convert the above matrix A into reduced row-echelon form by using the Gauss-Jordan elimination procedures (using Maple, this is done by using the "gaussjord" command):

> B:=gaussjord(A);

B :=	[ 1	0	0	0 ]
	[ 0	1	0	0 ]
	[ 0	0	1	0 ]
	[ 0	0	0	1 ]
	[ 0	0	0	0 ]

From the fourth row of matrix B, it is evident that there will be no solutions to the system of 5 linear equations in 3 variables. This is the case because the fourth row states the following:

0x + 0y + 0z = 1

and, as the knowledge of basic beginning algebra shows, the above is impossible. 0 is not equal to1

Solution: Thus, given the 4 vectors from R⁵ (A, B, C, and D), there exists no set of values for x, y, z that satisfy the given equality

xA+yB+zC=D

The solution is the empty (null) set

Problem 2

2. Two continuous functions f,g:[0,1] --> R are called orthogonal if the inner product f*g is 0. Find a non-zero continuous function f(x) which is orthogonal to x₂. Find infinitely many such functions. (Use the Maple operation "int" to compute integrals of functions.)

What we need to develop is a function f(x), such that the following is true:

0 Ù 1 (x₂*f(x)) dx = 0

First, we used trial and error to find one such function f(x) that made the previous expression true. That function was:

f(x) = (x-2) -3

From above, we realized that we could multiply both (x-2) and -3 by a constant, which we call "a", and still have the integral be equal to zero.

Therefore, our derived equation:

f(x) = a(x-2) - 3a **where a is any non-zero real number**

will yield infinitely many such functions as required by the problem. We check our answer with Maple to be sure that we have not made an error, this is also to check that we have arrived at the correct solution:

> int(((x^2)*((a*(x^(-2)))-3*a)),x=0..1);

0

Since our answer checks out with Maple, we have arrived at a correct solution.

Solution: Infinitely many non-zero continuous functions f(x) that are orthogonal to x₂ are yielded by the following equation:

f(x) = a(x-2) - 3a **where a is any non-zero real number**