Is there only one form of a reduced matrix? Why?

Question

If you and a friend are given a matrix, and you both go through different operations to reduce it, will you both have the exact same reduced matrix? If so, why is that? Be as clear as possible, state properties and laws if you'd like.

Answer 1

Yes, the reduced row echelon form of a matrix is unique. Here's why:

Note that the elementary row operations do not change the kernel of a matrix. Since any rref of any matrix M can be obtained from M by a series of elementary operations, it follows that two different rrefs of M must have the same kernel. Thus, to prove the uniqueness of the rref, it suffices to show that two distinct matrices of the same dimension in reduced row echelon form cannot have the same kernel. This we do as follows - let A and B be two distinct matrices in rref. Let n be the number of the first column in which they differ. We consider two cases:

Case 1: Both columns contain a leading 1. In this case, by the rules of rref, the leading 1 is the only entry in each column. Further, which row it appears in is completely determined by how many leading 1s have appeared in previous columns - since n is the first column that differs, all the previous columns are the same in both matrices, and thus the number of leading 1s that have appeared prior to the nth column is the same. It follows that the nth column of both matrices is the same, contradicting the assumption that they are not.

Case 2: One of the columns does not contain a leading 1. Assume without loss of generality that this column is in matrix A. By the rules of rref, only those entries with are in rows that have leading 1s in prior columns can be nonzero. Let the vector v be the vector defined thus:

The nth element of v is -1
If the pth column of A contains a leading 1 in the qth row, then the pth element of v is the qth element of the nth column of A.
All other elements of v are zero.

A moments contemplation reveals two things about this vector: first, that all elements past the nth are always zero (any nonzero element of v other than the nth would correspond to a column containing a leading 1. If the leading 1 appeared in the pth column and qth row with p>n, then necessarily the qth element of the nth column is zero, since otherwise it would precede the leading 1 for its row, which is impossible by definition. Therefore, that element would have to be zero, per the definition of v, and it follows that for every p>n, the pth element is zero).

The second thing you could deduce is that this vector is in the kernel of A (indeed, we can find a basis for the kernel of A by considering precisely vectors of this type). Now, by our knowledge that A and B are rrefs of the same matrix, this vector must also be in the kernel of B. Now, since all elements of v past the nth are zero, we can write the product Av as v_1*A_1 + v_2*A_2... + v_n*A_n, where v_m indicates the mth element of v and A_m indicates the mth column of A. Now, since all columns prior to the nth are the same, it follows that v_1*A_1 + v_2*A_2... + v_(n-1)*A_(n-1) = v_1*B_1 + v_2*B_2... + v_(n-1)*B_(n-1). Now, since Av=Bv=0, this implies that v_n*A_n also equals v_n*B_n. Since v_n is by definition -1, this means that -A_n=-B_n, which means A_n=B_n, so the nth columns are the same in this case as well.

Now clearly, if A and B were distinct rrefs of M, they would have to differ in at least one column, and there would be a first such column. But as we have just seen, that is impossible. Therefore, the rref of M is unique. Q.E.D.

Answer 2

You will have the exact same matrix if you go through the same type of reduction (types include Gauss-Jordan and Gaussian). If you pick the same type of reduction, there will be only one answer. This is due to the fact that there are no "choices" to be made at each step of the reduction (for example: you may multiply a row by 1/8 to get the first non-zero entry to be a 1 as per Gauss-Jordan. There is no other number to multiply the row by).

However, you should note that multiplying a row by a real number and adding rows to other rows does not change the solutions to the system of equations. Therefore, you could do a type of non-standard reduction and end up with a completely different (reduced) matrix that still has the same set of solutions as your friend's.

So, you see, it's really not that the "reduced" matrix is unique, but rather that the methods to reduce the matrix are standardized.

Answer 3

YES IF U R NOT GETTING THE SAME ANSWER THAT JUST MEANS THAT 1 OF U IS CLOSER THAN THE OTHR N THE MOST SIMPLIFIED FORM IS THE 1 WITH THE MOST NO. OF ROWS WITH ALL 0 OK