Question

Has anyone looked at q5 of Prob Set 3.3 in Intro to Linear Algebra 4th ed? The question is to find the conditions under which ref(A1+A2)=ref(A1)+ref(A2), where ref means “row echelon form” (*not* “reduced row echelon form”), and A1 and A2 are both (2,3). Prof Strang seems to think that the only solutions are A1=ref(A1) and A2=ref(A2), but I am seeing additional solutions when either col1 of A2 is multiple of col1 of A1 or row1 of A2 is multiple of row1 of A1. Any views?

Answer 1

I don't think your answers are true generally, you might have accidentally picked a special case. I tried A1 = [2 3 4; 3 5 9] and A2 = [4 6 8; 3 5 9] so row1 of A2 is a multiple of row1 of A1 and I don't get the proper equality.

Answer 2

Thanks for replying merkaba8! I was beginning to think there was no-one out there!

But I don't see your example as a counterexample. Don't forget I'm looking at row echelon forms not reduced row echelon forms. You have
A1 = [2 3 4; 3 5 9] ==> ref(A1) = [2 3 4; 0 1/2 3]
A2 = [4 6 8; 3 5 9] ==> ref(A2) = [4 6 8; 0 1/2 3]
A1+A2 = [6 9 12; 6 10 18] ==> ref(A1+A2) = [6 9 12; 0 1 6] = ref(A1)+ref(A2)

Best wishes

Josh

Answer 3

Sorry, you're right I did the elimination by hand late last night and just mixed up a sign somewhere. It seems that your formulation works. At the very least, a single example is proof that A1=R1 and A2=R2 is not implied, and I am guessing with a bit of algebra you could show that the proportional row case works generally.

Answer 4

merkaba8, thanks for taking a look. I did do the algebra but I thought maybe I had gone wrong.

Answer 5

I don't think you should assume the proposition in the question is supposed to be true. I think you are just supposed to determine that it is not.