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franz32
Dec2-03, 08:01 AM
Hello everyone! Can anyone help me here in this theorems (prove)?
(Or solve)

1. Suppose that A and B are square matrices and AB = 0. (as in zero matrix) If B is nonsingular, find A.

2. Show that if A is nonsingular symmetric matrix, then A^-1
is symmetric.

I hope these won't bother...

selfAdjoint
Dec2-03, 12:12 PM
For the first one, study the definition your book or notes gives for "nonsingular". Look at it again. How does that definition bear of the product AB = 0?

For the second one, do they give you a method for finding the inverse of a square matrix? What happens in that method when aij = aji for all i and j?

franz32
Dec4-03, 02:53 AM
Hello!

I got the answer... for the first, A must be a zero matrix. But
there are times when a matrix multiplied by a nonzero matrix can still result to a product of zero matrix. When is that?

For no. 2, I got it already. it's a matter of direct substitution using the properties given in my lecture.

HallsofIvy
Dec4-03, 08:30 AM
I got the answer... for the first, A must be a zero matrix. But
there are times when a matrix multiplied by a nonzero matrix can still result to a product of zero matrix. When is that?
Well, you know that a non-zero matrix multiplied by the zero matrix gives the zero matrix so I assume you mean "a nonzero matrix multiplied by a non-zero matrix" can give a zero matrix.

You have just determined that if one of the matrices is non-singular, then the other must be the zero matrix. What happens if you multipy two singular matrices?

For no. 2, I got it already. it's a matter of direct substitution using the properties given in my lecture.
I think you will find that true for many problems! Amazing that a lecture (and text book) can actually be useful isn't it?

franz32
Dec5-03, 01:22 AM
Hello.

You mean that the product of 2 singular (square ) matrices must result to either another matrix or a zero matrix?

Oh, thank you.

HallsofIvy
Dec6-03, 07:38 AM
"either another matrix or the zero matrix"???


Well that's always true isn't it! [:)]

What I meant was that, since you have already proved that if AB= O with A non-singular then B=0, the ONLY way you could get AB= 0 without either A or B 0 is to multiply two singular matrices.
The product of two singular matrices is not always 0 but is always a singular matrix.

franz32
Dec12-03, 11:21 AM
Hi again.

Thank you very much! =)