Can a non-zero matrix multiply itself to become zero matrix?

1. Nov 18, 2004

kakarukeys

In real number, only zero multiplies itself to become zero.
Can a non-zero matrix multiply itself to become zero matrix?
equivalent question: Can the row space of a matrix be orthogonal to the column space?

I'm sorry if the question looks stupid.

2. Nov 18, 2004

Muzza

Yeah, an example is:

Code (Text):

(1 -1/2)
(2 -1)

.

3. Nov 18, 2004

jcsd

Yep this is an example of a nilpotent (xn = 0 for some n) element of a ring (iu this case the ring of 2 x 2 matrices) and hence a zero divisor.

Last edited: Nov 18, 2004
4. Nov 18, 2004

kakarukeys

Thanks
I also found that if we restrict ourselves to symmetric matrices (Hermitian matrices in complex case).
No such matrix can be found.

5. Nov 20, 2004

Greg Bernhardt

No, based upon the definition of multiplication, the only way to have a product of zero is if one of the factors are zero. ie. if both factors are non-zero, the product must be non-zero.

6. Nov 21, 2004

Muzza

You must've missed the part where kakarukeys said this was about matrices.

7. Nov 21, 2004

jcsd

I think that's true for fields, but it's not true for rings in general, infact a non-zero memenber of a ring that has the property of being (right) multiplied by some other non-zero member to give zero is called a zero divisor.

Last edited: Nov 21, 2004