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

[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.

 

[Muzza]

Yeah, an example is:

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

.

 

[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.

 

[kakarukeys]

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

 

[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.

 

[Muzza]

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.

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

 

[jcsd]

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.

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.