# rank(A)=rank(cA), where c is scalar

Discussion in 'Undergraduate Math' started by Jen, Nov 2, 2005.

1. ### JenGuest

How can I prove that the rank(A)=rank(cA) where c is
a scalar and A is an mxn matrix?

I know that when you row reduce a matrix you can divide
the rows by that common "c" number, but how can I formally
write it up????

Jen, Nov 2, 2005

2. ### Arturo MagidinGuest

You can't... unless you require that c be a ->nonzero<- scalar.
Multiplying a matrix by an invertible matrix does not change the rank:
that is, if C is an invertible mxm matrix, and D is an invertible nxn

So set C to be the diagonal matrix whose diagonal entries are all 1/c
(here is where you need c is nonzero); this is exactly the same as
multiplying each row by 1/c.

--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================

Arturo Magidin

Arturo Magidin, Nov 2, 2005

3. ### JenGuest

How can I prove that the rank(A)=rank(cA) where c is
You can't... unless you require that c be a ->nonzero<- scalar.
Multiplying a matrix by an invertible matrix does not change the rank:
that is, if C is an invertible mxm matrix, and D is an invertible nxn

So set C to be the diagonal matrix whose diagonal entries are all 1/c
(here is where you need c is nonzero); this is exactly the same as
multiplying each row by 1/c.

If c is a nonzero scalar, and I want rank(A)=rank(cA),
then why do you have 1/c? Can I have C be the diagonal
matrix whose diagonal enteries are all c and not 1/c?
This way rank(A)=rank(cA).
Is this permitted?

Jen, Nov 3, 2005
4. ### Arturo MagidinGuest

Because that's the matrix that does what you said you wanted: "divide
each entry by c", starting with cA.
Yes: if you want to start with A, then you would do that.
A little bit of thought would have answered your question. If you
really do not see that what you are asking was exactly the same as I
said, then you did not understand the answer and it will be worthless
to you. Try to understand what is being said.

--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================

Arturo Magidin

Arturo Magidin, Nov 3, 2005