Proof: If k ≠ 0, Then A and kA Have Same Rank

[georgeh]

Prove: If k !=0, then A and kA have the same rank.
Given:
k is in the set of all Reals.

proof:
WOLOG let A be an M X N matrix
Assume m < n
=> the max rank(A) can be is m. (i.e. the row space and the column space)
=> the max the rank(kA) can be is also m, because multiplying a matrix by a constant does not change the dimensions of the matrix
=>
# of leading variables stays the same for A and KA
=>
Rank(A)=Rank(KA)..

 

[AKG]

How can you say that m < n WOLOG? Also, what are the "leading variables"? Anyways, there are a number of ways to define rank. What definition do you have? Some hints: If you have a set of vectors that are linearly independent, when you multiply them each by the scalar k (non-zero), aren't they still linearly independent? Same for if you have a set of linearly DEpendent vectors? Also, if you have a subspace of some vector space, and you multiply each vector in that subspace by a non-zero scalar, don't you get the same subspace?

 

[georgeh]

By leading variables i meant the leading 1's. Also i edited, i meant to assume m < n.
How does stating it is L.I. help?

 

[squenshl]

I have a problem.
Suppose that {u1,u2,...,um} are vectors in R^n. Prove, directly that span{u1,u2,...,um} is a subspace of R^n.
How do I go about doing this. Thanks.

 

[quasarpulse]

To address georgeh's question to AKG, the rank of a matrix is the number of linearly independent vectors it contains. So let's say you have an MxN matrix of rank R. Then it contains R linearly independent columns (or rows), and the rest, if any, are dependent on that set. All you have to show is that multiplying a set of linearly independent vectors by a constant doesn't make them linearly dependent, and that multiplying a vector dependent on that set by the same constant doesn't make it independent.

To squenshl: you need to show four things:
- that span {u1, u2,...um} is a subset of R^n
- that that subset contains the zero vector
- that that subset is closed under addition (if u and v are elements of the span, then u+v is an element)
- and that it is closed under scalar multiplication (if u is an element of the span, than k*u is an element).

 

[squenshl]

Okay.
I've done the first 2 but how do I show that it is closed under addition & multplication.