Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The question is:

Give a formal proof for the following statement: Given a matrix A and a scalar c, show that rank(cA) = rank(A)

Here are the steps that I took to go about the proof:

(1) Prove this claim: Let v1, v2, ..., vN be vectors

then {v1, v2, ..., vN} is linearly independent <==> {c* v1, c*v2, ..., c * vN} is also lin. ind.

I don't type out the whole thing here, but the proof is trivial by playing around with the coefficients

(2) Let (c * A_ij) where i = 1, 2, ..., m; and j is fixed where j belongs to {1, 2, ..., n} denotes a linearly independent column in matrix cA

(3) Then I let S = { (c * A_ij)} be the set of all linearly independent columns in matrix cA, where each element of S satisfies (2)

(4) By how I define the set S, all elements in S are lin. ind. columns in matrix cA.

Then I use the claim (1) to say that columns A_ij of matrix A must also be lin. ind.

I also note that by definition of the rank, it's the maximum number of lin. ind. columns (or rows) in a matrix. So I think rank(cA) is basically the cardinality of the set S. Then by (4), I conclude that when I "move" from each lin. ind. column of matrix cA to each lin. ind. column in matrix A, I didn't change the number of lin. ind. columns. Thus, rank(cA) = rank(A).

Would someone please help me check if there is anything missing or wrong in my proof ? Somehow I feel a bit shaky on how I define the indices for the linearly independent columns in matrix cA. Thank you very much ^_^

share|improve this question
6  
You might want to assume $c\ne0$. –  Chris Godsil Feb 12 '13 at 22:34
3  
@Cecile I see this is your sixth question here, yet you have accepted $0$ answers. Whenever you get a satisfatory answer to one of your questions, you should accept your favorite answer. –  Git Gud Feb 12 '13 at 22:35
add comment

2 Answers

up vote 1 down vote accepted

The rank is the dimension of the range.

Now try to prove that for all $c\neq 0$, the range of $A$ is equal to the range of $cA$.

Hints: $$ cA(x)=A(cx)\quad\mbox{and}\quad A(x)=cA\left( \frac{1}{c}x\right). $$

This is easier like this.

share|improve this answer
add comment

Suppose $A$ is an $m \times n$ matrix. Then $cA = (cI_m)A$, where $I_m$ is the $m \times m$ identity matrix. Since $cI_m$ is full rank (assuming $c \neq 0$) we have $\mathrm{rank}(cA) = \mathrm{rank}( (cI_m)A ) = \mathrm{rank}(A)$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.