# Question on a proof about the Rank of a Matrix

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 ^_^

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You might want to assume $c\ne0$. – Chris Godsil Feb 12 '13 at 22:34
@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

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.

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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 $\mathrm{rank}(cA) = \mathrm{rank}( (cI_m)A ) = \mathrm{rank}(A)$.

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