Proof: rank(A) ≤ k if A is the sum of k matrices of rank 1

Question

Let A be an m × n matrix that is the sum of k matrices of rank 1.

Prove that rank(A) ≤ k.

I have learned this proof the other way around (given the fact that rank(A)=k, prove that A can be written as the sum of k matrices of rank 1). But I don't know how to tackle this question.

I know that the rank is the dimension of row(A), so that the matrices with rank 1 have a basis for their row space that consists of 1 vector. In addition, each row is in the row space of the matrix.

Can somebody help me further with this?

$rank(A)=rank(A_1+\ldots +A_k)\le rank(A_1)+\cdots +rank(A_k)=k\cdot 1=k$. — Dietrich Burde, Jun 28 '17 at 08:56

score 1 · Answer 1 · answered Jun 28 '17 at 08:58

The number $\operatorname{rank}(A)$ is the dimension of the image of the linear map $f$ whose matrix, with respect to some basis, is $A$. Since $f$ is the sum of $k$ linear maps $f_1,f_2,\ldots,f_k$ such that the image of each $f_j$ is generated by a single vector $v_j$, the image of $f$ is generated by $\{v_1,v_2,\ldots,v_k\}$ and therefore its dimension is smaller than or equal to $k$.

Proof: rank(A) ≤ k if A is the sum of k matrices of rank 1

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