Rank A Equal to Min k: Matrix in Mn(F) Proof

[brru25]

If A is a matrix in M_n(F), show that rank(A) = min(k: A = A₁ + ... + A_k, rank(A_i) = 1, A_i in M_n(F) for every i between 1 and k.

 

[Hurkyl]

Sounds like a homework problem. Have you worked on it at all? What lines of attack have you come up with, and where'd you get stuck? What sort of theory might be applicable to the problem?

 

[brru25]

Well I was thinking on the lines of rank-nullity theorem but I didn't see how that would apply here. I was also thinking of assuming each A_i was a matrix of entirely zero except for one row (or column), of which has only one entry of 1...e.g.

|0 0 0 0|
|0 0 0 0|
|1 0 0 0|
|0 0 0 0|

Each A_i would have to be distinct. I tried an example using three 3x3 matrices and got:

A = A_1 + A_2 + A_3 = I =

|1 0 0|
|0 0 0| +
|0 0 0|

|0 0 0|
|0 1 0| +
|0 0 0|

|0 0 0|
|0 0 0|
|0 0 1|

All matrices have rank 1 (I believe) and the matrix A would thus have rank 3.

How to show that rank(A) = the smallest possible k is tripping me up though.